SYSTEM OF PARTICLES AND ROTATIONAL MOTION
9.1 Centre of Mass
9.2 Collisions
9.3 Rotational Mechanics
9.1
CENTRE OF MASS
So far, we have discussed the kinematics and dynamics of a particle. We discussed the displacement, velocity, momentum, and acceleration of particles using Newton’s laws of motion and work energy theorem. We applied law of conservation of energy and law of conservation of linear momentum for a system of particle in solving problems.
In case of extended bodies, which is a group of particles, the instantaneous quantities are not same for all the particles if its motion is not translatory. There will be many sets of equations of motion. It will be practically impossible or extremely difficult to analyse the motion of a rigid body by using Newton’s laws particle-wise. Hence, we need to develop a method in which the entire system can be made equivalent to a particle and then conveniently apply the laws of motion, work energy theorem, etc. The concept of centre of mass is introduced for this purpose.
Centre of Mass (COM) is a point where the entire mass of a system of particles is supposed to be concentrated.
Mass Moment of a particle about a reference point is defined as the product of mass of the particle and its position vector about the reference point.
Fig. 1
In the figure, mr is the mass moment of a particle of mass m at P about ‘O’. (The direction of r is from point to the position of mass.)
Key Insights:
■ Algebraic sum of moments of all masses about centre of mass is equal to zero.
9.1.1 Position of Centre of Mass
(a) Let us consider a system of particles of masses m1, m2 and m3 at position vectors 12 , rr
, and 3r
. Sum of their mass moments is 112233 . mrmrmr ++
Total mass of the system is M = m1 + m2 + m3. Centre of mass is a hypothetical point ‘C’ at position vector cm r such that mass moment calculated with
9: System of Particles and Rotational Motion
a particle of mass M at ‘C’ will be equal to sum of the mass moments of individual particles.
Hence, C represents equivalent position of total mass of the system of particles as a point mass.
(b) Let F be the sum of all the external forces acting on different point masses of a system. It makes no sense to tell that F acts on any particle of the system. Centre of mass is a hypothetical point in the system which moves with an acceleration F a M
It means that centre of mass behaves as a point where entire mass of system is concentrated as a point mass and all the external forces applied on the system are exerted on it.
Important points about centre of mass:
■ Mass is not actually concentrated at centre of Mass. It is a mathematical concept. It is just a point only.
■ For symmetric bodies with uniform mass distribution, centre of mass lies at the geometric centre.
Examples of centre of of mass for different shapes are given in the following table. Shape of the Body
or Solid cylinder
plate
= Centroid
■ Mass may or may not exist at centre of mass. It may lie outside the material of the body.
Example: ring, boomerang, hollow sphere, etc.
■ Centre of mass is defined for a system of particles but not for a particle.
■ If the mass of system is distributed nonuniformly, centre of mass does not lie at geometric centre but close to the region where more mass is concentrated.
Example: If we consider a classroom + benches + students as system, its centre of mass does not lie at the centre of the room but lies at some distance below it.
■ Before finding the position of centre of mass, first, we have to define the system for which we want to find the position of centre of mass. In the above example, if we consider only students as a system, their centre of mass lies within them. If the room alone is taken as a system, its centre of mass lies at its
centre. If the room + students + benches are taken as a system, their centre of mass lies at some other point.
■ Even though different particles of a rigid body move in complicated paths, centre of mass always moves in simplest path. Its trajectory may be a straight line or a curve.
Rolling wheel
For example, consider a wheel that rolls without sliding on a horizontal plane. If a particle on the rim of the wheel is considered, it will have a complicated path. However if a particle on the axis of rotation is considered, it moves along straight line path.
If an Indian club is projected into air, as shown in the figure, different particles in the Indian club move in complicated trajectories but a single particle moves in a simple parabolic path.
Fig. 4 Indian club motion
From the above two examples, we can say that a rigid body may have a complicated
motion but a certain point of the body has a simple pure translational motion. The motion of the single point represents the motion of the entire body. This point within the boundary of the body is called the centre of mass or centre of inertia.
■ The sum of mass moments of all particles of a system about its centre of mass is always equal to zero.
Here, C is the centre of mass of system.
■ Centre of mass always lies within the boundary of the system.
Coordinates of COM of a System of Particles
Position vectors of masses m 1 , m 2 , m 3 and centre of mass C are
are their position coordinates.
ycm, zcm)
On comparing both sides of components
ˆ , ij and ˆ k ,
cm mxmxmx x mmm ++ = ++
Position of centre of mass of a system of particles does not depend upon position, orientation of rectangular co-ordinate system with respect to which the co-ordinates of centre of mass is given.
1. Find the position of centre of mass of the system of 3 objects of masses 1 kg, 2 kg and 3 kg, located at the corner of an equilateral triangle of side 1 m. Take 1 kg mass object at the origin and 2 kg along the x-axis. Sol.
\ Co-ordinates of centre of mass ( x cm, ycm) 7 12 3 4 mm ,
Try yourself:
1. Four particles of masses 1 kg, 2 kg, 3 kg and 4 kg are placed at the four vertices A, B, C and D of the square of side 1 m. Find the coordinates of centre of mass of the particles. (0.5Ans: m, 0.7 m)
Centre of Mass of a Two Particle System:
a) Let us consider the origin of co-ordinate system to be at centre of mass and particles are on x-axis. Now xcm= 0, coordinates of m1, m2 are –x1, x2
Fig. 7
Then, 1122 12 cm mxmx x mm + = +
Masses of particles are at inverse proportion to their distances from centre of mass.
2. If the separation between carbon atom and oxygen atom in CO molecule is 1.4 Å, find the distance of its centre of mass from the C atom.
Sol. mC = 12 amu, m0 = 16 amu
Try yourself:
2. Two blocks of masses 10 kg and 30 kg are placed on x-axis. The first mass is moved on the axis by a distance of 2 cm towards right. By what distance should the second mass be moved to keep the position of centre of mass unchanged?
Ans: 2/3 cm, towards left
(b) Let 'x' be the distance between two masses
D x 1 , D x 2 then centre of mass moves by a distance D x cm here D x 1 , D x 2 and D x cm represent changes in x 1 , x 2 and x cm respectively. Then, ()() 1122 12 cm mxmx x mm ∆+∆ ∆= +
For many particle system, in general, ()()() 1122 12 . nn cm n mrmrmr r mmm ∆+∆+…+∆ ∆= ++…+ ()ii i mr m ∑∆ = ∑
If the centre of mass remains at the same position even after displacements of m1, m2, then D x cm =0.
()() 1122 0 mxmx ∴∆+∆=
()() 1122 mxmx ∆=−∆
–ve sign indicates that the two masses are displaced in opposite directions.
(d) Consider two groups of particles containing N1 and N2 number of particles. M1 and M2 are masses of two groups and their centres of masses are at points C 1 and C 2 such that their co-ordinates are C1(X1,Y1,Z1) and C2(X2, Y2, Z2) respectively. Then, coordinates of COM of system of two groups are 1122 12 cm MXMX X MM + = +
1122 12 cm MYMY Y MM + = +
1122 12 cm MZMZ Z MM + = +
(c) For two particle system, 1122 12 cm mxmx x mm + =
When m 1 , m 2 are moved by distances
Thus, if we know the centre of mass of parts of the system and their masses, we can get the combined centre of mass by treating the parts as point particles placed at their respective centres of mass.
3. When ‘n’ number of particles, each of mass ‘m’, are at distances x1 = 1, x2 = 2, x3 = 3, ...... x n = n units from origin on the x-axis, then find the distance of their centre of mass from origin.
Try yourself:
3. When ‘n’ number of particles of masses m, 2m, 3m,..... nm are at distances x1 = 1, x2 = 2, x3 = 3, ...... x n = n units, respectively, from origin on the x-axis, then find the distance of centre of mass of the system from origin. Ans: 1 3 21 n
9.1.2 Centre of Mass of Bodies with Cavity or Hole
Let M is the mass of the body before making a hole or cavity. Its centre of mass is 'C'. m1 is the mass that is removed in making the cavity and C1 is COM of m1 when it fills the cavity. m2 is mass of the remaining part of the body i.e. mass of the body with cavity and COM of m2 is at C2. Then ‘C ‘ will become COM of masses m1 and m2. Shift of COM is CC2 = x2
Here, M = m1 + m2 For two particle system, m1x1 = m2 x2
\ Shift in COM due to making of cavity is 11 22 2 mx xC m =⋅
COM always lies on the other side of the cavity w.r.t. ‘C’.
Example-1
From a uniform circular disc of radius R , centre C, a circular hole of radius ' b' is made with its centre C 1 at a distance ' a ' from 'C'. Shift of CM is CC2= S
C2
If t, ρ represent thickness, density of plate,
22 1 , mbtMRt =πρ=πρ and () 22 21 mMmtRb =−=πρ−
x1 = a and x2 = S
m1x1 = m2x2 ()() 2 22 ()btaRbtS ⇒πρ=π−ρ 2 22 ba S Rb ∴=
Example-3
From a uniform solid cylinder AB of radius R, length 4R, a hemispherical cavity of radius R is made at one edge.
Example-2
From a uniform solid sphere of radius R1, and centre C, a spherical cavity of radius b with centre at C1 is made and C C1= a. [Figure is similar the above case]
33 1 44 , 33 mbMR=πρ=πρ
323 1 2 (4)4 3 mRMRRR =πρ=πρ=πρ
33 2 210 4 33 mMmRr =−=πρ−=πρ
We know that BC1=3 R/8 11 313 2 88 RR xCCR ∴==−= C C2 = S = ?
3 1122 213 38 R mxmxR =⇒πρ 3 10 13 () 3 40 RsSR=πρ⇒=
4. A thin disc of radius R/2 is placed on another disc of radius R. The two discs are made of same material and have same thickness. Find the distance of centre of mass of the system from the centre of the larger disc. Assume that the discs are very thin so that vertical shift of COM can be ignored.
Sol. Let s be the mass per unit area of each disc.
Object 1 is a solid disc of radius R having surface density of mass + s . Object 2 is a solid disc of radius R/2, having same thickness as object 1 but surface density of ‘– s’. Object 2 is made of a hypothetical material.
When object 2 is placed on object 1, the given object is formed.
Now,
5. A circular hole of radius R/2 is punched in a thin circular disc of radius R as shown in figure. Find the coordinates of COM of the remaining disc.
cc R 6 0 and y
So the coordinates of COM are (R/6, 0)
Try yourself:
4. Identical blocks, each of mass M and length L , are placed one above the other, such that each extends out by maximum length, as shown in figure. Find the maximum extension of the nth block from the top, so that the blocks will not fall.
Sol. Let us consider two objects 1 and 2.
9.1.3 Centre of Gravity
We know tha t the earth attracts all bodies towards its centre (neglecting the effect due to earth’s spin motion). Thus, imagine a composite body made up of several particles of masses m1, m2, m3 etc. located at respective perpendicular distance of r 1, r 2, r 3 etc. from any axis AA’ (say). The moment due to the weights of individual particles put together is given by 112233 mgrmgrmgr+++…… (here, g is same for all particles of composite body).
For very extensive bodies, like mountains, oceans, etc., centre of mass and centre of gravity do not coincide as g will not be uniform. CG is shifted towards the regions where ‘g ‘ is more.
Example:
i) If we take all the water in an ocean as a system, its CG will be above its COM.
ii) For a mountain, CG will be below its COM. But for small bodies, both CG and COM. will coincide.
If a body is supported at or suspended from the centre of gravity, it will be in rotatory equilibrium in both uniform and non–uniform gravitational fields.
Determining the centre of gravity of a body of irregular shape:
If we replace all the particles by a single point mass at C equal to the mass of the body, at a distance r from the axis, such that the moment due to the weight of this point mass computed, yields the total moment due to individual particle masses. Then point C defines the centre of gravity of the body.
Thus,
which is same as the position of centre of mass.
Thus, centre of gravity of a body defines the position, where its whole weight could be imagined to be concentrated.
If you suspend the body from some point, like A, the vertical line through A passes through the CG. We mark the vertical AA1. We then suspend the body through other points, like B and C. The intersection of the verticals gives the CG. Since the body is small enough, the method allows us to also determine its centre of mass.
Table 2: Distinction between centre of mass and centre of gravity
Centre of mass Centre of gravity
(A) It is the point in the body where the total mass is supposed to be concentrated and at which, if an external force is applied, the body undergoes translatory motion.
(B) It refers to mass of the body.
(C) It depends on masses of the particles and their relative positions.
(I) It is the point in the body through which the weight of the body is supposed to be acting.
(II) It refers to weight of the body.
(III) It depends on the masses of the particles, their relative positions, and also the gravitational field.
9.1.4 Centre of Mass of Bodies of Continuous Mass Distribution
If we consider a body to have continuous distribution of matter, the summation in the formula of COM is replaced by integration. Suppose x, y and z are the coordinates of small element of mass dm; we write the co-ordinates of COM as
6. If the linear density of a rod of length L varies as ABx , find the position of its centre of mass.
Sol. Let the x -axis be along the length of the rod and origin at one of its ends. As rod is along x–axis, for all points on it, y and z coordinates are zero.
Centre of mass will be on the rod. Now consider an element of rod of length dx at a distance x from the origin. Then, dmdxxdx =A +B
The integration is to be performed under proper limits so that, as the integration variable goes through the elements covering the entire body.
Try yourself:
5. Find the distance of centre of mass of a uniform solid cone of height ‘ h ’ and base radius R , from the vertex on the line of symmetry is
Ans: 3h/4.
Proceeding in the similar manner, we can find the COM of certain rigid bodies. Centre of mass of some well known rigid bodies are given below:
1 Centre of mass of a uniform semicircular ring lies at a distance 2 hR = π from its centre, on the axis of symmetry, where R is the radius of the ring.
2. Centre of mass of a uniform semicircular disc of radius R lies at a distance of 4 3 hR = π from the centre of the axis of symmetry as shown in figure.
R 4R 3π
Fig.16
3. Centre of mass of a hemispherical shell of radius R lies at a distance of 2 hR = from its centre on the axis of symmetry as shown in figure.
5. Centre of mass of a uniform circular arc of radius R subtending an angle '2α' at the centre is at a distance sin R α α from centre of curvature 'O' and it lies on the bisector. ‘α‘ in denominator has to be substituted in radians.
Fig.17
4. Centre of mass of a solid hemisphere of radius R lies at a ditance of 3 8 hR = from its centre on the axis of symmetry.
Fig.19
6. COM of a uniform conical shell is located on its axis at a distance of 2 3 h from its vertex or 3 h from its base, where h is its altitude.
2h 3 h
Fig.20
7. COM of a uniform solid cone (or pyramid) of height h is at a height 4 h from its base.
8. h
Fig.21
Height of COM of a uniform triangular plate above the midpoint of the base is 3 h where h is the altitude of the triangle above this base. This is true from any one of the three bases.
7. L-shap ed piece is cut from a thin metallic sheet as shown in figure. Find the coordinates of its COM.
Try yourself:
6. A piece of sheet, whose shape is as shown in figure, is cut from a large thin sheet. Find the coordinates of its centre of mass.
Sol. Let us divide the object into two parts 1 and 2, as shown in the figure. If s be the surface density of mass,
9.1.5 Motion of the Centre of Mass
Let us consider the motion of the system of n particles of individual masses m1, m2 m n and total mass M. It is assumed that no mass enters or leaves the system during its motion, so that M remains constant. Then,
Differentiating this expression with respect to time t, we have
12 12 . CMn n drdr drdr Mmmm dtdtdtdt =++……
Since, = Velocity dr dt Therefore,
() CMnn Mvmvmvmvi =++……+……
Velocity of centre of mass is
Further, mv = momentum of a particle p . Therefore, equation (i) can be written as momentum of centre of mass
12 CMn pppp =++………+
1 n CMi i pp = =∑
Differentiating equation (i) with respect to time t, we get
12 12 CMn n dvdv dvdv Mmmm dtdtdtdt =++……+
(or) 1122 CMnn Mamamama=++…+…
Acceleration of centre of mass is
Further, in accordance with Newton’s second law of motion, Fma = . Hence, equation (ii) can be written as
Even though 12 , ..... nFFF
are resultants of internal and external forces acting on m1, m2, ... m n respectively, in the addition of the forces, all the internal forces will cancel each other as they exist in action - reaction pair. Hence, net external force acting on the system is the force acting on COM.
and Constant.
Thus, the cent re of mass of a system of particles moves as it was a particle of mass equal to the whole system with all the external forces acting directl y on it.
Parabolic path of the projectile
Explosion
Path of the COM of fragments
x x1
The centre of mass of the fragments of the projectile continues along the same parabolic path which it would have followed if there were no explosion.
A projectile, following the usual parabolic trajectory, explodes into fragments midway in air. The forces leading to the explosion are internal forces. They contribute nothing to the motion of the centre of mass. The total external force, namely, the force of gravity acting on the body, is the same before and after the explosion. The centre of mass under the influence of the external force continues, therefore, along the same parabolic trajectory as it would have followed if there were no explosion.
Fig.22
8. A thin rod AB of length 5 m and mass m inclined to a vertical wall is slipping. At an instant, end B is moving on the floor with a velocity of 2 m/s. Find the magnitude of velocity of COM of the rod at this instant. 3 m 2 m/s
Sol. Since the rod is uniform, its COM must lie at its centre. In this case, the rod can be replaced by two identical particles A and B, each of mass m/2, connected by a massless rigid rod AB of length 5 m.
m/2 u = 2m/s y
q
Velocity of particle A is given by Vsinq = 4cosq
[ \ The massless rod AB is rigid]
Try yourself:
7. An ob ject A is dropped from rest, from the top of a 30 m high building, and at the same moment, another object B is projected vertically upwards with an initial speed of 15 m/s from the base of the building. Mass of the object A is 2 kg while mass of the object B is 4 kg. Find the maximum height above the ground level attained by the centre of mass of the A and B system (take g = 10 m/s2).
Ans: 15 m.
Motion of a Man in a Boat
Suppose a man of mass m standing on a boat of mass M and the system is floating on water. When the man moves on the boat, the boat also moves in such a way that their centre of mass remains at the same position, i.e., the centre of mass does not move. This is because there is no external force acting on the system. The mutual forces between the man and the boat are internal forces which do not move the centre of mass.
If x1, x2 are the distances of M, m from the origin (shore) then distance of their COM is ()() 12 cm Mxmx x ....(1) Mm + = +
Suppose the man has moved a distance l w.r.t. the boat towards the shore. To make the position of COM constant, the boat moves away from the shore through a distance s w.r.t. the shore.
Their final distances will become 11 xxs ′ =+ and 22 xxls ′ =−+
and ()() 12 12 (2) cm Mxsmxls x mm ++−+ = … +
On equating (1) and (2) and simplifying, we get
ml s mM = +
(i) Final distance of man from the shore is 22 xxls ′ =−+
22 mlMlxlx mMmM =−+=− ++
Similarly final distance of the centre of boat from the shore is
111 ml xxsx mM ′ =+=+ +
(ii) If the man moves with velocity of magnitude V relative to the boat, then magnitude of velocity of boat w.r.t. the shore is Vb. dl V dt = and b ds V dt =
mldsmdl s mMdtmMdt =⇒= ++
bm m VV mM ⇒= +
Directions of V b and V m are always opposite.
(iii)If there are more-than one person on the boat, find the displacements of boat due to motion of each person separately, using the above formula. Algebraic sum of those displacements will be equal to the net distance moved by the boat.
Example: Let m1,m2 be masses of two persons standing on the boat. Suppose m 1 moves towards the shore through a distance l1 and m2 moves away from the shore through a distance l2 w.r.t. the boat,
11
1 12 ml s mmM = ++ away form the shore
22
2 12 ml s mmM = ++ towards the shore
Net distance moved by the boat s=s1 –s2
If s 1>s 2, the boat moves away from the shore
If s1 >s2, the boat moves towards the shore. If s1=s2, the boat remains at same position.
In this example, if both the persons move in the same direction, then s = s1+ s2 and the boat moves opposite to the direction of motion of persons.
9. A block of mass m is placed on the top of a wedge of mass M and the system is placed on a horizontal surface. Base length of the wedge is L. If all the surfaces are smooth, by the time the block reaches the bottom, Find the displacement of the wedge.
Sol. Here, resultant horizontal external force is zero. Hence, horizontal shift of centre of mass is zero.
Consider a wedge of mass M and base length L resting on a horizontal surface.
A small block of mass ‘m’ is on the top of the wedge. All surfaces are smooth. As the block slides down the incline plane, the wedge displaces backwards.
Here, centre of mass displaces only in vertical direction due to external gravitational force. However, since there is no external force along horizontal direction, centre of mass will not be displaced in that direction. So, by principle of moments,
m1x1 = m2x2 m[L – x] = Mxt
where x and (L – x) are horizontal displacements of block and wedge, respectively.
\ Displacement of wedge is x mL Mm
If L is the length of inclined plane and q is angle of inclination, then x mL Mm cos
Try yourself:
8. Find the acceleration of centre of mass of the blocks of masses m1 and m2 (m1 > m2) in Atwood’s machine.
Ans: mm mm g 12 12 2
Characteriestics of Centre of Mass
■ The centre of mass of a system of particles depends only on the masses of the particles and their relative positions. For a continuous distributed mass, it depends on the shape and its mass distribution.
■ Sum of the moments of masses of the system about the centre of mass is zero.
■ A mass particle need not be present at the position of the centre of mass. The centre of mass of a uniform ring lies at its centre, where no mass particle is present.
■ The location of centre of mass is independent of the reference frame used to locate it.
■ If no external force acts on the system, the centre of mass of the system may be at rest or moving with uniform velocity.
■ Position of centre of mass does not change due to the internal forces in the system.
■ The position of centre of mass changes in translatory only. It remains unchanged in rotatory motion.
■ Newton’s second law of motion can be used to determine the motion of centre of mass.
Laboratory
and Centre of Mass Frames of References
Laboratory frame of reference:
We know that the position of a point in the space is located by the three co-ordinates which are taken with respect to the origin of a reference system.
“When the origin of the reference system is a point rigidly fixed to the laboratory, it is known as laboratory frame of reference.
Centre of mass frame of reference:
When the origin of the reference system is rigidly fixed at the centre of mass of an isolated (a system of particles with no external force) system, it is known as centre of mass frame of reference or C - frame.
In this frame, position vector of CM is obviously zero. So the velocity of centre of mass is also zero. Consequently the linear momentum of the system is also zero. Due to this reason, this frame of reference is also called as “zero momentum frame”.
In the absence of any external force, the centre of mass frame is an inertial frame.
Relation between Velocities in Laboratory and C-frames:
In laboratory frame, position vector and velocity of COM for two particle system are 1122 12 cm mrmr r mm + = +
and 1122 12 cm mvmv v mm + = +
In C-frame, let the position vectors of two masses be ′′
12 and rr respectively.
Therefore, velocities of particles in the centre of mass frame of reference are 1212 1
Total KE of system of two particles in C-frame: If 12 , vv and cm V are velocities of two particles of masses m 1, m 2 and their centre of mass, respectively, in laboratory frame, velocity of the particles in C-frame are 11 22 cmcm vvvandvvv ′′ =−=−
where 1122 12 cm mvmv v mm + = +
KE of system in C-frame 22 1122 11 ' 22 mvmv ′ =+ 22 11 22 11 22cmcm mvvmvv =−+−
() 22 11 1 2 11 2 22 cmcm mvvvvm=+−⋅+
() 22 22 2 cmcm vvvv +−⋅
() 1122 12 121122 Using, cm cm mvmv v mm mmvmvmv + = + ⇒+=+
=+++
() 22 2 112212 111 222 cm mvmvmmv
momenta of the particles in C-frame are
() 1122 cm mvmvv −+⋅ () 222 112212 111 222 cm mvmvmmv=+++ () 12 cmcm mmvv −+⋅ () 222 112212 111 222 cm mvmvmmv=+−+ = KE of system in laboratory frame – KE of COM in laboratory frame
Key Insights: ■ KE of system 22 1122 11 22 mvmv=+ ()=+ 2 12 1 KE of COM 2 cm mmv
These two are not equal, i.e., sum of kinetic energies of particles of system is not equal to kinetic energy of COM of system.
■ In C-frame, the linear momenta of two colliding particles are equal in magnitude and opposite in direction. Their velocities are collinear. This is true before and after the collision.
In head-on collision, initial and final lines of motion are same. In oblique collision, the initial and final lines of motion are not same, as shown in figure.
10. Particle 1 of mass 1 kg is moving along positive x-axis with velocity 1 m/s and particle 2 of mass 2 kg is moving along positive y-axis with velocity 2 m/s. Using the COM as frame of reference, find the velocity of COM.
■ A block of mass m is placed on the top of a wedge of mass M and the system is placed on a horizontal surface. Base length of the wedge is L. If all the surfaces are smooth, by the time the block reaches the bottom, displacement of the wedge is mL x mM = +
■ A balloon of mass M , a light rope, and a monkey of mass m are at rest in air as shown in figure. If the monkey reaches the top of the rope of length L then the balloon descends by a distance mL Mm +
Try yourself:
9. Particle A of mass 2 kg is moving along negative x -axis with velocity 2 m/s and particle B of mass 1 kg is moving along positive x -axis with velocity 1 m/s. Using the COM of the system as frame of reference, find velocity of COM.
Ans: 1 m/s
TEST YOURSELF
1. P is the centre of mass of a system of four point masses A, B, C and D, which are coplanar but not collinear.
a) P may or may not coincide with one of the point masses.
b) P must lie within or on the edge of at least one of the triangles formed by taking A, B, C, and D, three at a time.
c) P must lie on a line joining two of the points A, B, C, D.
d) P lies outside the quadrangle ABCD.
(1) a and b are correct.
(2) c and d are correct.
(3) a and c are correct.
(4) b and d are correct.
2. A thin uniform rod of length L is bent at its midpoint, as shown in figure. The distance of centre of mass from the point ‘O’ is
3. A uniform wire of length L is bent in the form of a circle. The shift in its centre of mass is
(1) L/π (2) 2L/π
(3) L/2π (4) L/3π
Answer Key
(1) 1 (2) 2 (3) 3
9.2 COLLISIONS
A strong interaction between two bodies that occurs for a very short interval, during which redistribution of momenta occurs, ignoring the effect of other forces, is called a collision. During a collision, bodies are deformed considerably due to interaction.
In a collision, there need not be a physical contact between two entitites. A particle’s motion can be suddenly changed due to an electric field or a magnetic field. This sudden change in motion due to interaction of some field can also be treated as a collision.
There are three distinct identifiable stages in collision, namely before, during, and after. In the before and after stages the interaction forces are zero. Between these two stages, the interaction forces are very large and, often, the dominating forces govern the motion of
bodies. The magnitude of the interacting force is often unknown. Therefore, Newton’s second law cannot be used. The law of conservation of momentum is useful in relating the initial and final velocities.
9.2.1 Types of Collision
Types of Collisions on the Basis of Law of Conservation of Kinetic Energy
During collisions, there will be an exchange of momentum between the bodies. However, the kinetic energy of the bodies may remain constant or change. Accordingly, the collisions are classified into two kinds, namely are elastic collisions and inelastic collisions.
Perfectly Elastic Collisions
If, in a collision, kinetic energy of the system after collision is equal to kinetic energy before collision, the collision is said to be perfectly elastic. For perfectly elastic collision, both linear momentum and kinetic energy are conserved.
Examples of perfectly elastic collisions are i) Collision between atomic particles ii) Collision of α-particle with nucleus.
Inelastic Collision
If, in a collision, kinetic energy of the system after collision is not equal to kinetic energy before collision, the collision is said to be inelastic. For such collisions, linear momentum is conserved but, kinetic energy is not conserved. However, total energy is conserved.
Perfectly Inelastic Collision
If, in a collision, the two bodies stick together and move with a common velocity, then the collision is called perfectly inelastic collision. For perfectly inelastic collision, linear momentum is conserved, but kinetic energy is not conserved. Total energy is conserved.
An examples of a perfectly inelastic collision is the collision between a bullet and a block of wood into which it is fired, when the a bullet remains embedded in the block.
Classification of Collisions on the Basis of Direction of Motion of Colliding Bodies
On the basis of direction of colliding bodies, collisions are classified into two types.
(i) Head-on or one-dimensional collision
(ii) Oblique collision.
Head–On or One–Dimensional Collision
In a collision, if the motions of colliding particles before and after the collision are along the same straight line, then the collision is said to be head–on or one–dimensional collision.
Oblique Collision
In a collision, if the motion of colliding particles before and after the collision are not along the initial line of motion, then the collision is said to be oblique collision. In an oblique collision, if the particles before and after collision are in the same plane, then the collision is called two-dimensional, otherwise, the collision is called three-dimensional.
Line of Impact
When t wo bodies collide, their surfaces are in contact during collision. If a line is drawn perpendicular to the common tangent to the surfaces at the point of contact, the line is called as line of impact.
When the centres of mass of the colliding bodies are initially moving along the line of impact, then the impact is called head-on collision. When the centres of mass of the colliding bodies are not initially moving along the line of impact, then the impact is called oblique.
9.2.2 Analysis of Collision
When objects collide, there are some changes that occur with respect to force, energy, and velocity. The changes are explained below.
1) During impact, each of the colliding bodies experiences strong force. Therefore, during the period of impact, each of the bodies changes its momentum, but as a whole, the total momentum of the system does not change.
2) Impact is practically followed by light, sound, heat, etc. Therefore, mostly, during impact or collision, the mechanical energy of the system practically is not conserved.
3) In ideal case of collision, such as collision between gas molecules, atoms, electrons, protons, etc., the KE of the system of colliding particles remains constant before and after the impact. This type of collision is known as perfectly elastic collision. Remember that, during collision, the KE of the particles change due to large impulsive force.
4) In some collisions, the KE of the system changes. In this case, we cannot term the collision as perfectly elastic.
5) For perfectly elastic collision, the KE of the system at the begining and end of the impact must remain constant but not during the impact.
6) If the colliding particles/bodies stick together, they move together with the same velocity. This collision is known as perfectly inelastic. In this process, sometimes, KE is lost completely. For example, in dropping a stone into mud, the stone loses its total KE Sometimes, KE is fractionally lost e.g., the collision of a bullet and a hanging target.
7) Relative v elocity of the bodies just after the collision (velocity of separation) may or may not be equal to that before the collision along the line of impact (velocity of approach).
8) All the particles participating in a collision form a system.
It is postulated that the total period of impact D t is divided into period of deformation and period of recovery. The impulse of deformation and recovery are finite although the time interval of impact is extremely small.
Deformation period Recovery period
When two bodies collide against each other they remain in contact with each other for some time interval, after which they get separated from each other. The period of time during which the two bodies remain in contact with each other is known as period of impact
When two bodies with unequal velocities strike against each other, they start pressing against each other’s surfaces, and a local deformation results at the point of impact. The process of “pressing” continues until the two bodies attain the same velocities. This period of “pressing” against each other is known as period of compression.
After the two bodies attain equal velocities, due to transfer of momentum, they begin to separate from each other, “pushing” against each other. The period of time elapsed between the instant they attain equal velocites and the instant they separate from each other is known as period of restitution.
Key Insights:
■ The time of interaction should be very small compared to the time of motion (observation) of the interacting bodies, and during this time of interaction, the bodies must experience considerably larger forces compared to the forces acting on them in their general (or normal) state of motion.
Consideration of Line of Impact
It is most important to know the line of action of impact during the collision. The line of impact is the line along which the impulsive force acts on the bodies or the impact takes places. How do you find the line of impact using this concept?
Step 1: Let two bodies 1 and 2 collide. First of all, find the point of contact just before the collision.
Step 2: Draw a straight line passing through the point of contact of the bodies so that it does not touch any part of their surfaces. That is known as tangent to the surfaces in contact, or we can call it the common tangent.
9: System of Particles and Rotational Motion
Step 3: After drawing the tangent, draw a normal to the tangent at the point. This normal line is known as line of impact. This means that the impulsive force acts on the bodies along this line.
After finding the line of impact, calculate the components of their relative velocity just after and before the impact, along this line (common norm al) and set their ratio equal to e.
21 12 n n v e u = , where (u12)n and (v21)n are the magnitudes of components of relative velocity between them along the line joining them just before and just after the impact, respectively. ( u 12) n is known as velocity of approach and (v21)n is known as velocity of seperation.
9.2.3 Perfectly Elastic Head-on Collision between Two Bodies
Let us consider a n e lastic one-dimensional (head –on) collision between two particles. Let m1 and m2 be the masses of two particles. Suppose u 1, u 2 and v 1, v 2 be their respective velocities before and after the collision.
Applying the law of conservation of linear momentum, we have 11221122 mumumvmv +=+ or ()() 111222 ...(1) muvmvu −=−
According to law of conservation of kinetic energy, we have
+=+ or ()()() 2222 111222 ....2 muvmvu −=−
Dividing eq. (2) by eq. (1), we get () () () () 2222 1122 1122 uvvu uvvu = (or) () () () () () () 11112222 11 22 uvuvvuvu uvvu −+−+ = (or) 1122 uvvu +=+
(or) ()() 1212 ....()uuvva −=−−
Th us, in an elastic collision, in one dimension, the relative velocity of approach before collision is equal to the relative velocity of separation after collision.
From eq. =+− 1221 (),avvuu
Substituting this value of v1 in eq. (1), we get ()() 11221222 muvuumvu −+−=−
(or) 1112122222 2mumvmumvmu −−=− 1112221222 2mumumumvmv −+=+
(or) ()() 11212122 2mummummv +−=+ () () () 212 11 2 1212 2 () mummu vb mmmm ∴=+ ++
Further, from eq. (a), v2 = u1 + v1 – u2
Substituting this value of v2 in eq. (1), we get
(or) 1111212122 2
(or) 1121222111 2
(or)
Here, we consid er the following special cases:
1) When m1= m2 = m, then from eq. (b) and (c) v1 = u2 and v2 = u1
Hence, in case of an elastic collision of two bodies of equal masses, the bodies simply exchange their velocities after collision.
2) When u 2 = 0, i.e., the second body is at rest, we have
Now, in this case, we consider the following conditions:
■ If m1 = m2 = m, then v2 = u1 and v1= 0
In this case, the colliding particle comes to rest and the other particle (initially at rest) moves with the original velocity u1 of the colliding particle.
■ If m2>>m1, then v1= –u1 and 2 0 v
Thus whe n a lighter particle collides with a much heavier particle at rest, the
lighter particle rebounds with the same velocity while the heavier particle remains approximately at rest.
■ If m1>>m2, then v1= u1 and v2 = 2u1
Thus, when a heavy particle collides with a much lighter particle at rest, the velocity of the heavy particle remains practically unchanged but the lighter particle moves with double th e velocity of the heavier particle.
11. A sphere of mass 2 kg moving with a velocity collides elastically with another sphere which is at rest. If after the collision, the first sphere moves in the same direction with its velocity reduced to one third of its initial velocity, find the mass of the second sphere.
Sol. Here m1 = 2 kg, u1 = u1, u2 = 0, vu 11 1 3 = In elastic collisions
12. T wo perfectly elastic spheres of masses 2 kg and 3 kg moving in opposite directions with velocities 8 ms–1 and 6ms–1 respectively collide with each other. Find their velocities after the impact.
Sol. Here m1 = 2 kg, m2 = 3kg, u1 = 8ms–1, u2 = –6 ms–1
According to the law of conservation of linear momentum.
m1u1 +m2u2 =m1v1 +m2v2 i.e.,(2)(8) + (3) (–6) = 2 v1 + 3 v2
⇒ 2v1 + 3v2 = –2 .... (i)
Since the collision is elastic, u1 – u2 = v2 – v1 i.e., 8 – (–6) = v2 – v1
⇒ v2 – v1 = 14 .... (ii)
On solving the equations (i) and (ii) we get v1 = –8.8 ms–1 , v2 = 5.2 ms–1
Try yourself:
10. A particle of mass m moving with a speed v hits elastically another stationary particle of mass 2m on a smooth horizontal circular tube of radius r. The time in which the next collision will take place is equal to m
Key Insights:
■ Coefficient of restitution is termed degree of elasticity.
■ Practically, e lies between 0 and 1 .
■ For a perfectly elastic collision, e = 1
■ For a completely inelastic collision, e = 0
■ The equation of coefficient of restitution is applied in the direction of line of impact of the bodies.
■ The value of e depends on the nature of the colliding bodies.
■ e is a number. It has no units and it is dimensionless
Determination of Coefficient of Restituion
Try yourself:
11. A small bl ock of mass M moves with velocity 5 m/s towards another block of same mass M placed at a distance of 2 m on a rough horizontal surface. Coefficient of friction between the blocks and ground is 0.25. Collision between the two blocks is elastic, the separation between the blocks, when both of them come to rest, is (g = 10 m/s2)
Ans: 3 m
9.2.4 Coefficient of Restitution
Coefficient of restitutio is defined as the ratio of relative velocity of separation after collision to the relative velocity of approach before collision. = relative velocity of seperation after collision relative velocity of approach before collision
To d etermine the coefficient of restitution between two materials ,one of them is taken in the form of a very heavy plate and the other in the form of a small sphere. The small sphere is dropped onto the plate from a height ' h'.
It hits the plate with a velocity v1 (as shown in Fig. 4). Let the sphere rebound to a height 'h2 ' after collision with the plate. Assume the velocity of the plate before and after collision to be zero (u2 = v2 = 0) since the plate is very heavy.
12 2 vgh =−
v1 is in opposite direction to u1.
2 12 11 1 2 2 gh vh e uh gh ∴===
1. Equation for the height attained by a freely falling body after number of rebounds on the floor
A body is dropped from a height h0. The body strikes the ground and it rebounds to a height h1. It again falls from that height and rebounds to a height h2 and so on.
22 2 vuas −=
2 2 022()01 eghgh ⇒−=−
2 2201 eghgh⇒⋅=
2 10 heh⇒=
Similarly, after 2nd rebounce, the body rises to a height h2, which is given by
42(2)
200 heheh ==
Similarly,
62(3) 300 heheh ==
2 0 n n heh =
2. Let V, V1 be the velocities of ball just before and just after 1st impact with the ground.
11 1 0 0 VV eVeV VV ==⇒=
Fig. 5
22 2 vuas −=
22 0 02 vgh⇒−=
0 2 vgh =
The velocity with w hich it strikes the ground just before 1st impact is
0 2 vgh =
21 12 vv e uu =
1 0 0 20 v e gh =
10 2 vegh =−
Let the ball rebound to a height h1
If V2 is the velocity after 2nd impact, V2 = eV1 = e(eV)=e2V
Similarly, velocity of the ball just after nth impact if V n =enV
3. Total distance travelled by the body before it stops bouncing is 2 0 2 1 1 Hhe e + =
4. Total time taken by the ball to stop bouncing is 0 2 1 1 he T ge + =
13. A ba ll is dropped from a height h on to a floor. If, in each collision, its speed becomes e times of its striking value (a) Find the total change in momentum of the ball (b) Find the average force exerted by the ball on the floor.
9: System of Particles and Rotational Motion
Sol. (a) Change in momentum in I collision
mvmvmvv 10 10
Change in momentum in II collision, = m (v2 + v1)
Change in momentum in nth collision, = m (v n + vn–1)
Adding these all, total change in momentum of the ball is
(b) Now as so, We know, Fdp dt Fp T T h g e e av 21 1 f from , 12 1 1 2 1 12 1 1 pmghe e Fmghe e g h e av e mg
14. A body ‘ A ’ with a momentum ‘ P ’ collides with another identical stationary body ‘ B ’ one dimensionally. During the collision, ‘ B’ gives an impulse ‘J’ to the body ‘A’ . Then the coefficient of restitution is
Sol. According to law of conservation of linear momentum, m1u1 + m2u2 = m1v1 + m2v2 i.e., mu + m(0) = mv1 + mv2
⇒ P – P1 = P2 where P2 = J, given \ Coefficient of restitution, e vv u mvmv mu PP P PPP P PP P e JP P J P 21 21 21 2 22 1 22 2
Try yourself:
12. A steel ball is allowed to fall freely from a height of 36 cm onto a smooth floor. Find the height to which it rises after rebounding from the floor. Coefficient of restitution between the steel ball and the floor = 0.60.
Ans: 0.13 m
9.2.5 Inelastic Head-on Collision
Two bodies of masses m1 and m2 are moving with velocities u1 and u2 (u1 > u2). They collide as a result of this collision, their velocities change to v1 and v2 respectively.
u1 v1 v2
Before collision During collision After collision m1 m1 m2 m1 m2 m2 u2
Fig.6
11221122 ...(1) mumumvmv +=+
21 12 ; vv e uu = ()2112 vveuu −=−
2112 ....(2) vveueu =+−
1212 ....(3) vveueu =−+
Substitute v2 and v1 from (2) and (3) in (1) and on simplifying we have, 1222 11 2112 (1) memmeu vu mmmm
−+ =+ ++ 1121 22 1221 (1) meumem vu mmmm
+− =+ ++
Key Insights:
■ If m1= m2 = m and u2 = 0 11 12(1);(1)22 uu veve =−=+
Key Insights:
■ In the above case, if the two bodies move in opposite directions, then
9.2.6 Perfectly Inelastic Head-on Collision
In a collision, if the bodies stick together or one gets embedded into other and move with common velocity after collision, it is called a perfect inelastic collision. u1 u
Before
After
Applying the law of conservation of linear momentum, we have m
■ If 'e' is the co-efficient of restitution in a collision, then loss in KE is
when they move in same direction initially.
■ For any collision, general formula for loss in KE is
15. Ball 1 collides with an another identical ball 2 at rest as shown in figure. For what value of coefficient of restitution e, the velocity of second ball becomes two times that of 1 after collision? 1 2
Sol. Here, m1 = m2 and u2 = 0. After collision, v e u 2 1 2 and v e u 1 1 2 Given , v2 = 2 v1 on solving we get, e = 1 3
16. 6.65 A sphere of mass 50 × 10–3 kg moving with a velocity of 2 ms–1 hits another sphere which is at rest. Assuming the collision to be head–on collision and if they stick together after collision and move in the same direction with a velocity of 0.5 ms –1, find the mass of the second sphere
Sol. m1 = 50 × 10–3 kg, u1 = 2 ms–1, u2 = 0, v = 0.5 ms–1, v = v1 = v2 = 0.5 ms–1 m1u1 + m2u2 = (m1 + m2) v
Try yourself:
13. After perfectly inelastic collision between two identical particles moving with same speed in different directions, the speed of the combined particle becomes half the initial speed of either particle. The angle between the velocities of the two before collision is
Ans: q = 120°
Try yourself:
14. A particle of mass 0.1 kg moving with an initial speed v collides with another particle of same mass kept at rest. If after collision total energy becomes 0.2 J, then the minimum and maximum values of ‘v’ are
Minimum value of v is 2 m/s in case of perfectly elastic collision and maximum value of v is 22 m/s in case of perfectly inelastic collision.
Ans: q = 2 m/s
9.2.7 Ballistic Pendulum
Ballistic pendulum is an arrangement to determine the velocities of bullets. A log of wood of mass M is suspended by two strings, as shown in the Fig. A bullet of mass m is fired into the wooden block with a velocity u. The bullet gets embedded in the block.
The system rises to a height h, as shown in the figure. Applying the law of conservation of linear momentum, we have
KE of the system after collision is given by 2 1 KE() 2 mMv=+
PE at highest point =( m + M) gh
Using conservation of mechaical energy, KE at the lowest point = PE at highest point
Loss of KE = KEi – KEf
17. A bullet of mass 2 g travelling with a velocity of 500 ms–1 is fired into a block of wood of
mass 1kg suspended from a string of 1 m length. If the bullet penetrates the block of wood and comes out with a velocity of 100 ms–1, find the vertical height through which the block of wood will rise (assuming the value of g to be 10 ms–2).
Sol. Let the masses of the bullet and the block be ‘ m ’ and ‘ M ’ respectively. Let their velocities after the impact be v and V respectively. Let the initial velocity of the bullet be ‘u’.
According to the law of conservation of linear momentum. mu = mv + MV
Here, m = 2 × 10–3 kg, u = 500 ms–1, v = 100 ms–1
210 500 210 100 1 33 V
V = 0.8 ms–1.
When the block rises to a height of ‘h’, according to the law of conservation of energy.
Try yourself:
15. A wooden block A of mass m is suspended by a string of length ℓ. Another wooden block B moving horizontally strikes block A and sticks to it. If the combined mass rises to a height h above the initial position of A, find initial velocity of B.
Ans: 22 gh
9.2.8 Oblique Collision
Consider the situation in which a smooth sphere impinges upon a smooth fixed plane in a direction other than the normal at the point of impingement C.
Let 'u' and 'v' be the velocities of the sphere before and after the impact, making angles ‘ q ‘ and ' α ' respectively with the normal at C.
The components of u , along and perpendicular to the normal at C are ucosq and usin q respectively.
Similarly, the components of v along and perpendicular to the normal at C are v cos α and v sin α respectively.
Now, the components of velocities along the plane AB remain unchanged, as there is no friction.
Thus u sin q = v sin α ...(1)
However, the components of velocities normal to plane AB, change according the Newton’s experimental laws.
Thus, (cos0)(0cos) veuα−=−θ cos(cos)veu ⇒α=−θ or v cos α = eu cos q (numerically) ... (2)
From equations (1) and (2) 222 sincos....(3)ue =θ+θ and tan tan ....(4) e θ α=
Equations (3) and (4) give the magnitude and direction of the velocity after impact.
Special cases:
1) If the impact be a direct one, then q = 0
Equation (3) becomes v = eu and equation (4) gives α = 0 which have been already discussed.
2) If e = 1 (for an elastic collision) from equations (3) and (4) we get v = u and
α = q which shows that the velocity of approach and separation are same, as well as the angles of approach and separation are also same.
3) For a perfectly inelastic collision e = 0
Equations (3) and (4) yield that
v = u sin q and α = 90°
Thus, the component of velocity along the plane is retained, but the component of velocity normal to the plane “dies off”.
Loss in K.E. Due to the Impact
2 1cos22 2 u me
Special cases:
1) For a perfectly elastic collision e = 1 \ Loss in KE = 0
2) For a perfectly inelastic collision e = 0 θ
∴= 22 cos Loss in KE 2 mu
Impulse Due to Impact
Since the velocity components along the plane remains unchanged, therefore there is no impulse along the plane AB. However, the impulse of the impact, normal to the plane AB = Change in momentum along the normal to the plane AB
18. Two billiard balls of the same size (radius r) and same mass are in contact on a billiard table. A third ball also of the same size and mass strikes them symmetrically and remains at rest after the impact. The coefficient of restitution between the balls is
v u v sin; r r2 1 2 300
From conservation of linear momentum mumvv u e == = 230 3 0 cos or Now relative velocity of separation relattive velocity of approach in common normal direction Hence, e v u u u ===
Try yourself:
16. A ball of mass m hits a floor with a speed v making an angle of incidence q with the normal . The coefficient of restitution is ‘e’. Find the speed of the reflected ball and the angle of reflection of the ball.
Ans: evv e in’s s,co nta 22 21 an=t’
Applications of Oblique Impact
= mv cos α – m(–u cos q ) [in the upward direction] = mv cos α + u cos q ] = m[eu cos q + u cos q] = m(1 + e) u cos q.
Application-I: Consider the collision between block A and ball B. Block A is constrained to move in horizontal direction and ball B is free to move in the plane of the figure. All the surfaces have been considered smooth. At the instant of collision, action and reaction force F (internal force) is exerted on block A in vertical direction by ground. Following equations can be set up:
[t = tangential,n =normal]
VA, VB are initial velocities of block and ball respectively and V' A, V' B are final velocities.
1. The component along t -axis of the momentum of ball B is conserved since friction is absent. Hence the t component of the velocity of ball B remains unchanged, i.e., (VB)t = (V'B)t
2. The component along the horizontal x-axis of the total momentum of block A and ball B is conserved,
collides at some distance with a wall parallel to y -axis as shown in Fig. Let V x and V y be the components of its velocity along x and y directions at the time of impact with wall.
Co-efficient of restitution between the ball the wall is e. Component of its velocity along y direction (common tangent) V y will remain unchanged while component of its velocity along x-direction (common normal) V x will become eV x in opposite direction.
i.e., mAVA+ mB (VB)x= mAVA'+ mB(VB')x
3. The component along the n -axis of the relative velocity of block A and ball B after impact is obtained by definition of coefficient of restitution, i.e., (V'B)n –(V'A)n = e [(VA)n–(VB)n]
Application-II: Suppose a ball is projected with speed u at an angle q with horizontal. It
Further, since V y does not change due to collision, the time of flight (time taken by the ball to return to the same level) and maximum height attained by the ball will remain same as it would had been in the absence of collision with the wall.
Thus, tOAB = tCD+ tDEF and HA= HE
Fig.10
19. Prove that in case of oblique elastic collision of two particles of equal mass if one is at rest, the recoiling particles always move off at right angles to each other.
Sol.
In elastic collision momentum is conserved. So, conservation of momentum along x - axis yields mumvmv uvv 11 22 11 22 1 coscos ,cos cos i.e. and along g -axis yields y vv 02 11 22sinsin
Squaring and adding Eqns. (1) and (2), we get
uvvvv 2 1 2 2 2 12 12 2 cos .... (3)
As the collision is elastic 1 2 1 2 1 2 2 1 2 2 2 mumvmv
i.e., uvv 2 1 2 2 2 ... (4)
Put uvv 2 1 2 2 2 in Eqn (3) , we get 20 12 12 vv cos
As it is given that vv 1200 ≠≠ and
So, i.e cos, , 12 12 0090
Try yourself:
17. A steel ball A of mass 2 kg is rest on a smooth horizontal floor. Another steel ball B of mass 1 kg, moving with a velocity of 2 m/s striks A obliquely and clasifically. After the collision, direction of velocity of B makes an angle of 60° with its original direction of motion. Find velocity of A after the inpact.
1Ans: m/s
TEST YOURSELF
1. Choose the correct statements.
A) For any collision, momentum transfer takes place.
B) In any collision, energy transfer takes place.
C) Impulsive forces are involved in collisions.
(1) A and C are true.
(2) A and B are true.
(3) A, B and C are true.
(4) A, B, and C are false.
2. During a collision, a relatively large force acts on each colliding particle for a relatively _____time
(1) short
(2) moderately short
(3) large
(4) depends on a particular case
3. A heav y sheet of wood and a light ball is moving towards each other, as shown in the figure. What will be the speed of the ball after collision? M M>>>m 2v v m Wall e = 1/3
(1) 3v (2) 2v (3) v (4) v/2
9.3 ROTATIONAL MECHANICS
In gen eral, motion of an extended object, especially when its shape gets altered continuously, can be very complex. However, our interest is to study the motion of an object which retains its shape during the time of our observation. Any object may do so temporarily but no real object can maintain its shape or configuration permanently in all the circumstances. “An ideal object whose shape can never be changed, whatsoever may be the external influences, is called as a perfect rigid body or simply a rigid body.” From this very definition, it follows that the motions of different parts of a rigid body are not entirely arbitrary. ‘In all the possible motions of a rigid body as a whole, the separation between any two parts of the rigid body must be constant.’
Moreover, when any point in a rigid body is fixed then it can do nothing but turn or rotate about the fixed point. One important aspect in the rotation of a rigid body is that all the points in the rigid body rotate by same angle and in the same sense. Therefore, instantaneous angular velocities of all the particles in a rigid body must be same. In this chapterwe are going to deal only with rigid bodies (nonexistant though!). What types of motion can a rigid body have? Here we must remember that complete analysis of a rigid body motion is beyond our scope and we are going to restrict ourselves with the planar motion only. By planar motion we mean that the planes of motion of the particles in a rigid body are parallel or non-intersecting. Examples for planar motion are-pure translation, rotation and rolling.
9.3.1 Kinematic Analysis of a Rigid Body
Any kind of mo tion of a rigid body can be visualized in a two stage process (we can procede in any convenient order):
Stage-I: Find complete motion of centre of mass for a given time period.
Stage-II: Find instantaneous state of the body as a whole at any given time.
Combi ning t hese two stages vectorially gives us the complete motion of the rigid body. Stage-I can be computed from the fundamentals of previous chapter. Then what about stage-II ? Instantaneous state of a moving body is like a snap shot of the body taken at the required instant. StageII reveals rotational state of the body. At a given instant, a rigid body in planar motion can only have two possible motions: pure translation and rotation. When a body is in pure translation all the points in the body must have equal instantaneous velocity vectors. And if instantaneous state of the body is not of pure translational, all the particles in the body appear to be rotating in co-axial circles about a common axis.
Kinematic eq u ations of a rigid body in rotation are same as that of an isolated particle in circular motion.
Angular Velocity and Angular Acceleration
Consider one particle of a wheel rotating about its axis of symmetry, which is fixed in space. We can specify the position of the particle P i by the distance r i from the centre of the wheel and the angle q i between a line from the centre to the particle and a referene line fixed in space, as shown in Fig.
In a small time dt, the particle moves along the arc of a circle by a distance dsi, where,
dsi = vi dt ....(1)
where vi is the speed of the particle. The angle swept out by the line from the centre to the particle, in radians, is this distance divided by the radius ri:
From equ ation (2) the speed of the i th particle is related to its radius r i and the angular velocity of the wheel by
A wheel rotating about a fixed axis through its centre. The distance dsi moved by the ith particle in some time interval depends on i, but the angular displacement d q is the same for all particles on the wheel.
Althoug h t he distance ds i varies from particle to particle, the angle d q swept out in a given time is the same for all particles. For example, when one particle moves through a complete circle, so do all the other particles and Dq =2 π rad for all the particles. The rate of change of angle with respect to time, dq/dt, is the same for all the particles of the wheel. It is called the angular velocity w of the wheel, ()...3 d
The angular velocity w is positive or negative depending on whether q is increasing or decreasing. The units of angular velocity are radians per second. Since a radian is a dimensionless unit of angle, the dimensions of angular velocity are those of reciprocal seconds s-1). Although angular velocity is frequently expressed in units other than radians per second, it is important to remember that expressions like equations (3) and (4) and the other results we obtain for rotational motion are valid only when angles are expressed in radians.
In vector notation Vr =ω×
If a par ticle has velocity V while it is rotating in the plane of the paper with its radius vector r , then its angular velocity ω will be in a plane perpendicular to the plane of the paper along the axis of rotation.
Since angles are often expressed in degrees, angular velocity is sometimes expressed in degrees per second or degrees per minute. In such cases, the angular velocity can be expressed in radians by using 2 π rad =360°.
Angular velocity is also commonly expressed in revolutions per second and revolutions per minute. Conversion to radians is accomplished by using 2 π rad =1 rev.
20. A particle is moving anticlockwise in a circular path of radius 2 m with an angular velocity w = 5 rad/s. Find the vector expression of its velocity when the radius vector of the particle makes an angle of 60° with x-axis.
Vjims P P 53 55 3 k / /.
Angular acceleration ( α ): We define angular acceleration a as the time rate of change of angular velocity. Thus, d dt
Try yourself:
18. A particle is moving in a circular path, about z-axis with angular velocity 10 rad/s. Plane of the circular path is parallel to xy-plane. Find the velocity of the particle when its co-ordinates are (1 m, 1 m, 1 m)
Ans: 10(–î + ĵ) m/s
The rate of change of angular velocity with respect to time is called the angular acceleration α. For rotation about a fixed axis, 2 2 dd dtdt ωθ α==
The units of angular acceleration are radians per second squared. The relation between the tangential linear acceleration of the ith particle of the wheel and the angular acceleration is
obtained by taking the derivative of the speed (not velocity) vi in equation (4) with respect to t: i iti dvd ar dtdt ω ==
Thus ait = riα
Each particle of the wheel also has a radial linear acceleration or the centripetal acceleration which points inward along the radial line, and has the magnitude.
2 2 i ici i v ar r ==ω
The three quantities angular displacement q, angular velocity w, and angular acceleration α are analogous to the linear displacement x, linear velocity v x, and linear aceleration a x respectively in one-dimensional motion. Because of the similarity of the definitions of the rotational and linear quantities, much of what we learned in previous Chapters will be useful in treating problems of rotation about a fixed axis. For example, if the angular acceleration is constant, we can develop constant angular-acceleration expressions. If, for example, we have
0 tan d const dt ω α==α=
then w = w 0 + α0 t
where w0 is the angular velocity at t = 0, and
2 000 1 2 ttθ=θ+ω+α
As with the constant-linear-acceleration formulae, we can eliminate time from these equations to obtain an equation relating the angular displacement, angular velocity, and angular acceleration:
9.3.2 Kinetic Energy of a Body about a fixed Axis-Moment of Inertia
Consider a rigid body which rotates with an angular velocity w about a fixed axis. And consider an elementary particle of mass 'dm' located at a distance 'r' from the axis of rotation. The particle traces a circular path about the axis of rotation and radius of this circle is 'r'. Its velocity vector lies in the plane of the circle traced and is perpendicular to radius of the circle. And speed of the particle (say v) is: v = r w
and for a system of descrete particles, 2 ...(17) ii Irm =∑
The new physical quantity I is called second ordered moment of mass or moment of inertia.
If we look at the Eq (15) for kinetic energy of a rotating body it reminds us the similar expression for kinetic energy in “Particle Dynamics”:
Its kinetic energy becomes
Then total kinetic energy of the body is
But ω is same for all the particles in a rigid body.
Try to get anology between these two expressions- “Role of ω in rotation is same as that of v in translation”. Moreover we know that mass is the measure of inertia in translation and so I should be the measure of inertia in rotation.
Moment of inertia of a body depends on (a) the mass of the body (b) the position of axis of rotation and (c) the distribution of the mass of the body about the axis.
If we consider a heavy body and a lighter body of same size and shape rotating about the same axis, heavy body will have greater moment of inertia as its mass is more.
If a hollow body and a solid body of same mass and shape are rotated about the same axis, the hollow body will have greater moment of inertia as the particles of hollow body are at greater distances from the axis of rotation.
Consider a person rotating on a turn table. If he folds his hands his moment of inertia will be less and when he stretches his hands it will be more as distance of particles will be more from the axis on stretc hing hands.
21. A rigid body is rotating about a fixed axis and its kinetic energy of rotaion is 100 joule. If its angular momentum changes by 50%, what will be its final kinetic energy?
Sol. Initially, joule
. joule = 225 joule.
Try yourself:
19. If angular momentum of a body spinning about a fixed axis changes by 1%, find the corresponding change in its kinetic energy.
Ans: 2%
9.3.3 Radius of Gyration
From the abo ve e xample it is evident that in general, moment of inertia depends not only on the total mass but also on how the mass is distributed with respect to the axis of rotation. It also depends on orientation of the axis. When a body rotates, various points in the body rotate in coaxial circles of different radii. Then r.m.s value of these radii (K) can be calculated as follow:
a ‘fly wheel’. Because of its large moment of inertia, the fly wheel resists the sudden change in speed of the vehicle. It allows a gradual change in the speed and prevents jerky motions, thereby ensuring a smooth ride. There are two very useful theorems which will enable us to find easily moment of inertia of many objects about a given axis of rotation. They are: Perpendicular axis theorem and Parallel axis theorem.
22. Four particles, each of mass 100 g, are placed at the corners of a square of side 10 cm. Find the moment of inertia of the system about an axis passing through the centre of the square and perpendicular to its plane. Also find the radius of gyration of the system.
Sol. Mass = = 100 g, radiuscm Moment of inertia mr Imr 52 4 41 2 0 00 52 210 210 04 0077 23 2 3 () . kg cm m kI M m
Try yourself:
20. The radius of gyration of a system of particles about z -axis is 0.5 m. If total mass of the system is 2 kg, find momentum of inertia of the system about z-axis.
0.5Ans: kg 2m
9.3.4 Theorems on Moment of Inertia
K is called r.m.s radii of the body or radius of gyration of the body about the given axis.
Radius of gyration of a rigid body depends on the position of axis of rotation and distrbution of mass about the axis of rotation.
Moment of inertia as a property of rotational inertia has been used in many practical applications. The machines, such as steam engine and the automobile engine, etc., that produce rotational motion have a disc with a large moment of inertia, called
Moment of inertia theorems play a crucial role in the analysis of rotational motion and distribution of mass within a rigid body. These theorems, such as the parallel axis theorem and the perpendicular axis theorem, provide mathematical tools to simplify calculations of the moment of inertia for various shapes and axis orientations.
Perpendicular Axes Theorem:
Let us consider a planar body which lies in XY plane as shown in Fig. Let P be an elemental
| CHAPTER 9: System of Particles and Rotational Motion
point of mass ‘dm’. ‘P is located at a distance y from X-axis, x from Y-axis and 22 xy + from Z-axis.
Y P (x,y)
Moment of inertia of the body about x-axis is
2 x Iydm =∫
Moment of inertia of the body about y-axis is
2 y Ixdm =∫ and moment of inertia of the body about Z-axis is
22
zyx IxdmydmII =∫+∫=+
zxy III∴=+
This is known as perpendicular axis theorem. Note that this theorem is applicable only for planar bodies.
23. Radius of gyration of a disc of mass 5 kg about a transverse axis passing through its centre is 14.14 cm. Find its radius of gyration about its diameter and hence, calculate its moment of inertia about its diameter.
Sol. Radius of gyration of a disc about a transverse axis passing through its centre
K I M MR M R I MR 2 2 2 2 14 14 2 . ;. cm
Radius of the disc, R 14 14 220 .cm
Radius of gyration of the disc about its diameeter, cm
Moment of inertia about its d K I M MR M R 2 42 20 2 10 i iameter kg m
MR 22 22 4 50 2 4 50010 (. ) .
Try yourself:
21. Radius of gyration of a thin ring of mass m passing through its centre and perpendicular to the plane of the ring is R. Then find its moment of inertia about a tangent lying in the plane of the ring.
1.5Ans: 2mR
Parallel Axes Theorem
This theorem allows us to find I even if the axis does not pass through centre of mass.
Consider two parallel axes 'A-A' and 'B-B' as shown in Fig. 'A-A' passes through centre of mass C and 'B–B' passes through origin O. B B A A d O C P (x,y) (x0,y0)
For simplicity let us assume that the axes are parallel to Z-axis.
Then the separation between the axes 'A' and 'B-B' is 22 00dxy =+ . Assume that I of the body about the axis 'AA' is known to us and we want to find the same about the parallel axis 'BB'.
Consider an elementary mass P at (x,y). This elementa ry mass is at a distance 22 rxy =+ from the axis 'BB' and at a distance ()() 22 100 rxxyy =−+− from the axis 'AA'.
Moment of inertia of the body about the axis 'BB' is,
()222 0 Irdmxydm =∫=∫+
And Moment of inertia of the body about the axis 'AA' is,
2 1 C Irdm =∫ ()() 22 00 xxyydm
()() 2222 00 xydmxydm=∫++∫+ 2200 xxdmyydm−∫−∫ ()() 2 0 220000 IddmxMxyMy =+∫−−
() 00 xdmxMandydmyM ∫=∫= ()222 0 00 2 IMdMxy =+−+
222 00 2 IMdMdIMd =+−=−
2 0 .....(19) C IIMd∴=+
This is known as parallel axis theorem. The body we have taken for deriving this theorem is planar or laminar in xy-plane. This is only for the sake of simplicity. But parallel axis theorem is true for a general three dimensional body. Hence note the difference in applicability of these two theorems.
24. The radius of gyration of a body about an axis at a distance of 12 cm from its centre of mass is 13 cm. Find its radius of gyration about a parallel axis through its centre of mass.
Sol. By parallel axes theorem, M (13)2 = I 0 + M (12)2, I 0 = M (132-122) = M(25)
Its radius of gyration about a parallel axis through its centre of mass
Try yourself:
22. ____ of mass 5 kg moment of inertia of a rigid body about an axis is 12 kg m2. Distance of the COM of the body from the given axis is 0.5 m. Find moment of inertia of the body about an axis passing through its COM and parallel to the given axis.
10.75Ans: kg 2m/s
Application of Theorems of Moment of Inertia
Let us study the applications of parallel and perpendicular axes theorems for a few contnuous bodies.
Moment of Inertia of Thin Rod
Consider a thin uniform rod of length l and mass m. Let AB be an axis of rotation passing through the centre of the rod C and perpendicular to the length of the rod.
Conside r an element of length dx at a dis tance x from C. Mass of that element is dm=(m/l) dx.
Now moment of inertia of that element about AB is dI= (dm)x 2
Moment of inertia of the entire rod about AB is given by 2 /22 0 IdIxdxmm 12 =∫=∫=
If we consider an axis PQ parallel to AB and passing through one end of the rod, moment of inertia about PQ can be found using parallel axes theorem
Moment of Inertia of Circular Ring
Consider a thin circular ring of mass m and radius R. Consider an axis passing through centre of the ring. Ring can be divided into a large number of elements each of mass dm.
About the given axis, moment of inertia of that element is dI = (dm)R 2
O
In the first case, radius of gyration is 12 and in the second case it is 3
About an axis passing through an end and inclined to the rod at an angle α :
In the diagram, AB is the rod of length and mass M.
= sin
About the same axis moment of inertia of the entire ring is 22IdIdmRmR =∫=∫=
Radius of gyration K=R
If we consider an axis along its diameter we can find moment of inertia of the ring using perpendicular axis theroem. Here XX' and YY' be two axes along the diameters of the ring. These two axes are perpendicular to each other and intersect at the centre O.
Moment of inertia of the ring about XX ' is IX Moment of inertia of the ring about YY' is IY
But due to symmetry IX = IY
By perpendicular axes theorem IX+ IY = I
Where I is moment of inertia of the ring about an axis passing through centre and perpendicular to plane of the ring. We have proved that I = mR2 2 2 and 2 XYXY mR IImRIorI ⇒+==
Radius of gyration /2 KR =
Consider a tangent in the perpendicular plane to the circular ring. The moment of inertia of the ring about a parallel axis through the centre of mass is I C=MR2
Moment of Inertia of a Circular Disc
Consider a uniform thin circular disc of mass m and radius R. Consider axis passing through centre of the disc normal to plane of the disc. Moment of inertia of the disc about that axis is I. Divide the disc into a large number of smaller rings. Consider one such ring of radius x and thickness dx. Mass of that ring is 2 2 m dmxdx R =π π Moment of inertia of that ring is dI = (dm)x2
Moment of inertia of the entire disc is . IdI =∫ Limits of x are from zero to R. After integration we get 2 2 mR I =
Moment of inertia of the disc about a tangent perpendicular to its plane is 2 3 2 mR I = (use parallel axis theorem). If that tangent is in the plane of the disc, moment of inertia about that tangent is 2 5 4 mR
By parallel axis theorem, moment of inertia about a tangent perpendicular to the plane of the ring is 2 2 2 IMRandKR ==
Consider a tangent in the plane of the circular ring parallel to the diameter. We know that moment of inertia of the circular ring about a diameter is 2 2 MR By parallel axis theorem, moment of inertia about a tangent in the plane of the ring is I = I C + Md2
2 2 () 2 MRMRdR =+=
2 33 and 22 MR IKR⇒==
About a diameter moment of inertia of the disc is 2 mR 4
25. A hollow thin disc has mass m inner radius R, and outer radius 2R. Find its moment of inertia about a tangent lying in its plane. Sol. Moment of inertia of the disc about an axis passing through its centre C and perpendicular to the plane of the disc is,
1 2 2 5 2 2 2 2
By perpendicular axis theorem,
1 2 5 4 2
By parallel axis theorem, d mRmRmR 2 2 2 2 5 4 2 21 4
Try yourself:
23. A thin disc of mass 1 kg has inner radius 2 cm and outer radius 4 cm. Find its radius of gyration about an axis passing through its COM and perpendicular to the plane of the disc.
Ans: 10 m.c
Moment of Inertia of a Rectangular Lamina
Consider a rectangular lamina of length l and breadth b. Mass of that lamina is m. Moment of inertia of that lamina about an axis passing through its centre and perpendicular to its length is ml2 /12 (same as that of the moment of inertia of a thin uniform rod of length l about an axis passing through its centre and perpendicular to its length)
Similarly moment of inertia of that thin lamina about an axis passing through its centre and perpendicular to its breadth is mb2/12.
Moment of inertia of the rectangular lamina about an axis of rotation passing through its centre and perpendicular to its plane can be obtained by using perpendicular axis theorem.
Where IX =mb2/12 and IY =ml2/12
26. A bar magnet of length 5.0 cm and breadth 1.2 cm is rotated about an axis passing through its centre and perpendicular to its plane. Find its moment of inertia if the mass of the magnet is 200 g.
Sol. Moment of inertia I
Try yourself:
24. A squar thin sheet has moment of inertia I about one side. What is its moment of inertia about on axis passing through its centre and perpendicular to its plane?
Ans: I/2
Moment of Inertia of a Uniform Solid Sphere
a) About an axis passing through its diameter
Consider a solid sphere of mass M and radius R. Its moment of inertia about an axis of rotation passing through its diameter is 2 2 5 IMR =
b) About an axis passing through its tangent
Let A1B1 the tangent to the solid sphere. A parallel axis through its centre of mass is AB. By parallel axes theorem,
Moment of inertia about the tangent = Moment of inertia about a diameter +Mr2
I
Moment of Inertia of a Hollow Sphere (or) Spherical Shell
a) Moment of inertia about an axis passing through the diameter of a hollow sphere of mass M and radius R is
27. Four spheres, each of diameter 2a and mass m, are placed with their centres on the four corners of a square of the side b. Calculate the moment of inertia of the system about any side of the square.
b) Moment of inertia about an axis passing through its tangent can be obtained by applying parallel axes theorem. It is given by
iner t tia of the system,
CHAPTER 9: System of Particles and Rotational Motion
Try yourself:
25. Moment of inertia of a solid sphere about a diameter is I. What is its moment of inertia about an axis whose distance from its centre is 2R?
Ans: 11 I
Moment of Inertia of a Uniform Solid Cylinder
a) About its own axis i.e., axis of symmetry: A solid cyulinder of mass M can be supposed to be consisting of a large number of circular discs placed one over the other. The moment of inertia of each such disc of mass m is 2 2 mR and the moment of inertia of the solid cylender is the sum of the moments of inertia of such discs.
\ Moment of inertia of a solid cylinder about its own axis
22 22 RR mm=∑=∑
2 2 MR I ∴= I = 2 MR2
It is same as that of a circular disc.
b) Moment of inertia about an axis perpendicular to its geometrical axis and passing through its centre of mass is 22
Moment of Inertia of a Uniform Hollow Cylinder
a) Its moment of inertia about its own axis is I=MR2 . It is same as that of a circular ring.
b) Moment of inertia about an axis perpendicular to its geometrical axis and passing through its centre of mass is
28. A solid cylinder has mass m , radius R and length R. Find its moment of inertia about an axis passing through its centre and perpendicular to its length.
Sol. Im RR mR 22 2 12 4 1 3
Try yourself:
26. A thin cylindrical shall has mass m, radius R and length 2R. Find its moment of inertia about an axis passing through one of its ends and perpendicular to its own geometrical axis.
Ans: 11 6 2 mR .
9.3.5 Torque What makes an object to turn?
To open the door of the cupboard, one applies the force at the free ends so that it turns about the hinged end.
Consider the door of your class room and try to turn or to rotate it by applying a force at various points on the door. We find it difficult if the point of application of force is near to the axis of rotation or hinges. And if the force is applied exactly on the hinges, the door cannot move.Have you observed that a tight wheel nut cannot be unscrewed bare hand but the same can be done quite easily using a jockey rod or a spanner? These examples suggest that rotating effect is decided not only by the magnitude of applied force but also by the way it is applied. Consider below figure (a) & (b), when lever arm length is doubled it is seen that the effort or force required to turn the nut is halved.
“turning effect = Lever arm length × Component of force perpendicular to lever arm” (no rotation) nut
(only due to F1)
(lever arm length)
Moreover, if direction of force is reversed, the sense of rotation will also be reversed. Further analysis reveals another interesting point. If F is applied parallel to lever arm of the spanner, it is observed that there is no “turning effect”. And if F is applied at an angle ‘q‘ to lever arm, the component ‘F sin q ‘ alone participates in rotating the system.
From these observations, we can mathematically define the Turning effect as follows:
F11=FCos q F1=FSin q q l
Symbolically, lF τ=×⊥ where t is called “the turning effect” or technically, “torque”. This “turning effect” can be clockwise or anticlockwise. By default, we assigned positive sign for anticlockwise sense. Also note that clockwise sense as seen from one side of plane of rotation becomes anticlockwise sense if we observe the same from other side. To account for the sense of rotation, we define torque as a vector quantity. Hence vectorially torque can be defined as :
|| l is the separation of line of application of force F from point of rotation or fulcrum . Therefore, while calculating torque, line of application of force is as important as the force itself. Direction of torque vector is identified using right hand thumb rule. Mathematically, torque is called as first order moment of force or simply Moment of force.
Let a force F acts on a single particle at a point P in XOY plane whose position vector with respect to the origin O is r
The magnitude of moment of the force or torque of the force F about the point O is given by t = Fd
From figure, sin sin ddr r θ=⇒=θ
t = rFsin q
r is magnitude of position vector r
F is magnitude of the force F
q is the angle between r and F
Torque of the force F about a point O in vector form is given by rFτ=×
The dire ction of torque is perpendicular to the plane containing both r and F . Its direction can be determined by using right hand thumb rule.
Moment of force has dimensions [ML2T-2]. Its dimensions are the same as those of work or energy. It is however, a very different physical quantity than work. Moment of a force is a vector, while work is a scalar. The SI unit of moment of force is newton meter (Nm).
One may note that since is rF × a vector product. Properties of a vector product of two vectors apply to it. If the direction of F is reversed, the direction of the moment of force is reversed. If directions of both r and F are reversed, the direction of the moment of force remains the same.
If there are more than one force acting on the body, we can use principle of superposition to find net torque. Let 2 1 and FF be the forces acting on the body at points of position vectors
1r
and 2r respectively, then
121122 netrFrF τ=τ+τ=×+×
29. When a force of 10.0 N is exerted on the handle of a door, the door can be just opened. If the handle is at a distance of 50 cm from the hinges, find the torque applied on the door.
Sol. rF sin
= 0.50 × 10.0 sin 900 = 5 Nm.
Try yourself:
27. A thin rod AB of length 1.5 m is hinged at end A. A 10 N force is applied at end B as shown in figure. Find the torque of the applied force about end A.
9.3.6 Couple
Ans: 7.5 N-m
Couple is defined as a pair of antiparallel forces of equal magnitude whose lines of action do not coincide. Let 1F and 2F be the forces in a couple, then 12 FF =− . If these forces act on a body simultaneously and the lines of action of these forces do not coincide, we
say that a couple is acting on the body. Net force 12 net FFFzero =+= . Therefore, net force exerted by a couple is zero. Let 1r and 2r be the position vectors of points of application of 1F and 2F respectively, then net torque:
121 ( -)rrFmaybenonzero=−×
Therefore net torque exerted by a couple may be non-zero. The couple has another interesting property; let us look at the expression for net torque due to a couple. What does 12rr refer to? It indicates the separation between the line s of action of
Sol. Moment of couple = Smooth of moment of forces about any point = Moment of 15 N and 5 N forces about point A = (15 × 0.5 + 5 × 1) N–m = 12.5 N–m.
Try yourself:
28. Two equal and opposite forces, each of magnitude 10 N are applied at the ends of a diameter of a disc of radius 10 cm. Find the torque acting on the disc.
Ans: 2 N -m.
9.3.7 Rigid Body in Equilibrium
We now understand that force has a tendency to produce translational effect whereas torque, rotational effect. But the main ingredient of torque is again force. So, primarily, force is the cause of both translational and rotational effects.
12 FandF , which is independent of the choice of the fulcrum. Therefore, “net torque or the moment of the couple will be equal to product of magnitude of single force with the separation between the lines of force and it is independent of choice of fulcrum or of reference frame”.
30. A thin rod Ab of length 1 m is at rest on a smooth horizontal surface. AC is its midpoint. Forces 20 N, 15 N and 5 N are applied at points A, C and B. Find the moment of couple acting on the rod.
Our next task is to answer the question: “Suppose there are a number of forces acting on a rigid body whose lines of action may or may not coincide. Then how will the body move under the influence of these forces?’ It is difficult to answer this question all at once. We will try and answer this in bits and pieces as we procede section by section.
It is relatively easy to study a body which is in equilibrium. In general an object may be in translational equilibrium, rotational equilibrium or both. We know the condition for translational equilibrium: “Net external force on the system must be zero”. Then what is the condition for rotational equilibrium? We may be tempted to guess a similar relation for rotational equilibrium as that of translational equilibrium. At the same time we must keep in mind that rotation and translation are two independent motions. Therefore a body in translational equilirbium need not be in rotational equilibrium and vice versa.
Now, w e need to get a condition for rotational equilibrium quantitatively. For
9: System of Particles and Rotational Motion
the case of analysis let us consider a body which is in both translational and rotational equilibirum. In addition, if the body is at rest then this will be the simplest possible case we can have for a rigid body.Let us consider an example of common balance at rest. Shows a simplified model of common balance. AB is beam of the balance and ‘C’ is fixed point or fulcrum.
2211 dUWldWld =θ−θ
,
AC = l1 and BC = l2 are called arm lengths of the balance. Ideally for common balance, arm lengths must be equal. But let us consider a general case where l1 and l2 may not be equal and still have rotational equilibrium. Then obviously the loads W 1 and W 2 may not be equal. When the balance is at rest, force exerted by fulcrum 'C' on the beam makes sure that the common balance is in equilibrium. Assume that the beam and connecting strings are massless.
Torque due to the load W 1 is t 1= + W1l1
( counter clockwise sense)
and that due to the load W 2 is t 2= –W2l2
( \ clockwise sense)
Then the net torque acting on the beam
121122 . netWlWl τ=τ+τ=−
Now, let beam AB be rotated from the horizontal position slowly by a small angle d q in counter clockwise sense. As a result W1 moves down by a distance l 1 d q and W2 moves up by a distance l2 d q . Net change in gravitational potential energy is
This equation is similar to the equation x dU F dx −= . By applying the known condition that if a system is in equilibrium, its potential energy attains local maximum, local minimum or constant value, we can conclude that dU d θ must be zero at equilibrium.
Using this condition, when a rotating body is in rotational equilibrium, t net = zero.
Again, force can be divided into two groups internal forces and external forces. Internal forces always exist in action-reaction pairs and for each pair there is a single line of action. Hence torque due to action is exactly balanced by that due to reaction. When we add all the torques due to internal forces vectorially it sums up exactly to zero. Then, for any system, int .....(13) netextext τ=τ+τ=τ
And condition for a rotating body to be in rotational equilibrium is ....(14) extzeroτ=
31. A roller of ma ss 300 kg rests against a step of height 20 cm. If the radius of the roller is 50 cm, find the minimum horizontal force to be applied passing through its centre of mass to take the roller on to the step.
Sol. Taking the moments of forces about ‘O’, h = 0.2 m, R = 0.5 m, xRRh 22 2 xRRh 2 2 04 m
To climb up the step, moment due to horizontal force ≥ moment due to weight.
forces F 1 and F 2, parallel to each other, and usually perpendicular to the lever, as shown here, act on the lever at distances d 1 and d 2 respectively from the fulcrum.
FRhmgx
Fmgx Rh
F 300 98 04 03 3920 .. . N
Try yourself:
29. A force Fijk 23 4N is applied at point
P(1 m, 1 m, 1 m). Find the torque of the force about x-axis.
Ans: 1 N-m.
Principle of Moments
An ideal lever is essentially a light rod pivoted at a point called fulcrum along its length. Two
Table 3 Translatory and Rotatory Equilibrium
The lever is a system in mechanical equilibrium. If R be the reaction of support at the fulcrum. Then R=F1+F2 for translational equilibrium. For rotational equilibrium F1d1=F2d2. In the case of lever, F 1 represents some weight or load to be lifted and d1 is called the load arm. Force F2 is the effort to lift the load, and the distance d2 is called effort arm.
F2
The ratio F 1 /F 2 is called mechanical advantage (M.A).
12 21 Fd MA Fd ==
If d2 > d1, then M.A > 1 means small effort can be used to lift a large load.
32. PQR is a rigid equilateral triangular frame of side length ‘L’. Forces F 1, F 2 and F 3 are acting along PQ, QR and PR. Find the relation between the forces if system is in equilibrium.
Sol. If the system is in rotational equilibrium, find the relation between the forces.
Perpendicular distance of any force shown in the figure from centroid C of triangle is L 23
The forces F 1 and F 2 produce anti-clockwise turning effect whereas F 3 produces clockwise turning effect about ‘C’.
Since the system is in rotational equilibrium the total torque acting on the system about the centroid is zero.
9.3.8 Overturning of Vechicle on an Unbanked Path
When a vehicle negotiates a sharp turn on a level road, its driver should be careful about two possible mishaps:
1) skidding and
2) overturning
The cause of skidding and the speed limit imposed for its avoidance have already been discussed in topic laws of motion. Now we study why a vehicle overturns at a curved path and how that can be prevented.
Suppose the radius of the curved path is 'r' the speed of the vehicle is 'v' and the mass of the vehicle is 'm'. The centre of mass 'C' of the system is at a height 'h' above the ground. The distance between the inner and outer wheels of vehicle is '2l'
The forces acting on the vehicle are:
1) The force of gravity 'mg' acting vertically downwards through 'C'
2) The normal reactions N 1 and N 2 acting vertically upward through the inner and outer wheels respectively.
3) The forces of static friction f1 and f2 acting horizontally towards the centre of the curved path at the bottom of inner and outer wheels respectively.
Hence, F1 +F2 –F 3 = 0
\ F 3 = F1 + F2.
Try yourself:
30. The rod shown in figure has mass m and length L. Determine the value of F so that rod remains in equilibrium.
Ans: mgF in2s
Here, the two frictional forces provide the centripetal force. Then
Since the vehicle has no vertical acceleration, we have N1+N2=mg
The various torques acting on the vehicle about the centre of mass is
i) Torque due to N1 is t 1= N1l (clockwise)
ii) Torque due toN2 is t2= N2l (anticlockwise)
iii) Torque due to weight mg is t3= 0. (weight acts through centre of mass)
iv) Torque due to f1 = t 4 = f1h. (clockwise)
v) Torque due to f2= t 5 = f2h. (clockwise)
For the vehicle not to overturn the net torque about the centre of mass ‘C‘ is zero.
i.e. the sum of anticlockwise torques = the sum of clockwise torques.
\ t 2 = t 1+ t 4 + t 5
N2l = N1l +(f1 + f2) h or ()
On solving the equations, we get 22 12 and
Hence N 2 is always positive, but N 1 may become zero (or negative if ‘v‘ is sufficiently large). In that event, the inner wheels leave the ground and the car tends to overturn towards the outer edge of road. Then 2 gr mvh g0 or v 2rh
which should be the speed for the limiting case.
Precautions to avoid overturing at curved path
1) The speed (v) must be reduced before entering the curved path.
2) The road must be banked with the suitable angle q
3) The height of centre of gravity (h) above the ground should be lowered.
4) The wheel base (2l) should be large.
5) The radius of curvature must be large.
Key Insights:
■ In case of overturing of a vehicle on a level road, the inner wheels leave the ground first.
■ The driver of a car travelling with a velocity v suddenly sees a broad wall ahead of him at a distance a. Braking requires less friction than turning sharply in a circle. So braking is preferred.
33. A car is moving along a circular path of mean radius 100 m. Separation between inner and outer wheels of the car is 2 m and distance of centre of mass of the car with driver above the ground is 1 m. Assuming that there is sufficient friction between the wheels and the road surface, find the maximum speed of the car to avoid the risk of overturning.
Sol. VgR h max 10 100 2 1
10 20 m/s m/s
Try yourself:
31. A car is moving along a circular path of mean radius 150 m. Separation between inner and outer wheels of the car is 2 m. The car can move with a maximum speed of 15 10 m/s. There is sufficient friction between the wheels and the road surface to prevent skidding. Find the maximum height of the COM of the car with the driver to prevent over turning.
Ans: 1.33 m.
9.3.9
Work Done by Torque
Let us consider a body rotating about a fixed axis under an external torque τ
rFτ=× where F is force applied on the body. Let the body be rotated by a small angle dq. Then displacement of point of application is
() dldrvr =θ×=ω×
work done is:
()() dWFdlFdrrFd =⋅=⋅θ×=×⋅θ
Wd ∴=∫τ⋅θ
And power developed by the torque is
In a perfectly rigid body there is no internal motion. The work done by external torques is therefore, not dissipated and goes on to increase the kinetic energy of the body. The rate at which work is done on the body is to be equated to the rate at which kinetic energy increases. The rate of increase of kinetic energy is
2 2 22 dId I dtdt
We assume that the moment of inertia does not change with time. This means that the mass of the body does not change, the body remains rigid and also the axis does not change its position with respect to the body.
Since α= dω/dt. 2 we get 2 dI I dt
Equating rates of work done and of increase in kinetic energy.
tw = i
Just as force produces acceleration, torque produces angular acceleration in a body. The angular acceleration is directly proportional to
the applied torque and is inversely proportional to the moment of inertia of the body.
34. A disc of mass 5 kg and radius 1 m can rotate about a spindle passing through a point on its circuference and, initially, the disc is at rest. A torque is applied on the disc such that finally the disc starts rotating at 60 rpm about the spindle. Find the work done by the torque.
Sol. WI MR = Kinetic energy KE
Try yourself:
32. A solid sphere can rotate about fixed spindle passing through its centre. Its moment of inertia about that spindle is 2 kg m2. A torque starts acting on the sphere and after some time its kinetic energy becomes 16 joule. Find the angular velocity of the sphere at that instant.
Ans: 4 rad/s.
Work Energy Theorem
The work done by a constant torque acting on a body is equal to change in rotational kinetic energy.
35. What is the work done in increasing the angular frequency of a circular ring of mass 2 kg and radius 25 cm from 10 rpm to 20 rpm about its axis?
Sol. Work done = Increase in rotational kinetic
36. A solid cylinder of mass M and radius ‘ R ’ is mounted on a frictionless horizontal axle so that it can freely rotate about this axis. A string of negligible mass is wrapped round the cylinder and a body of mass ‘ m’ is hung from the string. The mass is released from rest. Find the tension in the string and the angular speed of the cylinder as the mass falls a distance h.
Sol.
The acceleration ‘a’ of the falling body is given by mg – T = ma... (1)
Torque on the cylinder is TRI
or T Ma = 2 ...(2) from (1) and (2), T mMg Mm 2
From conservation of energy, we have
Try yourself:
33. A thin rod AB of mass m and length ℓ is hinged at point O such that it can rotate in a vertical plane. End of A the rod is held in horizontal position. Now, the rod is released. Find the velocity of end B when the rod is vertical.
Hint: Find I 0 conserve energy.
Key Insights:
■ Two rigid bodies of moment of inertia I1 and I2 are rotating with angular velocities ww 12 and in the same direction about the common axes. If they couple, then:
a) The common angular velocity of the system LL12 = ;
Common frequency of the system,
b) Loss of energy of the system,
c) In the above case, if the two bodies are rotating in opposite directions before coupling, then w2 should be taken as negative,
Table 2 Comparison of Translational and Rotational Motion
Linear motion
Rotational motion about a fixed axis
4. Mass M Moment of inertia I
5. Force F = Ma Torque I
6. Work dW = Fds Work Wd
7. Kinetic energy KMv = 2 2 / Kinetic energy KI 2 2 /
8. Power P = Fv Power P =
9. Linear momentum pMv = Angular momentum LI
9.3.10 ANGULAR MOMENTUM
Now we are going to discuss about a very important physical quantity-angular momentum( L ). As we proceed to next chapters you will understand that concept of angular momentum is the prime output of this chapter to rest of Physics. Angular momentum () L to ‘Rotational Mechanics’ is linear momentum () P to ‘Particle Mechanics’. We know from Newton’s second law: “rate of change of linear momentum is equal to net external force” and we want to define L in such way that we get a similar principle- “rate of change of angular momentum is equal to net external torque”.
Angular Momentum of an Isolated Particle
It is best to start with a simple case in order to understand a relatively complex concept like that of L . We will first discuss about angular momentum of a single particle and analyze it thoroughly. And then we will try to extrapolate the results obtained to a rigid body. Consider a particle whose mass is m, position vector is r and velocity is v . Its linear momentum is Pmv = . Then, angular momentum of the particle about origin is defined as moment of momentum or mathematically:
is perpendicular to the plane containing randp
and its direction can be identified using right hand thumb rule.
Consider a particle rotating about origin O in a circle of radius r, as shown in figure. Momentum () P of the particle is always perpendicular to radius vector r . As the particle moves along the circular path, its position vector r rotates about O.
In a time interval ‘dt’, r rotates through an angle ‘d q ’, as shown in Fig. (b). Now, we are interested in finding out area of the region (shaded) swept by r . 2 1 2 dArd=θ (area of a sector) and 2 11 222 dAdL rvr dtdtm θ ===
Consider another example, a particle is moving along a straight line with a constant linear momentum Pmv = , as shown in Fig. (a). One interesting point in this case is that though the particle is in straight line motion its position vector r rotates with an angular velocity sin Vv rr
Hence, we can define angular momentum of the particle which is decided by the tangential component of linear momentum as
0 (sin) LPrmvr ⊥ ==θ . It means that, what counts for angular momentum is not how fast it is going away from origin (due to component of velocity along r , but how much it is turning around the origin (due to v ⊥ ). L0 can also be written as,
0 sin ()LmvrmvrPON ⊥ =θ==×
Here, ON is called lever arm of momentum which is similar to lever arm of force but is surely different from the lever arm of force. And to find area velocity, consider Fig. (b). Area of the triangle swept by radius vector in a time interval ‘dt’ is 1 ()() 2 dAvdtON =×
where, ‘vdt’ is base of the triangle and ‘ON’ is height.
To check the sign of angular momentum we may use the following idea: “If r rotates in counter clockwise direction, L 0 is taken as positive (out of the page) and if r rotates in clockwise sense, L0 is taken as negative (into the page)”.
Let us now derive a quantitative relation for the effect of choice of reference. Consider a particle of mass m moving with a velocity v , whose angular momentum is to be calculated about two fixed points O and O 1 . 1randr are the position vectors of the particle with respect to O and O1
Note that in this example, the particle continues to move with a constant speed ‘v’ along the line AB; and perpendicular distance ‘ON’ does not change with time. We thus see here that 0L and hence dA dt remain constant.
and 1 , OO r , O is the position vectors of O1 with respect to O.
So, () o Lmrv =×
and () 1 1 O Lmrv =×
(or) 1 ....(24) OOOO LLrP ′ −=×
This equation is true for any two fixed points O and O1 .
In Fig. (a), we considered a special case of time independent angular momentum. If L changes with time, we can find its rate of change.
Here, dr v dt =
and v is parallel to P ; and net dP F dt
Rate of change of angular momentum of an isolated particle is equal to net torque acting on the particle. We started off this section with a predetermined notion to establish the above equation and so far we succeded for the case of an isolated particle.
37. A particle of mass 0.1 kg is moving along a straight line 3x - 4y + 4 = 0 with velocity 10 m/s. Find the magnitude of angular momentum of the particle about the origin. Here x and y are measured in metre. Sol.
Principle of superposition: “Sum of angular momenta of the individual particles is the net angular momentum of the system of particles”.
The same principle can be applied even for continuous mass distribution. () ( ) ( ) .....(26) ii LrdPrvdm forcontinuousmatter rPfordiscreteparticles ∴=∫×=∫× =∑×
And applying equation (24) for system of particles
Try yourself:
34. A particle is projected at time t = 0 from a point ‘O’ with a speed ‘u’ at an angle 45o to horizontal. Find the angular momentum of particle at time t = u/g. Ans: mu g 3 22 /
Angular Momentum of System of Particles
So far we discussed about angular momentum of an isolated particle. Then, to find angular momentum of a group of particles, we can use
From the concept of centre of mass, c vdmMv∫= ; M is total mass of the system and c v
is velocity of centre of mass of the system.
Therefore,
()11 , ....(27) Oc OOO LLMrv −=×
Whenever we deal with a system of particles, centre of mass plays an important role. Keeping this in view, let us consider a special case of the above expression. Let O1 be a fixed point in space which coincides with centre of mass of the system. Then, 1 , , OOc rr =
position vector of centre of mass.
() 1 ....(28) Occ O LLMrv ∴=+×
Similarly, we can also apply the eq. 25 for system of particles.
() , , int ....(29) netnetextnet dL zero
This equation says that internal forces do not change angular momentum. This result matches with our common experience–bodies do not start spinning on their own.
This equation is known as “Newton’s law of rotational motion”.
We started this chapter with a hope to get analogy between translation and rotation. In fact, we have been quite successful in arriving at many important similarities between them, like: ‘translational equilibrium Vs rotational equilibrium, translational kinetic energy Vs rotational kinetic energy, mass Vs moment of inertia,force Vs torque, linear momentum Vs angular momentum’ and so on. All the equations arrived at in this section are applicable for any system of particles. However, our objective for the following sections is to apply these equations for a rigid body.
Angular Momentum of a Rigid Body(Pure Translation)
A rigid body in planar motion can either be in instantaneous translation or in instantaneous rotation. When a rigid body is in pure translation all the particles in the body must have identical instantaneous velocities. In this case angular velocity of the body is zero. Pure translation does not mean that the body must move in a straight line path. In the case of pure translation.
Angular Momentum of a Rigid Body (Rotation about a Fixed Axis)
Consider a rigid body which is rotating about z-axis with an angular velocity ω. We want to calculate component of angular momentum of the body along the axis of rotation(L z). If mass distribution of the body is symmetric about Z-axis, Lx and L y will vanish; then, net angular momentum about origin, L o =L z and, & zzzz LII=ωτ=α
In general () O Lrvdm =∫× ; where r is position vector of elementary mass ‘dm’. Its velocity , vr=ω×ω is angular velocity of the body and is directed along z-axis.
& xy rxiyjzkkvvivj =++ω=ω=+
And distance of the element from z-axis is 22 xy + . From the given data, it can be shown that ....(31) zz LI=ω
Hence to calculate angular momentum of a body in pure translation : “the whole body can be treated as a point mass located at its centre of mass”.
Remember that L x and L y may or may not be zero. However, in most of the cases we are only concerned about Lz-component of angular momentum along axis of rotation (z-axis). And if L z changes with time, we can find the time derivative of L z to calculate torque about z-axis. () ....(32) zzzz dd LII dtdt ω τ===α
Hence, net torque about axis of rotation is equal to product of moment of inertia of the object and its angular acceleration.
38. A thin spherical shell of mass 2 kg and radius 20 cm is spinning about its diameter with angular velocity of 30 rad/s. Find its angular momentum about the axis of rotation.
Try yourself:
35. A thin rod of mass 1 kg and length 2 m is spinning with angular momentum 2 kg m 2 /s about an axis passing through its centre and perpendicular to its length. Find its kinetic energy. Ans: 6 joules
Impulse–Momentum Equation
If an external force F changes the linear momentum P of a rigid body (i.e. linear momentum of the centre of mass), change in P is given by cm PorPFdt∆∆=∫
We know that Fdt ∫ over time interval D t is called linear impulse of the force F . Linear impulse is a vector quantity denoted by J Here F is net force if a number of forces act on it.
So, PFdt∆=∫ is known impulsemomentum equation.
If iP and fP denote initial and final linear momenta we can write if PFdtP +∫=
Similarly if an external torque τ changes the angular momentum L of the rigid body, we can write cm L or Ldt∆∆=∫τ
Here dt∫τ is called angular impulse during time interval D t about CM. It is also a vector quantity. Here τ is net torque if a number of torques act.
So Ldt∆=∫τ is known as impulse angular momentum equation.
So, if LdtL +∫τ= where iL and Lf respectively. Denote initial and final angular momenta
39. A disc of mass 2 kg and radius 0.4 m is rotating about a spindle passing through its centre
and perpendicular to its plane with angular velocity 5 rad/s. Due to an external torque acting on the disc for a short interval of time, its angular velocity is increased to 10 rad/s. calculate the angular impulse acted on the disc.
Sol. Angular impulse, Ifi 1 2 20 4105 2 .N-m-s
=0.8 J-s
Try yourself:
36. A solid disc is rotating about a spindle passing through its centre and perpendicular to its plane with kinetic energy 9 joules. An external torque acts on the disc for a short interval of time and, as a result, its kinetic energy increases to 16 joules. If moment of inertia of the disc about the axis of rotation is 2 kg m2, find the angular impulse acted on the disc.
Ans: 2 J/s
Conservation of Angular Momentum
How can we alter the total angular momentum of a system? “Net external torque acting on a system of particles about the axis of rotation is equal to rate of change of angular momentum of the system about the axis of rotation”.
Note that this equation is true whether the system forms a rigid body or not. From this equation it is very obvious that when net external torque acting on a system becomes zero then angular momentum of the system remains constant with time.
This is known as conservation of angular momentum which is similar to conservation of linear momentum.
Quantitatively, if t z = zero, Lz = I z ω= constant
This principle is very powerful and is valid even in relativistic mechanics and in quantum mechanics.
There are instances, for example nuclear reactions, where conservation of mass is not valid but conservation of linear momentum and conservation of angular momentum are valid.
40. If the radius of the earth is suddenly halved, keeping its mass constant, find its time period of rotation around its own axis.
Sol. When the radius of the earth gets reduced suddenly, keeping its mass constant, the angular momentum of the earth is constant. IMR T constant constant 2
Since mass ‘M’ is constant, RT 2
The radius changes from R to R/2 R RT T 2 2 4 24 6 hours hours
41. A horizontal disc is freely rotating about a vertical axis passing through its centre at the rate of 100 rpm. A bob of wax of mass 20 g falls on the disc and sticks to it at a distance of 5 cm from the axis. If the moment of inertia of the disc about the given axis is 2×10 –4 kg m 2 , find the new frequency of rotation of the disc.
Sol. I1 = Moment of inertia of the disc = 2 × 10–4 kg m2
I2 = Moment of inertia of the disc + Moment of inertia of the bob of wax on the disc
2102 10 20 10 0052 10 05 10 2
By the principle of conservation of angular
Try yourself:
37. A uniform rod AB of mass ‘m’ and length ‘2a’ is allowed to fall under gravity with AB in horizontal. When the speed of the rod is ‘ v ’, suddenly the end ‘A’ is fixed. Find the angular velocity with which it begins to rotate.
9.3.11 Rolling Motion
Consider a wheel moving along a straight track, the centre of the wheel moves forward in pure translation. A point on the rim of the wheel, however traces out a complex curve called cycloid. We can analyse the motion of a rolling wheel as a combination of pure translation and pure rotation.
Rolling can be of two types.
1) Rolling without slipping
2) Rolling with slipping (forward or backward)
In rolling without slipping the point of contact of the rolling body with the surface remains instantaneously at rest. Where as in rolling with slipping the point of contact of the rolling body slips either forward or backward.
Rolling without Slipping
Let v cm be the velocity of the centre of mass and therefore the translational velocity of a disc. Since the centre of mass of the rolling disc is at its geometric centre C,vcm is the velocity of C. It is parallel to the level surface. The rotational motion of the disc is about its symmetry axis,
9: System of Particles and Rotational Motion
which passes through C. Thus, the velocity of any point of the disc, like P0,P1 or P2, consists of two parts, one is the translational velocity v cm and the other is the linear velocity v r on account of rotation. The magnitude of vr is v r = rω, where ω is the angular velocity of the rotation of the disc about the axis and r is the distance of the point from the axis (i.e. from C). The velocity vr is directed perpendicular to the radius vector of the given point with respect to C.The velocity of the point P 2 (v2) and its components v r and v cm are shown; v r here is perpendicular to CP2. It is easy to show that v2 is perpendicular to the line P0P2. Therefore the line passing through P 0 and parallel to ω is called the instantaneous axis of rotation.
If aCM < Rα, then static friction arises and acts in the direction of motion of body to support translatory motion such that aCM is made equal to Rα to keep the body still in pure tolling motion. (Instantaneously V CM=Rω)
If the above two conditions fails then the body slips and the friction present is kinetic.
Some Important Points Related to Rolling
At P0, the linear velocity vr, due to rotation is directed exactly opposite to the translational velocity v cm . Further the magnitude of v r here is R w , where R is the radius of the disc. The condition that P 0 is instantaneously at rest requires v cm =Rω. Thus for the disc the condition for rolling without slipping is V cm =Rω.
Non–uniform Pure Rolling
If aCM > Rα, then static friction arises and acts opposite to motion of the body to support rotatory motion such that Rα is made equal to a CM to keep the body still in pure rolling motion. (Instantaneously V CM = Rα)
1) In one complete rotation of wheel, the centre of mass of the wheel moves a displacement s =2 α R. In this motion there is no relative motion between the point of contact of the moving body and the surface. If T is the time to complete a rotation, then we have
2) If Vcm= Rω, the wheel is in pure rolling. The velocity of the point of contact is zero.
3) If V cm > Rω, wheel slips forward. The wheel moves through a distance greater than 2πR in one full rotation.
4) If V cm < Rω, wheel slips backward. The wheel moves through a distance less than 2 π R in one full rotation.
Rolling motion = Pure translatory motion + Pure rotatory motion.
Fig.29
Velocity at any instantaneous point on the rolling body:
Consider a point x on a rolling body, the velocity of point x is the vector sum of velocity due to translation and rotation. Thus
or 2cos 2 RcmVV
Here q is the angle made by the radius vector with upward verticle in clockwise direction.
At ,180,0 R Avθ==
At 0 , , 02 R BvV θ==
At , , 902 R RvV θ==
At , , 603 R PvV θ==
Here, RV is resultant velocity at a given point on wheel.
42. A disc is rolling without slipping with velocity V on a horizontal surface. P is a point on its rim whose position is defined by angle q
Find the speed of P.
Velocities at different points on the rolling wheel:
Sol. Since the disc is rolling without slipping, velocity of point of contact O on the disc is zero. At the instant shown in figure, the whole disc is rotating with angular velocity w about an axis passing through O and perpendicular to the plane of figure.
We know V cm = V r =Rω =V.
At O, Vcm =V
Now, length OP = B RRRR
VOPPR 22 2 2 2 2 2 .. cos sin sin u ut VVOC
VV c P 2 2 sin.
CHAPTER 9: System of Particles and Rotational Motion
Try yourself:
38. A disc is rolling without slipping on a horizontal surface. AB is a horizontal diameter and speed of point A is V. What is the velocity of its centre at that instant?
Ans: V 2
3. Rolling bodies over moving platform
The rolling bodies do not slide on the surface on which they are moving. If they are rolling on a moving platform, the point of contact of the body with the platform should have same velocity as the platform.
Case-1: If points of contact of surface is moving with velocity u with respect to ground, then V cm – w R = u
43. A disc is rolling without slipping on a plank with angular velocity w = 10 rad/s. The plank is moving horizontally with velocity 2 m/s as shown in figure. Find the velocity of centre of disc relative to the ground. Radius of the disc is 200 cm. w =10 rad/s
Case-2: For no sliding on the moving platform, u = ωR-V
V cm – VP = w R ⇒ V cm = VP + w R = (2+10×02) m/s = 4 m/s.
Try yourself:
39. A disc is rolling without slipping on a plank with a velocity v relative to the plank. The plank itself is moving with velocity v relative to ground. Find the speed of A relative to ground.
Hint: VVVCgAAC // .
Ans: 5 V
4. Direction of friction in case of translation and rotation combined
The direction of friction cannot be found by direct observation in case of rotational motion, as the body is translating as well as rotating. The direction of friction force is determined after deciding the motion tendency of the point of contact of the body under consideration with the ground. A rolling object of mass M and radius R is placed on a rough horizontal surface. A force F is applied as shown.
Case-3: For accelerated surface, a cm – α R = a
Sol.
To find the direction of friction at P, let us consider the surface to be frictionless.
Acceleration of point P due to translation
t F a M = (towards right)
Acceleration of point P duet rotation only,
r FR aRRx II τ =α==
net acceleration of point p is ptr aaa =+
p FFRx a MI =− (Towards right) ....(i)
From the above equation it is clear that motion tendency at point P depends upon both x and I
Eq(i) can be written as, 2 1 p FRx a MK =−
....(ii)
If K2 > Rx : friction will act in backward direction.
If K2 = Rx : no friction will act
If K 2 < Rx : friction will act in forward direction
44. A force F is applied at the top of a ring of mass M and radius R placed on a rough horizontal surface. Friction is sufficient to prevent slipping. Find the frictional force acting on the ring and also find the acceleration of centre of mass. Sol. fFR R mR mR
FfMa 1 2 0 2 ; So,
Friction and acceleration faFM == 0.
Try yourself:
40. A horizontal force F is applied to the centre c of a solid disc of mass m and radius R. If the disc starts rolling without slipping, calculate its acceleration.
Ans: 2 F/3 m.
Kinetic energy of rolling body;
The kinetic energy of a rolling body can be separated into kinetic energy of translation and kinetic energy of rotation. This is a special case of a general result for a system of particles, according to which the kinetic energy of a system of particles (KE) can be separated into the kinetic energy of motion of the centre of mass (translation) (KEt) and kinetic energy of rotational motion about the centre of mass of the system of particles (KE r).
Thus a body rolling without slipping posses both translational KE and rotational K.E. KE = KE t + KE r
tcmcmr KEMVMVVVRV ====ω=
22 22 22 11 22 KK MVMV RR ==⋅ββ= 1122 22 tr KEKEKEMVMV ⇒=+=+β
2 1 (1) 2 MV =+β
hence, total kinetic energy of the rolling body without slipping:
2 1 .(1) 2 KEMV=+β
Where a R α= is a Dimensionless quantity. Its value vearies from body to body and axis to axis.
For example:
9: System of Particles and Rotational Motion
1) For ring 2 22 2 1 K IMRMK R ==⇒==β
2) For Solid sphere 2 22 2 22 55 K IMRMK R ==⇒==β
3) For disc 2 2 2 MR IMK == 2 2 1 2 K R ∴==β
4) For hollow sphere 22 2 3 IMRMK == 2 2 2 3 K R ∴==β
K.E of the rolling body without slipping can also be written as:
KErolling = KE t +KE r
22cmcmcm MVIVR =+ω=ω 11222 22 cm MRI =ω+ω 22 1 2 cm KEMRI =+ω
As we know MR2+I cm=IP is the moment of inertia of the wheel about the point of contact 'P' then 2 1 2 rollingP KEI=⋅ω
Rolling motion of a body may be treated as pure rotation about an axis normal to the plane passing through the point of contact with same angualar velocity.
1) Fraction of translational KE in the total kinetic energy 1 1 = +β
2) Fraction of rotational KE in the total kinetic energy 1 β = +β
3) Ratio of translational KE to rotational 1 KE = β
45. A uniform spherical shell rolls down a fixed inclined plane without slipping. Find the ratio of rotational kinetic energy to translational kinetic energy as it reaches the lowest point of the incline.
Sol. As there is no slip, vR is valid and rotational kinetic energy KImk R 1 2 1 2 22 2 cm and translational kinetic energy , KmvmR K K k R T R T 1 2 1 2 2 3 22 2 2 2 Therefore
Try yourself:
41. A solid disc of mass 1 kg is rolling without slipping with kinetic energy 3 joule. Find the velocity of its centre.
Ans: 2 m/s.
Angular momentum of rolling wheel: Angular momentum of a rolling wheel about an axis passing through the point of contact P and erpendicular to the plane of wheel: V cm
46. A solid cylinder of mass m and radius R is rolling without slipping with vlocity V on a horizontal plane (x-axis). P is a point whose coordinates are (0, h ), such that angular momentum of the cylinder about P is zero.
Sol. LmVhRkIk
R k PC 1 2 2 .
Since e we can write, L mVhRmRVhR P 0 1 2 0 3 2 , .
Try yourself:
42. A thin ring of mass 2 kg and radius 0.5 m is rolling slipping with velocity of 2 m/s. Find its angular momentum about the point of contact.
Ans: 2 kg /s.2m
Rolling of a body on an inclined plane:
Rolling of a body does not take place on smooth inclined plane, but it takes place on a rough inclined plane.
Consider a rigid round body of mass 'm' and radius ' R' is placed on a rough inclined plane making an angle q with the horizontal as shown in figure.
perpendicular to the inclined plane is 'mg cos q '.
The torque produced by the weight about the centre is t 1 = mg(0)=0. (Weight acts along the centre).
2) Normal reaction N. The torque produced by the normal reaction N is t 2=N(0)= 0. (Normal reaction passes through the centre).
3) The static frictional force 'fs' acts along the inclined plane upwards.
The torque produced by the frictional force 'f s ' is t 3 =f s R
\The resultant force normal to the inclined plane is zero because N balances mg cos q
\ N=mg cos q
The net force acting down the inclined plane is FR = mg sin q – fS
If α is the acceleration of the body down the inclined plane
sin .....(1) S mgf a m θ− ∴=
Net torque acting on the body is
123 τ=τ+τ+τ
00 S fR ∴τ=++
Due to this torque the angular acceleration of the body is given by S fR I α=
But a R α= and I = mK2 2
∴=⇒=
22 ....(2) SS afRfR a RmKmK
The forces acting on the body are:
1) Weight 'mg' acting vertically downwards. The component of the weight along the inclined plane is 'mg sin q ' and
[K is radius of gyration]
Substitute equation (2) in (1) sin f ag m =θ−
a) Condition for pure rolling without slipping is () cos SLLss fffNmg ≤=µ=µθ sin cos 1 mgs mg
Coefficient of static friction is
Ex: For solid spere 2 5 β= 2 tan 7 s ⇒µ≥θ
b) Velocity of the rolling body (without slipping) on reaching the bottom of the inclined plane is PEA= KEB 2 1 1 2 mghmv=+β
c) time taken by the rolling body to reach the bottom of the plane is 2 u0 12 gsin a 2 1 l utatt a =
or 12 (1) sin h t g =+β θ
Hence 1 t ∝+β
For sliding bodies β = 0, 212 sinsin lh t gg ⇒== θθ
Smaller the β , smaller the time taken by the rolling body to reach the bottom of the inclined plane.
Combination of Pure Translation and Rotation
Con sider the motion of a rigid body from position 1 to position 2. Here particle A translates by a distance SA. The straight line AB rotates through an angle q simultaneously.
d) Fraction of translational KE in the total K.E. is
Displacement of B relative to the ground is given by BBAA SSS =+
BBAA dSdSdS dtdtdt ⇒=+
(or) BBAA VVV =+
But BABABAVr =ω×
So, () BBABAA VrV =ω×+
e) Fraction of rotational KE in the total K.E. is
BBABAA ara =α×+
Similarly, we can obtain ()
Note: Consider a rigid body rotating with angular velocity 'ω' about its centre of mass. Assume that centre of mass moves with a constant velocity CV . Now velocity of a point P in that rigid body will be given by
PPCC VVV =+
2cosPPCCPCC VVVVV=++θ
Here, q is the angle beween V PC and VC
Kinetic Energy of a Rolling Body (Pure Rolling)
a) Translational K.E. is
trans KEMV = (where V is velocity of centre of mass)
9.3.12 Toppling and Sliding
Consider a block of mass m resting on a rough horizontal surface. Let F be the force applied on it at a height h above the base as shown.
b)
c) Total kinetic energy
If friction f is sufficient to prevent sliding, the block is in translational equilibrium. So, we have F=f and N=mg where N is normal reaction. If N also passes through can unbalanced torque
(F and f form a couple) acts which has a tendency to topple the block about point P.
To cancel this effect, N shifts towards right say a distance x. Such that, net torque becomes zero.
If F or h (or both) increases, distance x also increases. But x can not go beyond right edge at P. So, the extreme case is N passes through P as shown.
47. A force F is applied on the top of a cube, as shown in figure. The coefficient of friction between the cube and the ground is m . If F is gradually increased, find the value of m for which the cube will topple before sliding.
Sol. Let m be the mass of cube and ‘a’ be the side of cube.
The cube will slide if Fmg ... (1) and it will topple, if torque of F about P is greater than torque of mg about P, i.e.,
Fa a mg . 2 ; or Fmg > 1 2 ... (2)
From equations (1) and (2), we see that the cube will topple before sliding if 1 2
48. A uniform cube of side a and mass m rests on a rough horizontal table. A horizontal force F is applied normal to one of the faces at a point directly above the centre of the face, at a height 3a/4 above the base. What is the minimum value of F for which the cube begins to topple about an edge?
Sol. In the limiting case, normal reaction will pass through O. The cube will topple about O if torque of F exceeds the torque of mg. In this case, the normal force shifts to the edge O.
So, the minimum value of F is 2 3 mg
Try yourself:
43. A uniform cylinder of height h and radius r is placed with its circular face on a rough inclined plane and the inclination of the plane to the horizontal is gradually increased. If m is the coefficeint of friction, then under what conditions will the cylinder a) slide before toppling b) topple before sliding?
Ans: (n4 – 1)x
9.3.13 Instantaneous Axis of Rotation
1. When a rigid body undergoes a combined translation and rotation, if there is at least a point inside or outside the body whose velocity is zero, the body is said to have pure rotation about that point. Then axis passing through that point and perpendicular to the plane of rotation is known an instantaneous axis of rotation.
2. When a uniform disc is rolling on a surface its centre of mass and all the points on the axis of rotation undergo pure translation. But the remaining particles undergo rotation about that axis.
3. When the instantaneous point of contact of a rolling body does not move, the body is said to have pure rolling.
If the location of the instantaneous centre of rotation (IC) is unknown, it can be located as explained below.
We know that the relative position vector extending from IC to a point is always perpendicular to the velocity of the point.
There are three possibilities.
a) If v and w are known, there IC is located along the line drawn perpendicular to V at P such that v r = ω
b) If AV and BV are velocities of two points A and B of the rigid body, draw perpendiculars at A and B to the lines of action of VAB and V . The point of intersection of these perpendiculars gives the location of IC as shown,
Here AAVr=ω and BBVr=ω
c) If VA and BV are parallel, then IC can be located by proportional triangles as shown.
CHAPTER REVIEW
Centre of Mass
■ Motion of mass centre of mass describes the translational motion of a body or a system of particles.
TEST YOURSELF
1. A thin rod of mass m and length 2L is made to rotate about a normal axis through centre of rod. If its angular velocity changes from ‘0’ to ‘ω’ in time t, the torque acting on it is
(1) mL2 w /12t
(2) 4mL2 w /3t
(3) mL2 w /t
(4) mL2 w /3t
2. A child is standing with folded hands at the centre of a platform rotating about its central axis. The KE of the system is K. The child now stretches his arms so that the MI of the system is doubled. The KE of the system now is
(1) 2K (2) K/2
(3) 4K (4) K/4
3. A motor rotates a pulley of radius 25 cm at 20 rpm. A rope around the pulley lifts a 50 kg block. The power output of the motor is (1) 220 W (2) 520 W
(3) 350 W (4) 261 W
Answer Key
(1) 4 (2) 2 (3) 4
■ If the line of action of a force applied to a stationary body, which is free to move, passes through its centre of mass, the body will make pure translational motion.
■ Centre of mass may lie within the body or outside the body.
9: System of Particles and Rotational Motion
■ Position of centre of mass relative to the body does not depend upon frame of reference.
■ If net external force on a system is zero, its centre of mass will be at rest or may move with constant velocity.
■ Linear momentum of a system in a frame of reference attached to the centre of mass is zero.
■ Sum of static moments mrii 0 of the mass of particles of a system about its centre of mass is zero.
■ Position of centre of mass of a body relative to the body depends upon the shape of the body and distribution of mass in the body.
■ For a two particle system, distance of COM from mi is r md mm 2 1 12 .
■ Shift in COM when a small portion of mass is removed from a uniform body is x md shMm ift removed totalremoved .
Collisions
■ Collision is the strong interaction among bodies in a short interval of time.
■ When lines of motion of colliding bodies before and after the impact lie along the line of action of impact force, the collision is said to be head on collision.
■ When lines of action of colliding bodies before and after thee impact do not lie along the line of action of colling bodies, the impact is said to be oblique collision.
■ In any type of collision, momentum of the system of colliding bodies remains conserved.
■ Om perfectly elastic collision kinetic energy of the system before the collision must be equal to the kinetic energy of the system after the collision.
■ During the course of collision (period of contact), total kinetic energy of the colliding bodies does not remain constant.
■ In inelastic or partially elastic collision kinetic energy of the system of colliding bodies before collision is not equal to the kineic energy of the system after the collision.
■ In perfectly inelastic collision the two bodies stick together and move with a common velocity.
■ Coefficient of restitution(e) is defined as the ratio of velocity of separation after collision along the line of impact to the velocity of approach before collision along the line of impact.
For head on collision, e vv uu 21 12
■ For head on collision loss of energy is given by e mm mm euu 1 2 1 12 12 2 12 2 .
■ For perfectly elastic collision e = 1, for perfectly inelastic collision e = 0, for partially elastic collision 0 < e < 1.
■ A ball is released from height H 0 onto a horizontal surface. Coefficient of restitution between the ball and the surface is e.
(a) The height to which the ball will rise after n th impact is given by, hHe e n 0 2 2 1 1 .
(b) The velocity of the ball just after the n th impact is given by v n = e n, where u is the velocity of the ball just before the first impact.
(c) Total time taken by the ball to stop bouncing Te e H g 1 1 2 0
■ A ball moving with a velocity u strikes a smooth surface, making an angle α with the normal to the surface and rebounds with a velocity v, making an angle β with the normal. tan tan sincos . e e and V=u 22 2
u v αβ
■ A ball is projected from point O on a horizontal surface. It strikes the surface successively at points A, B, C, D, etc. Let e be the coefficient of restitution between the ball and the surface.
Vr , where r is the position vector of the point about any fixed point on the axis of rotation.
■ If a force F is applied at point A whose positon vector about the origin is
0 rF A
■ If a force F is applied at point A (position vector rA ), then moment of the force about point B (position vector rB ) is given by
■ For rotational equilibrium of a body, moment of all forces acting on the body about any point must be zero.
(a) If R n be the horizontal range after the nth impact, R n = en R0.
(b) If T n be the time of flight after the nth impact, T n = e n T0.
(c) If H n be the maximum height attained by the ball after.
■ If two balls having masses m 1 and m 2 , moving with velocities u1 and u2 (u1 > u2) in the same direction, make head-on inelastic collision, then after the collision,
■ Moment of a force FFiFjFk xyz acting at point Arxiyjzk about the origin is,
rFyFzFizFxFj xFyFk zyxz yx . This moment has three components; along x-axis along y-axis and along z-axis.
■ To find the moment of a force about an axis, only the components of the force perpendicular to the axis need to be considered. Component of the force parallel to the axis need not be considered.
Rational Mechanics
■ ABABn sin. , where is the smaller angle between AB and ; n is a unit vector
⊥ to the plane containing AB and
■ Direction of n is determined by right hand thumb rule or rotation of right handed screw.
■ Velocity V of a point on a body rotating with angular velocity w is given by
■ About the centre of gravity, total gravitational torque on the body is zero.
■ Three vectors ABC , and are coplanar if
■ Moment of inertia is the rotaional analogue of inertia (mass) in translatory motion.
■ MI is neither a vector quantity nor a scalar quantity. It is a tensor. Its magnitude is different about different axes.
9: System of Particles and Rotational Motion
■ MI of a body depends upon its mass, shape and size, distribution of mass about the axis and position of axis.
■ Parallel axis theorem, I=I c +md2 .
■ Perpendicular axis theorem, I z =I x +I y this theorem is applicable to laminar (Two dimensional) objects only. Plane of the object lies in xy plane.
■ Dimensional formula of MI is [ ML2T0].
■ For a two particle system, minimum possible MI about an axis perpendicular to the line joining the particles is µ.d2 where d = separation between the particles and µ = reduced mass of the system mm mm 12 12 .
■ With increase in temperature, MI of a body increases.
■ For a triangular sheet, if m be the mass of the triangle, h be the altitude with respect to a side and I be the moment of inertia of the sheet about that side, then Imh = 1 6 2 .
■ Equations of circular motion and comparision with equations of translational motion
■ Net torque acting on a rigid body, its moment of inertia about the axis of rotation and angular acceleration are related by the equation netI
■ If L be the angular momentum, then net dL dt
■ For a rigid body LI
■ Angular momentum of a moving particle about a fixed point O,
LrmV 0 .
■ Angular impulse dtLL fi .
■ If I be the MI of a rigid body about the axis of rotation, then its KE = (1/2)I w 2
■ If net external torque acting on a system is zero, then its angular momentum remains conserved, Lf = Li. But its linear momentum may not reamin conserved.
■ If net external force on a system is zero, its linear momentum remains conserved but its angular momentum may not conserved reamin.
■ For equilibrium of a system,
Fnetnet 00 and .
■ If a rigid body is under the action of a number of external forces, then Fmnea tc and netcmm
For constrained motion, aC is related to α by a third equation.
■ Work done by torque, Wd . 1 2
■ Power, P
■ Work energy theorem for totational motion, WdI..
■ Rolling motion is a combination of pure translation and pure rotation about the centre of the round object.
■ In rolling without slipping, point of contact of the rolling body must have the same velocity as the surface on which the body is rolling.
■ For pure rolling of a body on a surface, angular velocity VV Radiusofbody centresurfae .
■ When a rolling wheel slips forward, V centre > w R.
■ When a rolling wheel slips backward, V centre < w R
■ For forward slipping of a rolling wheel, direction of friction acting on the wheel is backward.
■ For backward slipping of a rolling wheel, direction of friction acting on the wheel is forward.
■ Velocity of instantaneous centre of rotation is zero but acceleration of instantaneous centre of rotation is not zero.
■ For pure rolling of a body, total energy = (1/2)I P w 2, where w = angular velocity of
centre of the body, which is the centre of mass of the body, IP = IC + mR2, R = radius and P is the point of contact.
■ For pure rolling, fraction of total energy is associated with rotational motion only 11 2 I mR C .
■ For pure rolling, fraction of total energy is associated with pure translation 11 2 mR I C .
■ Ratio of translational KE to rotational KE = mR I C 2
Exercises
JEE MAIN LEVEL
Level-I
Centre of Mass
Single Option Correct MCQs
1. Mass of a ring is non-uniformly distributed around its geometric centre. If R is radius of the ring, then
a) centre of mass does not coincide with geometric centre
b) position of centre of mass from the geometric centre will be x(0 < x < R)
c) centre of mass will be nearer to the greater mass distribution
d) centre of mass may lie outside the periphery
(1) Only a and b are correct.
(2) Only b and c are correct.
(3) a, b, and c are correct.
(4) a, b, c, and d are correct.
2. Two particles, each of the same mass, move due north and due east, respectively, with the same velocity ‘V’. The magnitude and direction of the velocity of centre of mass is
(1) NE 2 V
(2) 2 NE V
(3) 2V SW
(4) SW 2 V
3. Consider the two following statements.
A) Linear momentum of the system remains constant.
B) Centre of mass of the system remains at rest.
Choose the correct statement from the options given below.
(1) A implies B and B implies A.
(2) A does not imply B and B does not imply A.
(3) A implies B but B does not imply A.
(4) B implies A but A does not imply B.
4. Two particles of equal mass have velocities –1 1 ˆ 4i ms =
v and
First particle has an acceleration ()
5j ms =+
a , while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a path of (1) straight line (2) parabola (3) circle (4) ellipse
5. A shell of mass m in flight explodes into three equal fragments. One of the fragments reaches the ground earlier than the other fragments. The acceleration of the centre of mass of the shell system just after the first one reaches the ground is (1) g (2) g/3 (3) 3g/4 (4) 2g/3
6. A 50 kg man is standing at one end of a 25 m long boat. He starts running towards the other end. On reaching the other end of the boat, his velocity is 4 ms–1. If the mass of the boat is 250 kg, final velocity of the boat is (in ms–1)
(1) 2/5 (2) 2/3
(3) 8/5 (4) 8/3
7. Particles of masses 1 g, 2 g, 3 g, and 4 g are placed at x =1 cm, x = 2 cm, x = 3 cm, and x = 4 cm, respectively. Then, x cm = (1) 1 cm (2) 2 cm (3) 3 cm (4) 4 cm
8. Four cubes of side a each, of mass 40 g, 20 g, 10 g, and 20 g, are arranged in x-y plane as shown in the figure. The coordinates of centre of mass of the combination with respect to point O is
(1) (19a/18, 17a/18)
(2) (17a/18, 11a/18)
(3) (17a/18, 13a/18)
(4) (13a/18, 17a/18)
9. Three particles of masses 1 kg, 2 kg, and 3 kg are placed at the vertices A, B, and C of an equilateral triangle ABC. If A and B lie at (0, 0) and (1, 0) m, the coordinates of their centre of mass are
(1) 37mandm 26
(2) 73mandm 64
(3) 73mandm 124
(4) 77mandm 1212
10. The centre of mass of a non-uniform rod of length L, whose mass per unit length 2 , λ= kx L where k is a constant and x is the distance from one end, is (1) 3 4 L (2) 3 L (3) 4 L (4) 2 3 L
11. Four particles A, B, C and D with masses m A = m , m B = 2 m , m C = 3 m, and m D = 4 m are at the corners of a square. They have accelerations as shown. The acceleration of centre of mass of the particles is
12. In the system shown in the figure m1 > m2 If another block is attached below the block of mass m1, then the acceleration of COM will m2 m1
(1) decrease (2) increase
(3) remains the same (4) may increase or decrease
Numerical Value Questions
13. A uniform wire of length 2L is bent to form an angle of 60°, as shown. The distance of the centre of mass from the vertex O is 3 , L n where n is . O 60°
14. Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the centre of mass is m/s.
15. A shell is fired from the ground at an angle θ with the horizontal, with a velocity V. At its highest point, it breaks into two equal fragments. If one fragment comes back through its initial line of motion with the same speed, then the speed of the second fragment will be xvcos θ. Here, x = .
16. Two rods of lengths l 1 = 1.5 m and l 2 = 1 m, and masses m1 = 1 kg and m2 = 2 kg, are joined end to end. If the centre of mass of the rod from the end of the first rod is 19 m, 6 y then y is .
17. Masses 1 kg, 1.5 kg, 2 kg, and M kg are situated at (2, 1, 1), (1, 2, 1), (2, -2, 1), and (-1, 4, 3). If their centre of mass is situated at (1, 1, 3/2), the value of M is x/2 kg. Here, x = .
18. The position of the centre of mass of a uniform semicircular wire of radius R, placed in x-y plane with its centre at the origin and the line joining its ends as x-axis, is given by 0,.
xR Then, the value of |x| is .
19. A boat of mass 40 kg is floating on water. A boy of mass 10 kg on the boat moves by 8 m (w.r.t. the boat) towards the shore. The distance by which the boat moves away from the shore is x/10. Then, x is m.
20. Two particles of masses 4 kg and 6 kg are at rest separated by 20 m. If they move towards each other under mutual force of attraction, the position of the point where they meet is (from 4 kg body) m
21. Three particles of masses 8 kg, 4 kg, and 4 kg, situated at (4, 1), (–2, 2), and (1, –3), are acted upon by external forces 6 N,6 N, and 14 N.
jii If the acceleration of centre of mass of the system is k/8, then k is ____m/s2
22. Two particles of mass 2 m are at a distance d apart. Under their mutual gravitational force, they start moving towards each other. The acceleration of their centre of mass, when they are d/2 apart, is
Collisions
Single Option Correct MCQs
23. A particle of mass m and momentum P moves on a smooth horizontal table and collides directly and elastically with a similar particle (of mass m) having momentum –2p. The loss (–) or gain (+) in the kinetic energy of the first particle in the collision is (1) + 2 2 p m (2) 2 4 p m (3) + 2 3 2 p m (4) zero
24. A heavy ball moving with speed v collides with a tiny ball at rest. The collision is elastic. Then, immediately after the impact, the second ball will move with a speed approximately equal to (1) v (2) 2v (3) v/2 (4) v/3
25. A body of mass 6 kg travelling with a velocity of 10 m/s collides head-on and elastically with a body of mass 4 kg travelling at a speed 5 m/s in the opposite direction. The velocity of the second body after the collision is (1) 0 m/s (2) 6 m/s(3) 8 m/s (4) 13m/s
26. A body of mass 5 kg makes an elastic collision with another body at rest and continues to move in the original direction after collision, with a velocity equal to 1/10th of its original velocity. Then, the mass of the second body is
(1) 4.09 kg (2) 0.5 kg
(3) 5 kg (4) 5.09 kg
27. A body of mass m moving with a velocity of 8 m/s, collides with another body of mass 15 kg moving with 4 m/s in the opposite direction. After collision, they stick and move together. The loss of KE is 270 J. Then, m is .
(1) 5 kg (2) 10 kg
(3) 15 kg (4) 20 kg
28. Given below are two statements.
Statement I : In an elastic collision between two bodies, the relative speed of the bodies after collision is equal to the relative speed before the collision.
Statement II : In an elastic collision, the linear momentum of the system is conserved.
In light of the above statements, choose the correct answer from the options given below.
(1) Both statement I and statement II are correct.
(2) Both statement I and statement II are incorrect.
(3) Statement I is correct but statement II is incorrect.
(4) Statement I is incorrect but statement II is correct.
29. Given below are two statements. One is labelled Assertion (A) and the other is labelled Reason (R).
Assertion (A) : Collision between two particles is not necessarily associated with physical contact between them.
Reason (R) : Only in physical contact, momentum transfer takes place.
In light of the above statements, choose the correct answer from the options given below.
(1) Both (A) and (R) are true and (R) is the correct explanation of (A).
(2) Both (A) and (R) true but (R) is not the correct explanation of (A).
(3) (A) is true but (R) is false.
(4) (A) is false but (R) is true.
30. A ball of mass 0.4 kg, moving with a uniform speed of 2 ms–1 , strikes a wall normally and rebounds. Assuming the collision to be elastic and the time of contact of the wall as 0.4 s, find the force exerted on the ball.
(1) 1 N (2) 2 N (3) 3 N (4) 4 N
31. A body of mass 5 kg moving with a velocity of 10 m/s collides with another body of mass 20 kg, at rest, and comes to rest. The velocity of the second body after collision is (1) 2.5 m/s (2) 5 m/s
(3) 7.5 m/s (4) 10 m/s
Numerical Value Questions
32. In the arrangement shown, the pendulum on the left is pulled aside. It is then released and allowed to collide with the other pendulum which is at rest. A perfectly inelastic collision occurs and the system rises to a height of 1/4 h . The ratio of the masses of the pendulums is .
h
33. Two balls collide and bounce off each other, as shown in the figure. The 1 kg ball has a speed of 10 cm/s after collision. The velocity of the 0.5 kg ball will be____.
34. An object of mass 40 kg, having velocity 4 , ˆ m/s i collides with another object of mass 40 kg, having velocity 3j m/s . If the collision is perfectly inelastic, then the loss of mechanical energy (in SI unit) is .
35. A body of mass 1 kg collides head on elastically with a stationary body of mass 3 kg. After collision, the smaller body reverses its direction of motion and moves with a speed of 2 ms –1 . The initial speed of the smaller body before collision is ms–1
36. Two bodies of the same mass are moving with the same speed but in different directions in a plane. They have a completely inelastic collision and move together after the collision, with a final velocity which is half of the initial velocity. The angle between the initial velocities of the two bodies (in degrees) is
Rotational Mechanics
Single Option Correct MCQs
37. A thin uniform rod of mass M and length L is hinged at an end and released from rest in the horizontal position. The tension at a point located at a distance L /3 from the hinge point, when the rod becomes vertical, will be
(1) 11 13 Mg (2) 6 13 Mg (3) 22 27 Mg (4) 2 Mg
38. A uniform solid cylinder of mass m and radius R is placed on a rough horizontal surface. A horizontal constant force F is applied at the top point P of the cylinder so that it starts pure rolling. The acceleration of the cylinder is
(1) F/m (2) 4F/3m (3) 2F/3m (4) F/2m
39. A cylinder rolls up an inclined plane, reaches some height, and then rolls down (without slipping throughout these motions). The directions of frictional force acting on the cylinder are
(1) up the incline while ascending and down the incline while descending (2) up the incline while ascending as well as while descending (3) down the incline while ascending and up the incline while descending (4) down the incline while ascending as well as while descending
40. The speed of a homogeneous solid sphere after rolling down an inclined plane of vertical height h, from rest, without sliding, is
(1) gh (2) g 5 gh
(3) 4 3 gh
(4) 10 7 gh
41 An equilateral prism of mass m rests on a rough horizontal surface with coefficient of friction μ. A horizontal force F is applied on the prism, as shown in figure. If the coefficient of friction is sufficiently high so that the prism does not slide before toppling, the minimum force required to topple the prism is
(1) 3 mg (2) 4 mg
(3) 2 3 mg (4) 7 4 mg
42. In pure rolling motion of a ring,
a) it rotates about instantaneous point of contact of ring and ground
b) its centre of mass moves in translatory motion only
c) its centre of mass will have translatory as well as rotatory motion
(1) Only a is correct.
(2) a and c are correct.
(3) a and b are correct.
(4) a, b, and c are correct.
43. A disc of radius 10 cm is rotating about its axis at an angular speed of 20 rad/s. Find the linear speed of a point on the rim.
(1) 1 m/s (2) 4 m/s (3) 2 m/s (4) 6 m/s
44. Three thin rods, each of length L and mass M, are placed along x, y and z axes in such a way that one end of each rod is at the origin, as shown. The moment of inertia of this system about z-axis is
45. Two discs of same mass and different radii are made of different materials such that their thickness is 1 cm and 0.5 cm, respectively. Densities of materials are in the ratio 5 : 3. The moment of inertia of these discs, respectively, about their diameters will be in the ratio
(1) 2 : 3 (2) 5 : 6
(3) 3 : 10 (4) 4 : 10
46. The radius of gyration of a uniform rod of a uniform length ‘l’ , about an axis passing through a point 3 l away from the centre of the rod, and perpendicular to it is
47. A ring of mass M = 10 kg and radius R = 2 m is taken as shown in the figure. Calculate the moment of inertia of ring about the diametric axis
minimum value required for pure rolling. At any instant of time, let KT and KR be the translational and rotational kinetic energies of the ring. Then,
(1) KR = KT (2) KR > KT (3) KT > KR (4) KR = 0
51. A solid sphere of mass 1.5 kg rolls on a frictionless horizontal surface. Its centre of mass is moving at a speed of 5 ms −1. Then, it rolls up an incline of 30° to horizontal. The height attained by the sphere before it stops is (g =10 ms−2)
(1) 25 kg m2 (2) 30 kg m2 (3) 40 kg m2 (4) 20 kg m2
48. The moment of inertia of a uniform thin rod of length L and mass M about an axis passing through a point at a distance of L/3 from one of its ends and perpendicular to the rod is
(1) 2 7 48 ML (2) ML2 (3) 2 9 ML (4) 2 3 ML
49. A small pulley of radius 20 cm and moment of inertia 0.32 kg–m 2 is used to hang a 2 kg mass with the help of a massless string. If the block is released, for no slipping condition, acceleration of the block will be (Take g = 10 ms–2) 2 kg
(1) 2 m/s2 (2) 4 m/s2 (3) 1 m/s2 (4) 3 m/s2
50. A ring is kept on a rough inclined surface. But the coefficient of friction is less than the
(1) 5 m (2) 1.5 m (3) 1.75 m (4) 3.65 m
Numerical Value Questions
52. A body rotating with an angular speed of 600 rpm is uniformly accelerated to 1800 rpm in 10 s. The number of rotations made in the process is
53. Moment of inertia of a body about a given axis is 1.5 kg m 2. Initially, the body is at rest. In order to produce a rotational kinetic energy of 1200 J, the angular acceleration of 20 rad/s2 must be applied about the axis for a duration of s.
54. For the figure shown, a rod of mass 10 kg (of length 100 cm) has some point masses tied to it at different positions. Find the distance of a point (from A) at which, if the rod is picked over a knife edge, it will be in equilibrium about that knife edge (in cm).
20 kg 15 kg 5 kg 7.5 kg 15 cm
B A
50 cm 20 cm
100 cm
55. A uniform solid cylindrical roller of mass m is being pulled on a horizontal surface with force F parallel to the surface and applied at its centre. If the acceleration of the cylinder is a and it is rolling without slipping, then the value of = 3 F x ma. Find the value of x.
56. A solid sphere of mass 1 kg rolls without slipping on a plane surface. Its kinetic energy is 7 ×10–5 J. The speed of the centre of mass of the sphere is cms–1.
57. A wheel, which is initially at rest, is subjected to a constant angular acceleration about its axis. It rotates through an angle of 15° in time t seconds. The angle (in degrees) through which it rotates in the next 2 t seconds is
58. Two thin uniform circular rings each, of radius 10 cm and mass 0.1 kg, are arranged such that they have a common centre and their planes are perpendicular to each other. The moment of inertia of the system about an axis passing through their common centre and perpendicular to the plane at one of the rings is × 10–4 kg m2
59. The radius of gyration of a body about an axis at a distance of 3 cm from its centre of mass is 5 cm. The radius of gyration about a parallel axis through centre of mass is cm.
60. Particles, each of mass 1 kg, are placed at 1 m, 2 m, and 4 m on x -axis with respect to origin. Then, moment of inertia of the system about y-axis is 3n kgm2. Then, the value of n is equal to .
61. A light rope is wound around a hollow cylinder of mass 6 kg and radius 60 cm, which is free to rotate about its geometrical axis. The rope is pulled with a force of 50.4 N. The angular acceleration of the cylinder will be___ rad s–2
Level-II
Centre of Mass
Single Option Correct MCQs
1. Three particles of masses 1.0 kg, 1.5 kg, and 2.5 kg are placed at three corners of a right angle of triangle of sides 4.0 cm, 3.0 cm and 5.0 cm, as shown in the figure. The centre of mass of the system is at a point
(1) 0.6 cm right and 2.0 cm above 1 kg mass
(2) 0.9 cm right and 2.0 cm above 1 kg mass
(3) 2.0 cm right and 0.9 cm above 1 kg mass
(4) 1.5 cm right and 1.2 cm above 1 kg mass
2. A rod of length L has non-uniform linear mass density given by () 2 ,
where a and b are constants and 0
L. The value of x for the centre of mass of the rod is at (1) 4 323 ab L ab + +
(2) 32 43 ab L ab + + (3) 32 23 ab L ab + + (4) 3 22 ab L ab + +
3. The coord inat es of centre of mass of a uniform flag-shaped lamina (thin flat plate) of mass 4 kg (the coordinates of the same in SI system are shown in figure) is
(0,3)
(2,3)
(2,2) (1,2)
(1,0) (0,0)
(1) (1.25 m, 1.50 m)
(2) (0.75 m, 0.75 m)
(3) (0.75 m, 1.75 m)
(4) (1 m, 1.75 m)
4. As shown in fig, when a spherical cavity (centred at O ) of radius 1 is cut out of a uniform sphere of radius R (centred at C), the centre of mass of remaining (shaded) part of the following sphere is at G, i.e., on the surface of the cavity. R can be determined by ehich of the equation? 1
(1) (R2+R−1)(2−R)=1
(2) (R2+R+1)(2−R)=1
(3) (R2 –R+1)(2−R)=1
(4) (R2 –R−1)(2−R)=1
5. Two bodies of mass 1 kg and 3 kg have position vectors ˆ ˆ ˆ 2 ijk ++ and , ˆ ˆ ˆ 32 −−+ijk respectively. The magnitude of position vector of centre of mass of this system will be similar to the magnitude of vector
(3) 2 cm towards the 10 kg block (4) 4 cm away from the 10 kg block
7. A circular hole of radius 2 a is cut out of a circular disc of radius a as shown in fig. The centroid of the remaining circular portion with respect to point ‘ O’ will be Y axis x- axis 0 a/2 a (1) 2 3 a (2) 1 6 a (3) 10 11 a (4) 5 6 a
Numerical Value Questions
(1) ˆ
ˆ 2 ijk −+
(2) 32
ijk−−+
(3) ˆ 2 ˆ 2 ik−+
(4) ˆ 2 ˆ 2 ˆ ijk−−+
6. Two blocks of masses 10 kg and 30 kg are placed on the same straight line with coordinates (0, 0) cm and ( x , 0)cm, respectively. The block of 10 kg is moved on the same line through a distance of 6 cm towards the other block. The distance through which the block of 30 kg must be moved to keep the position of centre of mass of the system unchanged is (1) 4 cm towards the 10 kg block (2) 2 cm away from the 10 kg block
8. The distance of centre of mass from end A of a one-dimensional rod (AB), having mass density 2 0 2 1 x L ρ=ρ− kg/m and length L
(in metres) is 3L m α . The value of α is where x is the distance from end A.
9. A man of 60 kg is running on the road and suddenly jumps into a stationary trolly car of mass 120 kg. Then, the trolly car starts moving with velocity 2 ms–1. The velocity of the running man was__________ms–1 when he jumps into the car.
(1) 6 (2) 7
(3) 8 (4) 10
10. The position of the centre of mass of a uniform semicircular wire of radius ' R ' placed in x-y plane, with its centre at the origin and the line joining its ends as x-axis, is given by 0, xR π . Then, the value of |x| is
Collisions
Single Option Correct MCQs
11. It is found that, if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is pd ; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc . The values of pd and pc are, respectively, (1) (–89, –28) (2) (–28, –89) (3) (0, 0) (4) (0, 1)
12. A particle of mass m moving with velocity v, collides with a stationary particle of mass 2 m. After collision, they stick together and continue to move together with velocity (1) v (2) 2 v (3) 3 v (4) 4 v
13. A ball of mass 200 g rests on a vertical post of height 20 m. A bullet of mass 10 g, travelling in horizontal direction, hits the centre of the ball. After collision, both travel independently. The ball hits the ground at a distance of 30 m and the bullet at a distance of 120 m from the foot of the post. The value of initial velocity of the bullet will be (if g =10 m/s2)
(1) 120 m/s (2) 60 m/s (3) 360 m/s (4) 400 m/s
14. As per the given figure, a small ball P slides down the quadrant of a circle and hits the other ball Q of equal mass, which is initially at rest. Neglecting the effect of friction and assuming the collision to be elastic, the velocity of ball Q after collision will be:
(g = 10 m/s2)
(1) 0 (2) 4 m/s
(3) 0.25 m/s (4) 2 m/s
15. A particle of mass m with an initial velocity ˆ ui collides perfectly elastically with a mass 3m at rest. It moves with a velocity ˆ vj after collision. Then, v is given by (1) 2 3 vu = (2) 1 6 vu = (3) 3 u v = (4) 2 u v =
16. A mass ‘ m ’ moves with a velocity v and collides inelastically with another identical mass. After collision, the first mass moves with velocity 3 v in a direction perpendicu lar to the initial direction of motion. Find the speed of the second mass after collision. m
3 (1) v (2) 3v (3) 2/3 v (4) /3 v
17. A body of mass M, moving at speed V 0 , collides elastically with a mass 'm ' at rest. After the collision, the two masses move at angles θ1 and θ2 with respect to the initial direction of motion of the body of mass M The largest possible value of the ratio M/m, for which the angles θ1 and θ2 will be equal, is (1) 4 (2) 1 (3) 3 (4) 2
Numerical
Value Questions
18. A ball with a speed of 9 m/s collides with another identical ball at rest. After the collision, the direction of each ball makes an angle of 30º with the original direction. The ratio of velocities of the balls after collision is x : y, where x is .
19. A ball of mass 10 kg, moving with a velocity of 103 m/s along the x-axis, hits another ball of mass 20 kg which is at rest. After the collision, the first ball comes to rest while the second ball disintegrates into two equal pieces. One piece starts moving along y-axis with a speed of 10 m/s. The second piece starts moving at an angle of 30° with respect to the x-axis. The velocity of the ball moving at 30° with the x-axis is x m/s. The value of x to the nearest integer is . The configuration of pieces after collision is shown in the figure below.
pieces. One of the pieces starts moving along the y-axis at a speed of 10 m/s. The second piece starts moving at a speed of 20 m/s at an angle θ (degrees) with respect to the x-axis. The configuration of pieces after collision is shown in the figure. The value of θ to the nearest integer is
Collision
20. A ball of mass 10 kg, moving with a velocity 1 103ms along the x-axis, hits another ball of mass 20kg, which is at rest. After the collision, the first ball comes to rest and the second one disintegrates into two equal
Rotational Mechanics
Single Option Correct MCQs
21. Seven identical circular planar disks, each of mass M and radius R, are welded symmetrically, as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is
(3) 2 73 2 MR (4) 2 181 2 MR
22. From a uniform circular disc of radius R and mass 9 M, a small disc of radius R/3 is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is:
(3) 7.5 × 102 kg/m3 (4) 1.49 × 102 kg/m3
24. Moment of inertia of a square plate of side l about the axis passing through one of the corner, and perpendicular to the plane of square plate is given by (1) 2 6 Ml (2) Ml2 (3) 2 12 Ml (4) 2 2 3 Ml
(1) 10 MR2 (2) 2 37 9 MR
(3) 4 MR2 (4) 2 40 9 MR
23. The solid cylinder of length 80 cm and mass M has a radius of 20 cm. Calculate the density of the material used, if the moment of inertia of the cylinder about an axis CD parallel to AB, as shown in figure, is 2.7 kg m2
25. The radius of gyration of a uniform rod of length , about an axis passing through a point L/4 away from the centre of the rod and perpendicular to it is (1) 7 48 L (2) 3 8 L (3) 1 8 L (4) 1 4 L
26. An object of mass 8 kg is hanging from one end of a uniform rod CD of mass 2 kg and length 1 m pivoted at its end C on a vertical wall as shown in figure. It is supported by a cable AB such that the system is in equilibrium. The tension in the cable is (Take g = 10 m/s2) B
cm wall C A
° 30° D 8 kg
(1) 14 9. kg/m3
(2) 7.5 × 101 kg/m3
(1) 240 N (2) 30 N (3) 300 N (4) 90 N
27. Shown in the figure is rigid and uniform one metre long rod AB held in horizontal position by two strings tied to its ends and attached to the ceiling. The rod is of mass ‘m’ and has another weight of mass 2 m hung at a distance of 75 cm from A. The tension in the string at A is
(1) 2 mg (2) 1 mg (3) 0.5 mg (4) 0.75 mg
28. A slender uniform rod of mass M and length is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod, when it makes an angle θ with the vertical, is
29. A cord is wound round the circumference of a wheel of radius r. The axis of the wheel is horizontal and the moment of inertia about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance ‘h’, the square of angular velocity of wheel will be (1) 2 2mgh Imr + (2) 2gh (3) 2 2 2 mgh Imr + (4) 2 2 gh Imr +
30. Consider a uniform ro d of mass M = 4 m and length L , pivoted about its centre. A mass m, moving with velocity ν and making angle 4 θ=π to the rod’s long axis, collides with one end of the rod and sticks to it. The angular speed of the rod-mass system, just after the collision, is
(1) 32 7 L ν (2) 3 7 L ν (3) 3 72 L ν (4) 4 7 L ν
31. A thin circular ring of mass M and radius R is rotating with a constant angular velocity of 2 rads −1 in a horizontal plane about an axis vertical to its plane and passing through the centre of the ring. If two objects, each of mass m, are attached gently to the opposite ends of a diameter of the ring, the ring will then rotate with an angular velocity (in rads−1) of
(1) () M Mm + (2) () 2 2 Mm M + (3) () 2 2 M Mm + (4) ()22Mm M +
32. Consider a situation in which a ring, a solid cylinder, and a solid sphere roll down on the same inclined plane without slipping. Assume that they start rolling from rest and they have identical diameter. Choose the correct statement for this situation.
(1) The sphere has the greatest and the ring has the least velocity of the centre of mass at the bottom of the inclined plane.
(2) The ring has the greatest and the cylinder has the least velocity of the centre of mass at the bottom of the inclined plane. (3) All of them will have same velocity.
(4) The cylinder has the greatest and the sphere has the least velocity of the centre of mass at the bottom of the inclined plane.
33. A uniform sphere of mass 500 g rolls without slipping on a plane horizontal surface with its centre moving at a speed of 5.00 cm/s. Its kinetic energy is (1) 1.13 × 10−3 J (2) 6.25 × 10−4 J (3) 8.75 × 10−3 J (4) 8.75 × 10−4 J
34. A solid cylinder and a solid sphere, having same mass M and radius R, roll down the same inclined plane from top without slipping. They start from rest. The ratio of velocity of the solid cylinder to that of the solid sphere, with which they reach the ground, will be (1) 5 3 (2) 4 5 (3) 3 5 (4) 14 15
35. Solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is (1) 2 5 (2) 2 7 (3) 1 5 (4) 7 10
Numerical Value Questions
36. A uniform solid cylinder with radius R and le ngth L has moment of inertia I 1, about the axis of the cylinder. A concentric solid cylinder of radius 2 R R ′ = and length 2 L L ′ = is carved out of the original cylinder. If I2 is the moment of inertia of the carved out portion of the cylinder, then 1 2 = I I (Both I 1 and I 2 are about the axis of the cylinder)
37. I CM is the moment of inertia of a circular disc about an axis (CM) passing through its centre and perpendicular to the plane of disc. IAB is its moment of inertia about an axis AB perpendicular to plane and parallel to axis CM at a distance 2 3 R from centre, where R is the radius of the disc. The ratio of IAB and ICM is x : 9. The value of ‘x’ is . 2 R M B 3 C R A
38. Two discs of the same mass and different radii are made of different materials such that their thickness is 1 cm and 0.5 cm, respectively. The densities of materials are in the ratio 3 : 5 . The moment of inertia of these discs, respectively, about their diameters will be in the ratio of 6 x . The ratio of x is .
39. Two identical solid spheres, each of mass 2 kg and radii 10 cm, are fixed at the ends of a light rod. The separation between the centres of the spheres is 40 cm. The moment of inertia of the system about an axis perpendicular to rod passing through its middle point is × 10–3 kg-m2
40. If a solid sphere of mass 5 kg and a disc of mass 4 kg have the same radius, then the ratio of moment of inertia of the disc about a tangent in its plane to the moment of inertia of the sphere about its tangent will be 7 x . Then, the value of x is .
41. A solid sphere of mass 2 kg is making pure rolling on a horizontal surface with kinetic energy 2240 J. The velocity of centre of mass of the sphere will be ms–1
42. A thin uniform rod of length 2 m, cross-sectional area ‘ A ’, and density ‘ d ’ is rotated about an axis passing through the centre and perpendicular to its length, with angular velocity ω. If value of ω, in terms of its rotational kinetic energy E, is , αE Ad then the value of α is
43. A pulley of radius 1.5 m is rotated about its axis by a force F = 12 t –3 t 2 N applied tangentially (while t is measured in seconds). If moment of inertia of the pulley about its axis of rotation is 4.5 kg m 2, the number of rotations made by the pulley before its direction of motion is reversed will be K π The value of K is .
44. A solid disc of radius 20 cm and mass 10 kg is rotating with an angular velocity of 600 rpm, about an axis normal to its circular plane and passing through its centre of mass. The retarding torque required to bring the disc to rest in 10 s is _____ π × 10 –1 Nm.
47. The position vector of 1 kg object is () ˆ 3 ˆ rijm =− and its velocity is
1 v3j ˆ ms ˆ k =+ . The magnitude of its angular momentum about the origin is x Nm , where x is
48. A hollow spherical ball of uniform density rolls up a curved surface with an initial velocity of 3 m/s (as shown in figure). Maximum height with respect to the initial position covered by it will be cm. (take g = 10 m/s2)
m 3 S
49. For a rolling spherical shell, the ratio of rotational kinetic energy and total kinetic energy is 5 x . The value of x is
50. A solid sphere is rolling on a horizontal plane without slipping. If the ratio of angular momentum about axis of rotation of the sphere to the total energy of moving sphere is π : 22, then the value of its angular speed will be _________ rad/s.
is 1 x . The value of x will be .
45. A rod of mass M and length L is lying on a horizontal frictionless surface. A particle of mass m travelling along the surface, hits one end of the rod with a velocity u, in a direction perpendicular to the rod. The collision is completely elastic. After collision, the particle comes to rest. The ratio of masses m M
46. A circular plate is rotating in horizontal plane, about an axis passing through its centre and perpendicular to the plate, with an angular velocity ω. A person sits at the centre having two dumbbells in his hands. When he stretches out his hands, the moment of inertia of the system becomes triple. If E is the initial kinetic energy of the system, then the final kinetic energy will be E x . The value of x is
51. A uniform disc of mass 0.5 kg and radius r is projected with velocity 18 m/s at t = 0 s on a rough horizontal surface. It starts off with a purely sliding motion at t = 0 s. After 2 s it acquires a purely rolling motion (see figure). The total kinetic energy of the disc after 2 s will be . J (given: coefficient of friction is 0.3 and g = 10 m/s2).
18 m/s t=0 t=2 s
Level-III
Single Option Correct MCQs
1. Four point masses m , 2 m , M 1 and M 2 are placed on a plane, as shown in figure. If the coordinates of the centre of mass of the system are 67 ,, 510 then the value of M1 and M2 are, respectively,
(1) 3m and 4m (2) 4m and 3m (3) 2943 and 44 mm (4) 29 and6 4 mm
2. A circular plate of uniform thickness has a diameter of 28 cm. A circular portion of diameter 21 cm is removed from the plate, as shown. O is the centre of mass of complete plate. The position of centre of mass of the remaining portion will shift towards left from O by (in cm)
3. A paraboloid shaped solid object has equation y = Kx2 about y-axis, as shown in figure. If the height of the body is h, then the position of the centre of mass from the origin (assume density to be uniform throughout) is X Y O h
(1) 2.5 (2) 3.5 (3) 4.5 (4) 5.5
(1) h/m (2) h/4 (3) 2h/3 (4) 2h/4
4. Shown in the figure is a system of three particles of mass 1 kg, 2 kg, and 4 kg, connected by two springs. The acceleration of A, B, and C at any instant is 1 ms–2, 2 ms–2, and 1/2 ms–2, respectively, directed as shown in the figure. External force acting on the system is m1 m2 m 3
a2 a1 (1) 1 N (2) 7 N (3) 3 N (4) 10 N
5. Six identical balls are lined in a straight groove made on a horizontal frictionless surface, as shown. Two similar balls, each moving with a velocity v, collide elastically with the row of 6 balls from left. What will happen?
(1) One ball from the right rolls out with a speed 2v and the remaining balls will remain at rest.
(2) Two balls from the right roll out with speed v each and the remaining balls will remain stationary.
(3) All the six balls in the row will roll out with speed v/6 each and the two colliding balls will come to rest.
(4) The colliding balls will come to rest and no ball rolls out from right.
6. A body of mass m, having an initial velocity v , makes head on elastic collision with a stationary body of mass M . After the collision, the body of mass m comes to rest and only the body having mass M moves. This will happen only when (1) m >> M
(2) m << M
(3) m = M
(4) 1 2 mM =
7. A ball is dropped onto a floor. If the coefficient of restitution is 3 , 2 the percentage loss of energy of the ball on rebounding from the floor is
(1) 12.5 % (2) 50% (3) 25% (4) 75%
8. A sphere of mass m, moving with a constant velocity u, hits another stationary sphere of the same mass. If e is the coefficient of restitution, then the ratio of the velocity of two spheres after collision will be
(1) 1 1 e e + (2) 1 1 e e +
9. A body falling from a height of 10 m rebounds from hard floor. If it loses 20% energy in the impact, then coefficient of restitution is
(1) 0.89 (2) 0.56
(3) 0.23 (4) 0.18
10. A ball is given velocity 3 ugd = at an angle of 45° with the horizontal. It strikes a wall at a distance d and returns to its original position. Find the coefficient of resistitution e.
(1) e = 0.8 (2) e = 0
(3) e = 1 (4) e = 1/2
11. A thin wire of length L and uniform linear mass density ρ is bent into a circular loop with centre at O, as shown. The moment of inertia of the loop about the axis xx' is
12. Particles of masses 1 g, 2 g, 3 g, ……., 100 g are kept at the marks 1 cm, 2 cm, 3 cm,……, 100 cm, respectively, on a meter scale. Find the moment of inertia of the system of particles about a perpendicular bisector of the meter scale.
(1) 0.34 kg –m2
(2) 0.43 kg –m2
(3) 4 kg –m2
(4) 2 kg –m2
13. Which of the following graphs represents the relation between log K and log I, where K is the radius of gyration and I is the moment of inertia?
(1) log I
14. A solid cylinder is suspended symmetrically, through the massless strings, as shown in the figure. The distance of centre of mass from the initial rest position, for the cylinder to fall by unwinding the strings to achieve a speed of 6 ms–1, is cm.
(1) 120 (2) 210 (3) 160 (4) 270
15. A ‘T’ shaped object, with dimensions shown in the figure, is lying on a smooth floor. A force F is applied at the point P parallel to AB, such that the object has only the translational motion without rotation. Find the location of P with respect to C.
(1) 6 sin g l θ (2) 6 sin 2 g l θ (3) 6 cos 2 g l θ (4)
17. A uniform disc of mass m and radius R is projected horizontally with velocity V0 on a rough horizontal floor, so that it starts off with a purely sliding motion at t = 0. After t0 seconds, it acquires a purely rolling motion as shown in figure. The velocity of the centre of mass of the disc at t0 is
(1) 2 3 (2) 3 2 (3) 4 3 (4) ℓ
16. A uniform rod of length ‘ ’ is free to rotate about a fixed horizontal axis through O. The rod begins rotating from rest from its unstable equilibrium position. When it is turned through an angle θ , its angular velocity ω is given as
18 A solid cylinder and a hollow cylinder, both of the same mass and same external diameter, are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first?
(1) Both reach together when the angle of inclination is 45°
(2) Both reachtogether
(3) Hollow cylinder
(4) Solid cylinder
19. A bullet of mass m, moving with a speed v, hits a stationary block of mass M at the topmost point and gets embedded in it. If friction is sufficient to prevent slipping, then the block
(1) must topple down
(2) must not topple down
(3) may topple down (4) will remain at rest
Numerical Value Questions
20. A rod AB of is such that its linear density (mass per unit length) μ varies as =, µ a bx where x is the distance of the section from end A (and b > l). The distance of the centre of mass from end A is ,
where z is
21. Two blocks of masses m = 2 kg and M = 4 kg are inter connected by an ideal spring of stiffness 4/3 Nm–1. If m is pushed with a velocity v0 = 3 ms–1, Then the velocity of the centre of mass of the system will be ms–1.
2 kg v0 4 kg
22. A system consists of two particles. At t = 0, one particle is at the origin and the other particle, having a mass of 600 kg, is on the y-axis at y = 80 m. At t = 0, the centre of mass of the system is on the y -axis at y = 24 m and has a velocity given by 2 6, ktj where k =1 ms –3. The acceleration of the centre of mass, in ms −2, at time t = 2 s is
23. A particle of mass m, moving with speed π m/s, hits elastically another stationary particle of mass 2 m on a smooth horizontal circular tube of radius 5 m. The time in which the next collision will take place is s.
24. A body of mass m1, moving with uniform velocity of 40 m/s, collides with another mass m2, at rest, and then, the two together begin to move with uniform velocity of 30 m/s. The ratio of their masses 1 2 m m is
25. A metal ball falls from a height of 32 m on a steel plate. If the coefficient of restitution is 0.5, to what height (in m) will the ball rise after the second bounce?
26. A ball of mass 10 kg, moving with a velocity of 103 m/s along the x–axis, hits another ball of mass 20 kg, which is at rest. After the collision, the first ball comes to rest while the second ball disintegrates into two equal pieces. One piece starts moving along the y-axis with a speed of 10 m/s. The second piece starts moving at an angle of 30° with respect to the x -axis. The velocity of the ball moving at 30° with x-axis is x m/s. The value of x to the nearest integer is The configuration of pieces after collision is shown in the figure below.
y–axis
V1 = 10 m/s
x–axis 300 Piece-1
27. Consider two uniform discs of the same thickness and different radii, R 1 = R and R2 = a R made of the same material. If the ratio of their moments of inertia, I1 and I2, respectively, about their axes is I1 : I2 = 1 : 81 then the value of ' a ' is
28. I1 is the moment of inertia of a solid cylinder of radius R and length l about its geometrical axis. A concentric cylinder having radius R/2 and length l /2 is carved from the original cylinder and I 2 is the MOI of the carved cylinder. Find 1 2 . I I
29. Three spheres, each of mass m and radius R are kept in touch with each other such that their centres form an equilateral triangle. The moment of inertia of the system about
a median of triangle is 2 . 5 x MR The value of x is .
30. A shell of mass m is projected with a velocity v at an angle 60° to horizontal. When it reaches the maximum height, its angular momentum with respect to point the of projection, is 3 3mv xg The value of x is
31. A rod of mass M and length L is lying on a horizontal frictionless surface. A particle of mass m travelling along the surface hits at one end of the rod with a velocity u, in a direction perpendicular to the rod. The collision is completely elastic. After the collision, the particle comes to rest. The ratio of masses 1 is m Mx
The value of x will be .
32. A shaft rotating at 6000 rpm is transmitting a power of 2π kW. The magnitude of the driving torque is equal to Nm.
33. A rod of length L is held vertically on a smooth horizontal surface. The top end of the rod is given a gentle push. At a certain instant of time, when the rod makes an angle of 37° with the horizontal the velocity of COM of the rod is 4 ms–1. The velocity of the end of the rod in contact with the surface at that instant is m/s.
THEORY-BASED QUESTIONS
Single Option Correct MCQs
1. Among the following, correct one in the case of two particle system is
(1) CM lies nearer to lighter body
(2) CM always lies at midpoint of line joining the bodies
(3) CM lies nearer to heavier body
34. A thin rod of mass 12 m and length 6 L is bent into a regular hexagon. The MI of the hexagon about a normal axis to its plane and through centre of system is KmL2. Then, the value of K is .
35. Moment of inertia of a body about a given axis is 1 kgm2. Initially, the body is at rest. In order to produce a rotational kinetic energy of 800 J, the constant angular acceleration of 10 rad/s2 must be applied about the axis for a duration of s.
36. Moment of inertia of a thin rod of mass M and length L, about an axis passing through its centre, is 2 . 12 ML Its moment of inertia about a parallel axis at a distance of L/4 from this axis is given by 2 48 KML . The value of K is
37. A particle of mass m is moving in time t on a trajectory given by () 2 , ˆ 1055 ˆ rtitj =α+β− where a and b are dimensional constants. The angular momentum of the particle becomes the same as it was for t = 0 at time t equal to s.
38. A force 34 ˆ 4 ˆ ˆ Fijk =++ N is applied on an intersection point of x = 2 m plane and x-axis. The magnitude of torque of this force about a point (2 m, 3 m, 4 m) is _____Nm.
(4) CM always lies outside the system
2. A cylinder is completely filled with water. If 1/4 of the volume of water leaks out, its centre of mass
(1) moves up
(2) moves down
(3) does not change
(4) moves towards vertical surface
3. Three identical spheres, each of radius R, are placed touching each other on a horizontal table. The centre of mass of the system is located
(1) at one of the centres of the spheres
(2) at the midpoint joining the centres of any two spheres
(3) at the point of intersection of the medians of the triangle formed by the centres of the three spheres
(4) at the midpoint of a median of the triangle formed by the centres of the three spheres
4. Choose the wrong statement.
(1) In the case of small bodies [uniform gravitational field], centre of mass and centre of gravity coincide with each other.
(2) In non-uniform gravitational field, centre of mass and centre of gravity will not coincide with each other.
(3) Centre of mass is independent of acceleration due to gravity, but centre of gravity depends on acceleration due to gravity.
(4) Centre of mass describes the stability of the body whereas centre of gravity describes the nature of motion of the body.
5. A shell is thrown vertically up. The shell at the highest point explodes into two equal fragments. The centre of mass of the two fragments
(1) goes further up and then falls (2) falls down with an initial speed
(3) falls down with zero inital velocity (4) comes to rest
6. A uniform metre stick is placed vertically on a horizontal frictionless surface and released. As the stick is in motion, the centre of mass moves
(1) vertically up
(2) vertically down
(3) in a parabolic path
(4) horizontally
7. When an external force is applied at the centre of mass of a system of particles, then it undergoes
(1) Only translatory motion
(2) Only rotatory motion
(3) Both translatory and rotatory motion
(4) An oscillatory motion
8. If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that
(1) linear momentum of the system does not change in time
(2) kinetic energy of the system does not change in time
(3) angular momentum of the system does not change in time
(4) potential energy of the system does not change in time
9. A thin rod has a non-uniform density. It is mounted on an axle passing perpendicular to it, through its centre of mass, as shown, and is then rotated about the axle.
The axle divides the rod into two parts, one on each side of it. Which of the following must be true, no matter how the mass in the rod is distributed?
(1) The two parts have the same mass.
(2) The magnitudes of the momentum of the two parts are equal.
(3) The magnitudes of the angular momentum of the two parts, about the centre of mass, are equal.
(4) The kinetic energies of the two parts are equal.
10. Choose the incorrect statement.
(1) In the process of explosion, some changes may occur in momentum of individual fragments due to internal forces but the motion of the centre of mass is unaltered.
(2) Motion of centre of mass depends upon the external force.
(3) The location of centre of mass relative to the body depends on the reference frame used to locate it.
(4) The position of centre of mass depends upon the shape of the body and the distribution of mass.
11. A body of mass M1 collides elastically with another mass M2 at rest. There is maximum transfer of energy when
(1) M1 > M2 (2) M1 < M2
(3) M1 = M2
(4) same for all values of M1 and M2
12. Which of the following statements is wrong?
(1) KE of a body is independent of the direction of motion.
(2) In an elastic collision of two bodies, the momentum and energy of each body is conserved.
(3) If two protons are brought towards each other, the PE of the system increases.
(4) A body can have energy without momentum.
13. A ball hits the floor and rebounds after inelastic collision. In this case,
(1) the momentum of the ball just after the collision is the same as that just before the collision
(2) the mechanical energy of the ball remains the same in the collision
(3) the total momentum of the ball and the earth is conserved
(4) the total energy of the ball and the earth is conserved
14. Choose the incorrect options.
(1) Kinetic energy is not conserved in an elastic collision.
(2) Momentum of the system is conserved during a collision.
(3) Kinetic energy of the system is conserved during an elastic collision.
(4) Coefficient of restitution for perfectly inelastic collision = 0.
15. If two balls collide in air while moving vertically, then momentum of the system is conserved because
(1) gravity does not affect the momentum of the system
(2) force of gravity is very less compared to the impulsive force
(3) impulsive fore is very less than the gravity
(4) gravity is not acting during collision
16. During elastic collision, (1) law of conservation of momentum is valid
(2) law of conservation of kinetic energy is valid
(3) both (1) and (2)
(4) neither (1) nor (2)
17. Which of the following is not a perfectly inelastic collision?
(1) Striking of two glass balls
(2) A bullet striking a bag of sand
(3) An electron captured by proton (4) A man jumping on to a moving cart
18. Coefficient of restitution depends upon (1) the relative velocities of approach (2) the masses of the colliding bodies
(3) the nature of the colliding bodies
(4) all of the above
19. Moment of inertia of a body depends upon (1) distribution of mass of the body
(2) position of axis of rotation
(3) temperature of the body
(4) all of the above
20. In a rectangle ABCD, AB = 2l and BC = l. The centre of rectangle is at the origin of co-ordinate system. AB and CD are parallel to x-axis. The MI is least about
(1) BD (2) BC
(3) x-axis (4) y-axis
21. The graph between rotational energy E r and angular velocity (ω) is represented by which of the following curve?
(1) Y X E r 0 ω (2) Y X E r 0 ω
(3) Y X E r 0 ω (4) Y X E r 0 ω
22. When temperature of a body made of brass increases, its moment of inertia (1) increases (2) decreses (3) remains the same (4) becomes zero
23. If I is the moment of inertia of a solid sphere about an axis parallel to a diameter of the sphere and at a distance x from it, which of the following graphs represents the variation of I with x?
(1) I X (2) I X
(3) I X (4) I X
24. Choose the correct statement (1) The total torque on a system is independent of the origin if the total external force is zero.
(2) The angular momentum L and angular velocity ω are always anti-parallel vectors.
(3) We cannot have a situation in which total external torque is non-zero while net external force is zero.
(4) The centre of gravity always coincides with centre of mass.
25. When a constant torque is applied on a rigid body, then
(1) the body moves with linear accelaration
(2) the body rotates with constant angular velocity
(3) the body rotates with constant angular accelaration
(4) the body undergoes equal angular displacement in equal intervals of time
26. A solid uniform sphere rotating about its axis with kinetic energy E1 is gently placed on a rough horizontal plane at time t = 0. Assume that, at time t = t 1, it starts pure rolling and, at that instant, total KE of the sphere is E2. After some time t = t2, KE of sphere is E3. Then,
(1) E1 = E2 = E 3 (2) E1 > E2 = E 3
(3) E1 > E2 > E 3 (4) E1 < E2 = E 3
27. In both the figures, all other factors are same, except that in figure (i), AB is rough and BC is smooth while in figure (ii), AB is smooth, and BC is rough.
Kinetic energy of the ball on reaching the bottom:
(1) is same in both the cases
(2) is greater in case (i)
(3) is greater in case (ii)
(4) information insufficient
28. A sphere is rolled on a rough horizontal surface. It gradually slows down and stops. The force of friction tries to
(1) increase the linear velocity
(2) increase the angular velocity
(3) increase the linear momentum
(4) decrease the angular velocity
29. A disc is rolling (without slipping) on a rough surface. C is its centre and Q and P are two points equidistant from C. Let Vp, VQ, and VC be the magnitudes of velocities of points P, Q, and C, respectively. Then,
S-II : Centre of gravity of a body is the point through which the whole weight of the body acts.
32. S-I : In a collision between two objects, the centre of mass of the system always remains stationary if there are no external forces acting on it.
S-II : The centre of mass of a system of particles will remain at rest or move with a constant velocity unless acted upon by an external force. This is a consequence of the law of inertia and is consistent with the concept of conservation of momentum.
(1) VQ > VC > VP
(2) VQ < VC < VP
(3) CP 1 ,V 2 QP VVV ==
(4) VQ < VC > VP
Statement Type Questions
Each question has two statements: statement I (S-I) and statement II (S-II). Mark the correct answer as
(1) if both statement I and statement II are correct,
(2) if both statement I and statement II are incorrect
(3) if statement I is correct but statement II is incorrect,
(4) if statement I is incorrect but statement II is correct.
30. S-I : The centre of mass of a body may lie where there is no mass.
S-II : The centre of mass has nothing to do with the mass.
31. S-I : Centre of mass is that fixed point of a system of particles or a rigid body within the boundaries of the system where the entire mass is concentrated.
33. S-I : Two bodies A and B, initially at rest, of masses 2m and m, respectively, move towards each other under mutual gravitational force of attraction. At the instant when the speed of A is v and that of B is 2v, the speed of the centre of mass of the system is 4v/3.
S-II : The speed of the centre of mass of a system changes if an external force acts on the system.
34. S-I : The initial kinetic energy is equal to the final kinetic energy but the kinetic energy during the collision time is not constant. Such a collision is called an elastic collision.
S-II : A collision between two particles in which the two particles move together after the collision is called a completely inelastic collision.
35. S-I : If a uniform metal disc is remoulded into a solid sphere, then the moment of inertia about the axis of symmetry increases than that before.
S-II : For a given body and for a given plane, the moment of inertia is minimum about an axis passing through the centre of mass.
36. S-I : A rigid body executes both translatory and rotatory motions.
S-II : A rigid body executes pure rotational motion when acted upon by unbalanced torque.
37. S-I : If there is no external torque on a body about its centre of mass, then the velocity of the centre of mass remains constant.
S-II : The linear momentum of an isolated system(Fext= 0) remains constant.
Assertion and Reason Type Questions
In each of the following questions, a statement of Assertion (A) is given, followed by a corresponding statement of Reason (R). Mark the correct answer as
(1) if both (A) and (R) are true and (R) is the correct explanation of (A), (2) if both (A) and (R) are true but (R) is not the correct explanation of (A), (3) if (A) is true but (R) is false, (4) if both (A) and (R) are false.
38. (A) : In a perfectly inelastic one-dimensional collision between two spheres, velocity of both spheres just after collision, is equal.
(R) : For any collision between finite bodies, momentum is always conserved.
39. (A) : Two particles moving in the same direction do not lose all their energy in a completely inelastic collision.
(R) : Principle of conservation of momentum holds true for all kinds of collisions.
40. (A) : Moment of inertia of a body is same, whatever be the axis of rotation.
(R) : Moment of inertia does not depends on mass but not the size of the body.
41. (A) : If a rod is thrown up vertically in the absence of air drag, it returns back to the ground without any change in its angular velocity.
(R) : In the absence of any torque, the angular momentum of system is conserved.
42. (A) : Two cylinders, one hollow (metal) and the other solid (wood), with the same mass and identical dimensions are simultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow cylinder will reach the bottom of the inclined plane first.
(R) : By the principle of conservation of energy, the rotational kinetic energies of both the cylinders are different when they reach the bottom of the incline.
43. (A) : A ring moving down on a smooth inclined plane will be in slipping motion.
(R) : Work done by friction in pure rolling motion is zero.
44. (A) : A solid sphere is rolling on a rough horizontal surface. Acceleration of contact point is zero.
(R) : A solid sphere cannot roll on the smooth surface.
45. (A) : Heavy boxes are loaded along with empty boxes on a cart. Heavy boxes should be kept at the centre of gravity of the cart, for more stability.
(R) : The vertical line passing through the centre of gravity should fall within the base of cart for stability.
46. (A) : The centre of mass of an electron and proton, when released, moves faster towards proton.
(R) : This is because proton is lighter.
47. (A) : A shell moving in a parabolic path explodes in mid air. The centre of mass of the fragments will follow the same parabolic path.
(R) : Explosion is due to internal forces, which cannot alter the state of motion of centre of mass.
48. (A) : The centre of mass of system of n particles is the weighted average of the position vector of the n particles making up the system.
(R) : The acceleration of the centre of mass of a system is independent of internal forces.
49. (A) : In an elastic collision of two billiard balls, both kinetic energy and linear momentum remain conserved.
(R) : During the collision of the balls, as the collision is elastic, there is no exchange of energy. Therefore, both energy and momentum are conserved.
50. (A) : Moment of inertia of a circular disc of mass M and radius R about x , y-axes (passing through its plane) and z -axis, which is perpendicular to its plane, were found to be I x, I y , and I z, respectively. The respective radii of gyration about all the three axes will be different.
(R) : A rigid body making rotational motion does not have fixed mass and shape.
51. (A) : If there is no external torque on a body about its centre of mass, then the velocity of centre of mass remains constant.
(R) : The linear momentum of an isolated system may not remains constant.
52. (A) : If polar ice melts, then day will be longer.
(R) : Moment of inertia of the earth increases
JEE ADVANCED LEVEL
Multiple Option Correct MCQ
1. An object comprises of a uniform ring of radius R and a uniform chord AB (not necessarily made of the same material), as shown. Which of the following cannot be the centre of mass of object?
B X Y
A
and, thus, angular velocity decreases.
53. (A) : A body is moving along a circle with a constant speed. Its angular momentum about the centre of the circle remains constant.
(R) : In this situation, a constant non-zero torque acts on the body.
54. (A) : For a particle moving on a straight line with a uniform velocity, its angular momentum is constant.
(R) : The angular momentum is zero when the particle moves with a uniform velocity.
55. (A) : The velocity of a body at the bottom of an inclined plane of given height is more when it slides down the plane compared to when it is rolling down the same plane.
(R) : In rolling down, a body acquires both kinetic energy of translation and rotation.
56. (A) : A solid sphere and hollow sphere, when released from the top of a smooth and fixed inclined plane, reach the ground simultaneously.
(R) : Acceleration while sliding is independent of mass and dimension of the body.
(1) , 33
(2) , 34
(3) , 22
(4) , 22
2. If a disc of radius r is removed from the disc of radius R, then which of the following are true?
(1) The minimum shift in centre of mass is zero
(2) The maximum shift in centre of mass cannot be greater than 2 . r Rr +
(3) Centre of mass must lie where mass exists.
(4) The shift in centre of mass is 2 . r Rr +
3. A long, thin, inextensible, and very flexible uniform wire is lying on the rough horizontal floor. One end of the wire is bent back and then pulled backward with constant velocity V such that, at any instant of time, the moving part of the wire, always remains just above the part of the wire which is still at rest at that instant on the floor, as shown in diagram. If the wire has unit length and unit mass, then
floor
(1) speed of centre of mass of the moving part at all the times will remain constant
(2) speed of centre of mass of the moving part at all the times will be 3/4 V
(3) minimum force needed to pull the moving part will be V2/2
(4) minimum force needed to pull the moving part will be 3/8 V2
4. A block is kept on a wedge, which is kept on a horizontal floor, as shown in figure. All the surfaces are frictionless. Then, M
(1) centre of mass will not shift horizontally
(2) centre of mass will shift horizontally
(3) centre of mass will shift vertically
(4) centre of mass will shift neither horizontally nor vertically
5. Which of the following is/are correct?
(1) If centre of mass of three particles is at rest and two of them are moving along different lines, then the third particle must also be moving.
(2) If centre of mass remains at rest, then net work done by the forces acting on the system must be zero.
(3) If centre of mass remains at rest, then the net external force must be zero.
(4) If speed of centre of mass is changing then there must be some net work being done on the system from outside.
6. A block of mass m slides down on an inclined plane of a wedge of same mass m (see fig.). Friction is absent everywhere. Which of the following statements is/are correct?
(1) Vertical component of acceleration of block is 2 2 cos 1sin g θ +θ downwards.
(2) Acceleration of centre of mass of (block + wedge) is 2 2 sin 1sin g θ +θ downwards.
(3) Acceleration of centre of mass of (block + wedge) is 2 2 cos 1sin g θ +θ downwards.
(4) Vertical component of acceleration of block is 2 2 2sin 1sin g θ +θ downwards.
7. A cannon and a supply of cannon balls are inside a sealed rail road car at rest. The cannon fires to the right, the car recoils to the left. The balls remain in the car after hitting the farther wall. Assume friction to be absent. If L is the length of the rail road car, D is the distance traversed by the rail road car relative to the ground, m is the total mass of cannon balls fired, and M is the total mass of remaining system (cannon + car), then
(1) D < L
(2) D > L
(3) ML D Mm = +
(4) mL D Mm = +
8. A block of mass m is placed at rest on a smooth wedge of mass M placed at rest on a smooth horizontal surface. As the system is released, M m
(1) The centre of mass of the system remains stationary
(2) The centre of the system has an acceleration g vertically downwards
(3) momentum of the system is conserved along the horizontal direction
(4) acceleration of centre of mass a cm < g is vertically downwards
9. In the figure, the block ‘B’ of mass m starts from rest at the top of a wedge 'W' of mass
M. All surfaces are without friction. 'W' can slide on the ground. ‘B’ slides down on to the ground, moves along it with a speed ‘V’, has an elastic collision with the wall, and climbs back on to ‘W’.
B(m) W(M)
(1) From the beginning, till the collision with the wall, the centre of mass of ‘B’ plus ‘W’ does not move horizontally.
(2) After the collision, the centre of mass of ‘B’ plus ‘W’ moves with the velocity 2 . mV mM +
(3) When ‘B’ reaches its highest position of W, the speed of W is + 2 . mV mM
(4) When B reaches its highest position of W, the speed of W is . mV mM +
10. Ball A of mass m strikes a stationary ball B of mass M and undergoes an elastic collision. After collision, ball A has a speed one-third of its initial speed. The ratio of M/m is
(1) 1/4 (2) 1/2
(3) 2 (4) 4
11. A particle moving with kinetic energy E makes a head-on elastic collision with an identical particle at rest. During the collision, (1) elastic potential energy of the system is always zero
(2) maximum elastic potential energy of the system is E/2
(3) minimum kinetic energy of the system is E/2
(4) kinetic energy of the system is constant
12. Two blocks A and B, each of mass m , are connected by a massless spring of natural length L and spring constant K. The blocks are initially resting on a smooth horizontal floor with the spring at its natural length, as shown in the figure. A third identical block C, also of mass m, moves on the floor with a speed v along the line joining A and B and makes a perfectly elastic collision with A. Then,
(1) the kinetic energy of the A-B system, at maximum compression of the spring, is zero
(2) the kinetic energy of the A-B system, at maximum compression of the spring, is mv2/4
(3) the maximum compression of the spring is () /2 vmK
(4) all of these
13. A set of n identical cubical blocks lies at rest parallel to each other, along a line on a smooth horizontal surface. The separation between the near surfaces of any two adjacent blocks is L. The block at one end is given a speed v towards the next one at time t = 0. All collisions are completely inelastic. Then,
(1) the last block starts moving at () 1 nL t v =
(2) the last block starts moving at () 1 2 nnL t v =
(3) the centre of mass of the system will have a final speed v
(4) the centre of mass of the system will have a final speed v/n
14. In a one-dimensional collision between two identical particles, B is stationary and A has
momentum p before impact. During impact, A gives impulse J to B. Then,
(1) the total momentum of A plus B system is p before and after the impact and (p–J) during the impact
(2) during the impact, B gives impulse J to A
(3) the coefficient of restitution is (2 J/p)–1
(4) the coefficient of restitution is (2 J/p)+1
15. A projectile is fired on a horizontal ground. Coefficient of restitution between the projectile and the ground is e . Let a , b and c be the ratio of time of flight 1 2 , T T
maximum height 1 2 , H H
and horizontal range 1 2 R R
in the first two collisions with the ground. Then,
(1) 1 a e = (2) 2 1 b e =
(3) 2 1 c e = (4) all of these
16. A ball strikes the ground at an angle α and rebounds at an angle β with the vertical, as shown in the figure. Then, ab
(1) coefficient of restitution is tan a /tan b (2) if α < β, the collision is inelastic (3) if α = β, the collision is elastic (4) if α > β, the collision is inelastic
17. Ball A of mass m moving with velocity V collides head on with a stationary ball B of mass m. If e is the coefficient of restitution, then which of the following statements are correct?
(1) The ratio of velocities of balls A and B after the collision is 1 1 e e +
(2) The ratio of the final and initial velocities of ball A is 1 2 e
(3) The ratio of the velocities of balls A and B after the collision is 1 1 e e +
(4) The ratio of the final and initial velocities of ball B is 1 . 2 e +
18. A ball of mass m moving horizontally at a speed v collides with the bob of a simple pendulum at rest. The mass of the bob is also m
(1) If the collision is perfectly inelastic, the height to which the two balls rise after the collision is v2/8g.
(2) If the collision is perfectly inelastic, the kinetic energy of the system immediately after the collision becomes half of that before collision.
(3) If the collision is perfectly elastic, the bob of the pendulum will rise to a height of v2/2g.
(4) If the collision is perfectly elastic, the kinetic energy of the system immediately after the collision is equal to that before collision.
19. A ball A collides elastically with an another identical ball B with velocity 10 m/s at an angle of from the line joining their centres C1and C2. Select the correct option(s).
(4) Kinetic energy will not be conserved here, because collision is not head on.
20. A uniform rod of mass m = 2 kg and length l = 0.5 m is sliding along two mutually perpendicular smooth walls with the two ends P and Q having velocities VP = 4 m/s and VQ = 3 m/s, as shown. Then,
(1) Velocity of ball A after collision is 5 m/s.
(2) Velocity of ball B after collision is 53m/s .
(3) Both the balls move at right angles after collision.
(1) the angular velocity of rod ω = 10 rad/s, counter-clockwise
(2) the angular velocity of rod ω = 5 rad/s, counter-clockwise
(3) the velocity of centre of mass of rod, V cm = 2.5 m/s
(4) the velocity of centre of mass of rod, V cm = 5 m/s
21. A rod of length l is moving in a vertical plane(x-y plane) when the lowest point A of the rod is moved with a velocity v. Which of the following statements are true?
(1) Angular velocity of the rod is
(2) Angular velocity of the rod is 2
(3) Velocity of the end B is 2 vtan θ .
(4) Velocity of the end B is 2 vcot θ .
22. If a sphere is rotating about a diameter at uniform angular speed, then which of the following options is/are incorrect:
(1) The particles on the surface of the sphere do not have any linear acceleration.
(2) The particles on the diameter mentioned above do not have any linear acceleration.
(3) Different particles on the surface have different angular speeds.
(4) All the particles on the surface have the same linear speed.
23. A ring type flywheel of mass 100 kg and diameter 2 m is rotating at the rate of (300/π) revolutions per minute. Then,
(1) the moment of inertia of the flywheel is 100 kgm2
(2) the kinetic energy of rotation of the flywheel is 5 kJ
(3) the flywheel, if subjected to a rotating torque of 200 Nm, will come to rest in 5 s
(4) the moment of inertia of the flywheel is 143 kgm2
24. ABCD is a square plate with centre O. The moments of inertia of the plate about the perpendicular axis through O is I and about the axes 1, 2, 3, and 4 are I 1, I 2, I 3, and I 4, respectively. It follows that
(1) the x-axis is 2 2 3 ml
(2) the z-axis is 2 4 3 ml
(3) an axis parallel to the z-axis and passing through a corner is 2 10 3 ml
(4) one side is 2 5 3 ml .
26. Consider a body of mass 1.0 kg at rest at the origin at time t = 0. A force ˆ ˆ Ftij =α+β is applied on the body, where α = 1.0 N–1 and β = 1.0 N. The torque acting on the body about the origin at time t =1.0 s is τ. Which of the following statements is (are) true?
(1) 1 N.m 3 τ=
(2) The torque τ is in the direction of the unit vector ˆ k + .
(3) The velocity of the body at t = 1 s is () 1 2m/s. ˆ ˆ 2 vij =+
(4) The magnitude of displacement of the body at t = 1 s is 1 m. 6
27. A force ˆ ˆ FAiBj =+ acts at a point whose radius vector with respect to origin is ˆ ˆ raibj =+ , where a , b , A, and B are constants. Then,
(1) torque about ‘O’ is () ˆ aBbAk
(1) I2 = I 3 (2)
25. Four identical rods, each of mass m and length l, are joined to form a rigid square frame. The frame lies in the x-y plane, with its centre at the origin and the sides parallel to the x and y axes. The moment of inertia about
(2) arm of force 22 aBbA AB + about origin
(3) torque about ‘O’ is zero
(4) torque about ‘O’ is () ˆ aBbAk +
28. The torque of a force () 537 N ˆ ˆ ˆ ijk +− about the origin is τ . If the force acts on a particle
whose position vector is () ˆ ˆ ˆ 2 m ijk ++ , then the torque acting on the particle is
(1) () ˆ ˆ 17127.m ˆ N ijkτ=−+−
(2) 482 N.m τ=
(3) () 17127 N.m ˆ ˆ ˆ ijk
(4) () 17127 N.m ˆ ˆ ˆ ijk ++
29. A small ball of mass m, suspended from the ceiling at a point O by a thread of length l, moves along a horizontal circle with a constant angular velocity ω . Which of the following statements is/are correct?
(1) speed of particle after collision is 0 6 v
(2) speed of centre of rod after collision is 0 5
(3) angular speed of rod just after collision is 3 2 o v l ω=
(4) change in kinetic energy of the particle is 0 k ∆=
31. A particle of mass m is projected with a velocity v, making an angle of 45° with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection, when the particle is at its maximum height h, is
zero (2)
(1) Angular momentum is constant about O.
(2) Angular momentum is constant about C.
(3) Vertical component of angular momentum about O is constant.
(4) Magnitude of angular momentum about O is constant.
30. A particle of mass m collides with the end of a spinning rod of mass 2m and length l, at the end of the rod. If the coefficient of restitution of collision is 2 3 e = , then
32. The potential energy of a particle of mass m at a distance r from a fixed point O is given by () 2 4 kr Vr = , where k is a positive
constant of appropriate dimensions. This particle is moving in a circular orbit of radius R about the point O. If v is the speed of the particle and L is the magnitude of its angular momentum about O, which of the following statements is (are) true?
(1) k vR m = (2) 2 k vR m =
33. A uniform meter stick of mass 200 g is suspended from the ceiling through two vertical strings of equal lengths fixed at the ends. A small object of mass 40 g is placed on the stick at a distance of 70 cm from the left end. Find the tension in the two strings.
(1) Tension in left end is 1.12 N.
(2) Tension in right end is 1.28 N.
(3) Tension in left end is 1.28 N.
(4) Tension in right end is 1.12 N.
34. A uniform circular disc of mass M and radius R rolls without slipping on a horizontal surface. If the velocity of its centre is v0, then the total angular momentum of the disc about fixed point ' p ' at a height 3 2 R above the centre C
that the spool is a uniform, solid cylinder that does not slip. M F
(1) increases continuously as the disc moves away
(2) decreases continuously as the disc moves away
(3) is always equal to 2MRV0
(4) is always equal to MRV0
35. If a thin ring of mass 1 kg and radius 1 m is pure rolling at a speed of 1 m/s, then
(1) its total kinetic energy is 1 J
(2) its translational kinetic energy is 1/2 J
(3) its rotational kinetic energy is 1/2 J
(4) its total kinetic energy is 1/2 J
36. A spool of wire of mass M and radius 2R is unwound under a constant force F. Assume
(1) The acceleration of the centre of mass is 4 . 3 F M
(2) The force of friction is to the right and equal in magnitude to 3 F
(3) The force of friction is to the left and equal in magnitude to . 3 F
(4) The force of friction is to the right and equal in magnitude to 4 3 F
37. Consider a cylinder of mass M and radius R lying on a rough horizontal plane. It has a plank lying on its top, as shown in figure. When a force F is applied on the plank, it moves and causes the cylinder to roll. The plank always remains horizontal. There is no slipping at any point of contact. Then,
(1) the acceleration of the cylinder is 4cos 83 F mM θ +
(2) the acceleration of the cylinder is 4sin 83 F mM θ +
(3) the frictional force between plane and cylinder is cos 38 FM Mm θ +
(4) the frictional force between plane and cylinder is sin 38 FM Mm θ +
38. A wheel of radius 3R and mass m is placed on an inclined plane of inclination θ and a massless, inextensible string is wound around an axle of radius R , as shown in figure. If the inclined plane is frictionless and string does not slip on the axle, then which of the following statements about the acceleration of the wheel and the tension in the string is/are true? (I = Moment of inertia of the wheel)
(1) Acceleration of the wheel is 2 I 1 gsin mR θ +
(2) Acceleration of the wheel is 2 g I 1 mR +
(3) Tension in the string is 2 . 1 I mgsin mR θ +
(4) Tension in the string is 2 1 I mg mR +
39. The figure shows a system consisting of
(i) a ring of outer radius 3R rolling clockwise without slipping on a horizontal surface with angular speed ω and
(ii) an inner disc of radius 2R rotating anticlockwise with angular speed ω /2.
The ring and disc are separated by frictionless ball bearings. The system is in the x-z plane. The point P on the inner disc is at a distance R from the origin, where OP makes an angle of 30° with the horizontal Then with respect to the horizontal surface,
(1) the point O has linear velocity 3
(2) the point P has velocity
(3) the point P has velocity
(4) the point P has velocity
Numerical Value Questions
40. A cylindrical rod of length 1.2 m is made of three materials. The first part of 0.5 m is made of iron, the second part of 0.3 m is of copper, and the last part of 0.4 m is of aluminium. If the densities of iron, copper, and aluminium are 7.9×103 kg/m3, 8.9 ×103 kg/m3, and 2.7 ×103 kg/m3, respecitively, the centre of mass of the rod from the right end of the aluminium rod is m.
41. Two rods of lengths l1 = 1.5 and l2 = 1 m, and masses m1 = 1 kg, and m2 = 2 kg, are joined end to end. The centre of mass of the rod from rod is 19/6y m. Then, y is
42. Four particles of masses 2, 2, 4, and 4 kg are arranged at the corners A, B, C, and D of a square ABCD of side 2 m, as shown in the figure. The perpendicular distance of their centre of mass from the side AB is
43. From a hemisphere of radius R, a cone of base radius R/2 and height R is removed as shown in figure. Calculate the height of centre of mass of the remaining object in cm, if R = 28 cm.
46. Two balls A and B of equal mass are projected upwards simultaneously, one from the ground with speed 50 ms −1 and the other from a tower of height 40 m above the first ball, with initial speed 30 ms−1. Calculate the maximum height attained by their centre of mass, in metres. [Take g = 10 ms–2]
44. A small particle of mass m starts sliding down from rest along the smooth surface of a fixed hollow hemisphere of mass M (4 m ). The distance of centre of mass of (particle + hemisphere) from centre O of hemisphere, when the particle is separated from the surface of hemisphere, is 69 . 5 R α Find the value of 'α'.
47. A platform of mass m and counter weight of M + m (where, M = 3m) are connected by a light cord, which passes over two smooth pulleys. A man of mass M is standing on the platform, which is at rest. If the man leaps vertically upwards (and out of platform) with velocity 10 ms–1, find the distance (in m) through which the platform will descend.
45. Two particles A and B of mass 5 kg and 2 kg are projected simultaneously from same ground level, as shown in figure. Find the maximum height (in cm) reached by the centre of mass. Assume that only gravitational force of attraction by the earth is present, and take g = 10 ms–2 .
48. A body of mass 1 kg makes an elastic collision with another body at rest and continues to move in the original direction after collision with a velocity equal to 1/5th of its original velocity. Find the mass of the second body in kg.
49. A steel sphere of mass 100 g, moving with a velocity of 4 m/s, collides with a dust particle
elastically moving in the same direction with a velocity of 1 m/s. The velocity of the dust particle after the collision is ____ m/s.
50. Ball 1 collides with another identical ball 2 at rest, as shown in figure. For what value of coefficient of restitution e , the velocity of second ball becomes two times that of 1 after collision?
51. A ball strikes a horizontal floor at an angle θ=45°. The coefficient of restitution between the ball and the floor is 1 . 2 e = The fraction of its kinetic energy lost in collision is
52. A ball of mass 0.2 kg is dropped down vertically from a height of 1 m above the ground. If it rebounds to a height of 0.64 m, find the coefficient of restitution between the surface of the ball and the ground.
53. A circular disc X of radius R is made from an iron plate of thickness t and another plate Y of radius 4R is made from an iron plate of thickness t/4. The ratio between moments of inertia y x I I is .
54. A ring and a solid sphere rotating about an axis passing through their centres have same radii of gyration. The axis of rotation is perpendicular to plane of ring. The ratio of radius of ring to that of sphere is 2 . x Then, the value of x is .
55. The length of a rod is 1012m. . The radius of gyration of a cylinder rod about an axis of rotation perpendicular to its length and passing through the centre will be m.
56. ABCDECA is a planar body of mass m of uniform thickness and same material. The dimensions are as shown in the figure. The moment of inertia of the body about an axis passing through point A and perpendicular to planar body is I1 and that of about an axis
passing through C and perpendicular to planar body is I2. Then, 1 2 I I is .
57. Three identical uniform rods, each of length 1 m and mass 2 kg, are arranged to form an equilateral triangle. The moment of inertia of the system about an axis passing through one corner and perpendicular to the plane of the triangle is ___kg–m 2.
58. A square plate of mass M and edge L is shown. The moment of inertia of the plate about the axis in the plane of plate and passing through one of its vertices, making an angle 15° from horizontal, is 2 11 , 8 ML x where x is
59. A light rod of length 2 m is pivoted at its centre and two masses of 5 kg and 2 kg are hung from the ends, as shown in figure. The initial angular acceleration of the rod, assuming that it was horizontal in the beginning, is α =____(rad/s2). Take g = 9.8 m/s2.
60. A uniform circular disc of mass 1.6 kg and radius 0.5 m is initially at rest on a horizontal frictionless surface. Three forces of equal magnitude 0.6 N are applied simultaneously along the three sides of an equilateral triangle XYZ with its vertices on the perimeter of the disc (see figure). One second after applying the forces, the angular speed of the disc (in rad/s) is .
through its centre and perpendicular to the plane of disc. A body of same mass as that of disc is fixed at the highest point of the disc. Now, the system is released. When the body comes to lowest position, its angular speed
will be 4 3 x R rad/s, where x = .
(g = 10 m/s2)
64. A uniform solid sphere of 1 kg rolls on a horizontal surface without slipping. If the velocity of centre of mass is 5 cm/s, the work done in stopping the sphere is N × 10–4 J. The value of N is .
61. A horizontal platform of mass 200 kg in the form of a circular disc rotates at 10 rpm along a vertical axis passing through its centre. A man weighing 80 kg is standing on its edge. The angular velocity (in rpm) with which the platform rotates, if the man moves to its centre, is
62. A horizontal circular platform of radius 0.5 m and mass 0.50 kg is free to rotate about its axis. Two massless spring toy-guns, each carrying a steel ball of mass 0.1 kg, are attached to the platform at a distance 0.25 m from the centre on its either side, along its diameter (see figure). Each gun simultaneously fires the ball horizontally and perpendicular to the diameter in opposite directions. After leaving the platform, the ball has horizontal speed of 9 ms −1 with respect to the ground. The rotational speed of the platform (in rad/s) after the balls leave the platform is
65. A solid sphere of mass 10 kg rolls without sliding with uniform velocity 2 m/s along a horizontal table. The total energy of the sphere is J.
Integer Value Question
66. Two blocks A and B are connected by a massless string, as shown in figure. Force of F = 30 N is applied on block (assuming no friction) B. Then, distance travelled by the centre of mass in 2 seconds, starting from rest, is x metres. Find the value of x
B F A 10 kg 20 kg
67. A ball of mass m moving with a certain velocity collides against a stationary ball of mass m. The two balls stick together during collision. If E is the initial kinetic energy, then the loss of kinetic energy in the collision is E/A. Find the value of A.
63. A disc of mass 4 kg and radius R is free to rotate about a horizontal axis passing
68. A ring and a disc are initially at rest, side by side, at the top of an inclined plane, which makes an angle of 60° with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reachi ng the ground is () 23 s, 10 then the height of the top of the inclined plane, in metres, is x/4. The value of x is _____. (Take, g = 10 ms–2)
Passage-based Questions
Passage-I:
A (2n + 1) identical bricks, each of length L and height h, are arranged, with h are arranged,with n bricks above and n bricks below the central brick AB, as shown in figure, for n = 3. Each brick is displaced with respect to the one in contact by L/10. The x–y axes are defined with respect to the lowest brick, as shown.
(1) 0, 2 L (2) ,0 2 L
(3) ,0 23 L (4) L/3
72. The position coordinates of the centre of mass of triangular and circular plates with respect to point O are (L is side of triangles)
(1) () 31 23 π− π+ right of O
(2) 2 3 π+ right of O
(3) 1 3 π+ left of O
69. The y-coordinate of the centre of mass of the system is
(1) () 1 2 Lhn + (2) 1 2 nh +
(3) 2 nh (4) () 1 1 2 nh +
70. The x-coordinate of the centre of mass for the arrangement shown in the figure is
(1) 16 45 L (2) 32 45 L
(3) 22 35 L (4) 35 44 L
Passage-II:
Equilateral triangular, square, and circular plates of same thickness but made up of different materials are arranged as shown in figure. O
(4) 3 3 π+ rigth of O
Passage-III:
Persons A, of mass 60 kg, and B, of mass 40 kg, are standing at rest on a horizontal platform of mass 50 kg. The platform is supported with wheels on a horizontal frictionless surface and is initially at rest. Both A and B jump from the platform simultaneously and in the same horizontal direction. Just after the jump, the persons A and B move away from the platform with a speed of 3 m/s relative to the platform and along the horizonta l.
A B
71. The coordinates of the centre of mass of the triangular plate of side L, (if O is the reference point) is
73. Just after both A and B jump from the platform, velocity of centre of mass of the system (A, B, and the platform) is ____m/s.
74. Final speed of the platform in situation (i), i.e., just after both A and B have jumped, will be nearly _____m/s.
9: System of Particles and Rotational Motion
Passage-IV:
A block of mass m moves with velocity v o towards a stationary block of mass M on a smooth horizontal surface.
75. Velocity of centre of mass of the system is
(1) 0 mv Mm + (2) 0 mv M
(3) 0 mv Mm (4) 0 2 v
76. Initial velocity of block of mass m w.r.t. centre of mass of the system is (1) o Mv mM + (2) 0 mv Mm +
(3) 0 2 v (4) 0 3 v
Passage-V:
In fig, block A of mass 2 kg is moving to the right with a speed of 5 m/s on a horizontal frictionless surface. Another block B of mass 3 kg, with a massless spring attached to it, is moving to the left on the same surface and with a speed of 2 m/s. Let us take the direction to the right as the positive x -direction. At some instant, block A collides with the spring attached to block B. At some other instant, the spring has maximum compression and then, finally, the blocks move with their final velocities. Assume that
(i) the spring force is conservative and, so, there is no conversion of kinetic energy to interval energy, and
(ii) no sound is made when block A hits the spring.
78. Final speed of centre of mass of the system of blocks A and B will be _____m/s.
Passage-VI:
Two pendulum bobs of mass m and 2m collide elastically at the lowest point in their motion. If both the balls are released from height h above the lowest point, answer the following.
79. Speed of the bob of mass m just after collision is 5 3 xgh Then, the value of x is .
80. If the height to which the bob of mass m rises after the collision is 9 yh , then the value of y is .
Passage-VII:
Two smooth spheres, A and B, of equal radius lie on a horizontal table. A is of mass m and B is of mass 3 m . The spheres are projected towards each other with velocity vectors 52 and 2, ˆˆˆˆ ijij +− respectively, and when they collide, the line joining their centres is parallel to the vector ˆ i . The coefficient of restitution between A and B is 1/3.
81. Their velocities after impact are (1) 22, ˆ ˆ ˆˆ 3 ijij +− (2) 22, ˆ ˆ ˆˆ 3 ijij (3) ˆˆˆˆ ,3 ijij −+ (4) ˆˆˆˆ ,3 ijij
82. The loss in kinetic energy caused by the collision is (1) m (2) 2m (3) 3m (4) 4m
Passage-VIII:
In the given figure, F = 10 N, R = 1 m, mass of the body is 2 kg, and moment of inertia of the body about an axis passing through O and perpendicular to plane of body is 4 kgm 2. O is the centre of mass of the body .
77. Final speed of block A will be ____m/s.
83. If ground is sufficiently rough, what is the kinetic energy of the body (in J) after 2 s?
84. If the ground is smooth, what is the total kinetic energy (in J) of the body after 2 s?
Passage-IX:
The minimum coefficient of static friction between an inclined plane and cylinder for the cylinder not slipping on the inclined plane is µ. The angle of inclination is 45° . Now the angle of inclination is changed to 30°, then the cylinder reaches the bottom with an acceleration a ms–2 (given g = 10 ms–2)
85. The value of µ is
86. The magnitude of a is
Matrix Matching Questions
87. A rectangular slab ABCD has dimensions a × 2 a , as shown in figure. Match the following two columns 1 2
(A) (B) (C) (D)
(1) II III III I
(2) II III IV I
(3) I IV II III
(4) II IV I III
88. Square frame is made of four uniform rods. Mass of each rod is m and length is l.
Column I (Axis of rotation)
Column II (MI about axis of rotation)
Column I
Column II
(A) Radius of gyration about axis-1 (I) 12 a (B) Radius of gyration about axis-2 (II) 2 3 a (C) Radius of gyration about axis-3 (III) 3 a (D) Radius of gyration about axis-4 (IV) 4 3 a
Choose the correct answer from the options given below:
(A) (B) (C) (D)
(1) III IV II I
(2) IV II I III
(3) IV III I II
(4) II I IV III
89. A uniform disc can rotate about a fixed horizontal axis passing through its centre and perpendicular to its plane. A light chord is wrapped around the rim of the disc (mass 2 kg and radius 1 m) and mass of 1 kg is tied to the free end. If it is released from rest, then match the followin g.
Column–I
Column–II
(A) The tension in the chord is ____ N. (I) 10
(B) In the first 4 seconds, the angular displacement of the disc is___rad.
(C) The work done by the torque on the disc in the first 4 seconds is _____J.
(II) 5
(III) 40
(D) Acceleration of mass is________m/s2 (IV) 200 (V) 50
Choose the correct answer from the options given below.
(A) (B) (C) (D)
(1) III II IV I
(2) II IV III I
(3) I III IV II
(4) II III IV II
90. A ring type flywheel of mass 200 kg and diameter 2 m is rotating at the rate of 300/π revolutions per minute. Answer the following w.r.t. the situation .
List-I
List-II
(A) Moment of inertia of the flywheel is___kg.m2 (I) 10
(B) The kinetic energy of rotation of the flywheel is _____kJ. (II) 200
(C) The flywheel, if subjected to a rotating torque of 200 N.m will come to rest in time t = s.
(III) 5
(D) Initial angular velocity is___rad/s. (IV) 15 (V) 50
Choose the correct answer from the options given below.
(A) (B) (C) (D)
(1) III I III II
(2) II III I III
(3) I II III III
(4) II I I I
91. A disc of radius R rolls without slipping on a rough surface with constant angular velocity ω . If v is the linear speed of the centre of mass of the disc, then match the speed of points.
List-I List-II
(A) A (I) 2Rω
(B) B (II) v
(C) C (III) 2v
(D) O (IV) 0
(V) v/2
Choose the correct answer from the options given below.
(A) (B) (C) (D)
(1) IV II I, III II
(2) IV I II, III II
(3) IV II II, III I
(4) I II IV, III II
BRAIN TEASERS
1. A uniform rod AB of mass M is attached to a hinge at one end A and released from the horizontal position. The rod rotates about A and when it reaches the vertical position, the rod strikes a sphere of mass m and radius r, initially at rest, on the smooth horizontal surface, as shown in the adjacent figure. The impact is along the horizontal direction and perfectly elastic. At the moment of impact, the lowest end of the rod is very close to the smooth horizontal surface. After the impact, the sphere moves along the horizontal and the rod subsequently rises to a maximum of 60° with the vertical. Choose the correct statement(s) from the following, taking into account the information given above. The length of the rod equals 62 2.m. 10 rr
(Take g = 10 m/s2)
(1) The ratio M/m is 3/2.
(2) The ratio m/M is 3/2.
(3) The ratio m/M is 2/9.
(4) The ratio m/M is 9/2.
2. A solid cone is placed on a horizontal surface with height h and radius R. Its apex angle is θ. To change the position of the cone from figure (A) to figure (B), no work is required. Then,
3. Particle like balls, each of mass m, are affixed at both the ends of a light rigid rod of length l. The composite body thus formed in suspended from a nail O, with the help of two inextensible cords affixed on each ball. Length of each chord is also l . Now, arrangement in pulled aside, bringing one ball in level with the nail and keeping both the chords straight and then released. Let θ be the angular position of centre of mass of the system from horizontal at any instant of time. Then, O m m
(1) its angular acceleration of the system as a function of θ is 3 cos 2 g l θ
(2) if T1 and T2 are the tensions in the two threads, then T1 –T2 = mgcos θ
(3) angular position θ at which tension becomes maximum is given by 10 tan 3 θ=
(4) maximum tension in a cord during subsequent motion 103 3 23 mg
(1)
4. Two identical uniform discs, each of mass m and radius r, are placed on a horizontal frictionless floor. One of the discs is given an angular velocity ω about its stationary vertical axis and the other is projected with a horizontal velocity υ without rotation for a head on collision to occur between them. If the collision is perfectly inelastic and slipping between the discs ceases in the end of the
9: System of Particles and Rotational Motion
collision. Angular velocities of the discs and speeds of their centres after the collisions are
(1)
5. A force F is applied on the plank such that the hollow hemispherical shell is in equilibrium, as shown in figure. The coefficient of friction μ is same between the with respect to plank hemispherical shell and the plank as its between the plank and the ground. Friction is just sufficient to prevent the slipping. (Take g = 10 m/s2 and m = 5 kg). Then, which of the following statements is/are true?
(1) The angular velocity vector of cone about O has magnitude 100 rad/s.
(2) The angular velocity vector of cone about O has magnitude 1002rad/s.
(3) The angle made by this angular velocity vector with vertical is 45°.
(4) The direction of this angular velocity vector about O is constant.
7. A uniform rod of mass M and length L is hinged at lower end on a table. The rod can rotate freely in vertical plane and there is no friction at the hinge. A uniform solid ball of mass M and radius = 3 L R is placed in contact
with vertical rod and a horizontal force F is applied at the upper end of the rod. There is sufficient friction between ball and ground for pure rolling. Mark the correct statement (s).
(1) The minimum magnitude of coefficient of friction μ = 1/2.
(2) The minimum magnitude of coefficient of friction μ = 1/4.
(3) The magnitude of applied force is 100 N.
(4) The acceleration of plank is 10 m/s 2 .
6. A round cone with half-angle α = 45° and the radius of the base R = 10 cm rolls uniformly without slipping over horizontal plane. The apex is hinged at the point O (see figure), which is at the same level as the point C, the centre of the base. The velocity of the point C is v = 10 m–1
(1) The acceleration of centre of mass of ball just after the application of force is 15 . 22 F M
(2) The angular acceleration of rod just after the application of force is 45 22 F ML
(3) The horizontal component of hinge reaction on the rod just after application of force is 43 . 44 F
(4) The horizontal component of hinge reaction on the rod just after application of force is 44 43 F
8. A stationary homogenous sphere of radius R and mass M is gently placed on the conveyor belt moving with a constant velocity v 0 towards the right. After a time t from the instant the sphere was placed on the belt, it starts pure rolling, and its COM moves with a velocity v relative to the ground. Coefficient of kinetic friction is μ k. Then,
(1) = µ 0 2 5 k v t g (2) = µ 0 2 7 k v t g
(3) = 0 2 5 v v (4) = 0 2 7 v v
9. A semi-cylinder is cut out of a block and the block is kept on the horizontal surface. A small body of mass m is released at the top of the cylinder shaped hole. (Friction is negligible.) If the block is able to move on the horizontal surface without friction, then the force exerted on the block by the small body at the lowest point is 7mg/2. Find the ratio of M/m. m M
11. Initial conditions have been set up so that a coin of radius r rolls around in a circle, as shown in fig. The contact point on the ground traces out a circle of radius R, and the coin makes a constant angle θ with the horizontal. The coin rolls without slipping (assume that the friction with the ground is as large as needed). What is the angular velocity, ω , of the circular motion of the contact point on the ground? (Where R = 6 m, r = 2 m, θ = 53°) (g = 10 m/s2)
10. The moment of inertia of the shaded area about y axis is 2 12 12 3 . MLhh nhh + + The mass of the shaded area is M. Then, n is______.
12. There are five balls at rest at equal distances in a straight line on a smooth horizontal surface. Their masses are in geometrical progression with common ratio 2 and their coefficients of restitution are each 2/3. If the first ball is started towards the second with velocity u, then the velocity communicated to 5th ball is () yx x u y
Find the value of x+y.
13. A T-shaped system is made by using two identical rods AB and CD, each of mass m and length ‘l’. A linear impulse J is imparted to the system at midpoint of rod AB. Just after impulse, angular speed of rod AB is (1) ω1 if midpoint of rod CD is connected to end B by a pin support so that rod CD
CHAPTER 9: System of Particles and Rotational Motion 118 can rotate freely about hinge at end B, and (2) ω2 if midpoint of CD is rigidly connected to end B.
a vertical wall while hinge B is connected to the centre of an oscillating disc (amplitude of centre = A) of mass m by a light rod. The disc is connected to two identical springs of spring constant k each. Friction between disc and horizontal surface in its contact is sufficient to prevent slipping of the disc. It is given that,
A = 0.2 mm, 1 3 mkg = , k = 90 N/m, θ = 45°
(when the disc is at mean position).
Passage-I:
A heavy steel ball is held in contact with an extremely light steel ball at height 53 h = above a hard horizontal floor. The line joining the centres of the balls makes angle θ=30° with a vertical. If all the collisions are elastic and sizes of the balls are negligible as compared to the height h , the horizontal range of the lighter ball is R (radius of the ball is much smaller than horizontal range R). Speed just after collision is u. Then, g = 10 m/s2 .
16. Speed of hinge C at the instant θ = 45° is mm/s.
17. Angular speed of rod AD of length = 2 mm L at the instant θ = 45° is rad/s.
Passage-III:
A cardboard strip, bent in the shape of the letter C, is put on a rough inclined plane, as shown in the figure.
14. The value of R (in m) is_____________
15. The value of u (in m/s) is ____________
Passage-II:
Twelve light rods are hinged to each other as shown in the arrangement in the figure geometrically resembling three rhombuses of sides in the ratio 3:2:1. Hinge A is fixed to
18. At what angle of inclination to the horizontal plane will it topple? (assume that it does not slide)
19. What should be the minimum coefficient of friction so that it does not slide before toppling?
(1) 0.66 (2) 0.75 (3) 1.0 (4) 0.33
FLASHBACK (P revious JEE Q uestions )
JEE Main
Numerical Value Questions
1. A solid circular disc of mass 50 kg rolls along a horizontal floor so that its center of mass has a speed of 0.4 m/s. The absolute value of work done on the disc to stop it is ______ J. (2024)
2. The identical spheres each of mass 2 M are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 4 m each. Taking point of intersection of these two sides as origin, the magnitude of position vector of the centre of mass of the system is 42 x where the value of x is (2024)
3. A disc of radius R and mass M is rolling horizontally without slipping with speed . It then moves up an inclined smooth surface as shown in figure. The maximum height that the disc can go up the incline is: (2024)
4. A uniform rod AB of mass 2 kg and Length 30 cm at rest on a smooth horizontal surface. An impulse of force 0.2 Ns is applied to end B (penpendicular to the rod). The time taken by the rod to turn through at right angles will be s x π , where x = ____. (2024)
5. Four particles each of mass 1 kg are placed at four corners of a square of side 2 m. Moment of inertia of system about an axis
perpendicular to its plane and passing through one of its vertex is _____ kgm 2 . (2024)
6. A ring and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of both bodies are identical and the ratio of their kinetic energies is 7 x where x is ________. (2024)
7. A cylinder is rolling down on an inclined plane of inclination 60°. It's acceleration during rolling down will be 2 / 3 x ms , where x = _________. (use g=10m/s2). (2024)
8. A body of mass 5 kg moving with a uniform speed 32 ms−1 in X–Y plane along the line y = x + 4. The angular momentum of the particle about the origin will be ______ = kgm2s−1. (2024)
9. A particle of mass m projected with a velocity ‘ u ’ making an angle of 30º with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height h is : (2024)
(1) 3 3 16 mu g (2) 2 3 2g mu (3) 3 2 mu g
(4) zero
10. Consider a Disc of mass 5 kg, radius 2 m, rotating with angular velocity of 10rad/s about an axis perpendicular to the plane of rotation. An identical disc is kept gently over the rotating disc along the same axis. The energy dissipated so that both the discs continue to rotate together without slipping is _______ J. (2024)
11. Two discs of moment of inertia I 1=4kgm 2 and I 2 =2kgm 2 about their central axes & normal to their planes, rotating with angular speeds 10 rad/s & 4 rad/s respectively are brought into contact face to face with their axe of rotation coincident. The loss in kinetic energy of the system in the process is_____J. (2024)
12. Two identical spheres each of mass 2 kg and radius 50 cm are fixed at the ends of a light rod so that the separation between the centers is 150 cm. Then, moment of inertia of the system about an axis perpendicular to the rod and passing through its middle point is 2 20xkgm , where the value of x is_____ (2024)
13. A body of mass 'm' is projected with a speed ‘ u’ making an angle of 45° with the ground. The angular momentum of the body about the point of projection, at the highest point is expressed as 3 2mu Xg . The value of ' X ' is _______. (2024)
14. In the figure shown, ABC is a uniform wire. If centre of mass of wire lies vertically below point A, then BC/AB is close to (2022)
(1) 1.85 (2) 1.37 (3) 1.5 (4) 3
15. A spherical shell of 1 kg mass and radius R is rolling with angular speed ω on horizontal plane (as shown in figure). The magnitude of angular momentum of the shell about the origin O is 2 3 ω a R The value of a will be (2022)
(1) 2 (2) 3 (3) 5 (4) 4
16. Consider a badminton racket with length scales as shown in the figure.
The mass of the linear and circular portions of the badminton racket are same ( M) and the mass of the threads are negligible. The moment of inertia of the racket about an axis perpendicular to the handle and in the plane of the ring at r /2 distance from the end A of the handle will be Mr2 (2022)
17. Three identical spheres, each of mass M, are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 3 m each. Taking point of intersection of mutually perpendicular sides as origin, the magnitude of position vector of centre of mass of the system will be m. x The value of x is . (2021)
18. Three blocks A, B, and C are lying on a smooth horizontal surface, as shown in the figure. A and B have equal mass m while C has mass M. Block A is given an inital speed v towards B, due to which it collides with B perfectly inelastically. The combined mass collides with C, also perfectly inelastically. 5/6th of the initial kinetic energy is lost in the whole process. What is the value of M/m? (2020)
(1) 3 (2) 4 (3) 2 (4) 5
19. A small particle of mass m is projected at an angle θ with the y x-axis with an initial velocity v0 in the x-y plane, as shown in the figure. At a time 0 sin , θ < v t g the anguIar momentum of the particle is
(2) continuously increases (3) first increases and then decreases (4) remains unchanged
21. Two particles of equal mass m have respective initial velocities ˆ ui and
They collide completely inelastically. The energy lost in the process is (2019)
2 3 4 mu (2) 2
22. Distance of the centre of mass of a solid uniform cone from its vertex is z 0. If the radius of its base is R and its height is h, then z0 is equal to (2016) (1)
where i,j, and k ˆ ˆˆ are unit vectors along x, y, and z-axis, respectively. (2020) (1) 2
23. A body A of mass M, while falling vertically downwards under gravity, breaks into two parts; a body B of mass 1/3 M and a body C of mass 2/3 M. The centre of mass of bodies B and C taken together shifts compared to that of body A towards (2015) (1) depends on height of breaking (2) does not shift (3) body C (4) body B
20. A thin horizontal circular disc is rotating about a vertical axis passing through its centre. An insect is at rest at a point near the rim of the disc. The insect now moves along a diameter of the disc to reach its other end. During the journey of the insect, the angular speed of the disc (2020) (1) continuously decreases
24. A body A of mass m = 0.1 kg has an initial velocity of –1 3 ms ∧ i It collides elastically with another body B of the same mass, which has an initial velocity of –1 . 5 ms j ∧ After collision, A moves with a velocity 4. ∧∧
=+
vij The energy of B after collision is written as 10 x J The value of x is . (2010)
25. Moment of inertia (MI) of four bodies having same mass M and radius 2R are as follows:
I1= MI of solid sphere about its diameter
I2 = MI of solid cylinder about its axis
I 3 = MI of solid circular disc about its diameter
I 4 = MI of thin circular ring about its diameter
If 2(I2 + I3) + I 4 = x. I1, then the value of x will be (2011)
26. A particle of mass m is dropped from a height h above the ground. At the same time, another particle of the same mass is thrown vertically upwards from the ground with a speed of 2. gh If they collide head-on completely inelastically, the time taken for the combined mass to reach the ground, in units of , h g is (2005)
3
JEE Advanced
(3) θ = 60° (4) = 1 23 d h
28. One end of a horizontal uniform beam of weight W and length L is hinged on a vertical wall at point O and its other end is supported by a light inextensible rope. The other end of the rope is fixed at point Q, at a height L above the hinge at point O. A block of weight a W is attached at the point P of the beam, as shown in the figure (not to scale). The rope can sustain a maximum tension of 22 W. Which of the following statements is(are) correct? (2023)
(1) The vertical component of reaction force at O does not depend on a.
(2) The horizontal component of reaction force at O is equal to W for a = 0.5.
(3) The tension in the rope is 2 W for a = 0.5.
(4) The rope breaks if a > 1.5.
29. A thin circular coin of mass 5 g and radius 4/3 cm is initially in a horizontal x-y plane. The coin is tossed vertically up (+z direction)
27. A slide with a frictionless curved surface, which becomes horizontal at its lower end, is fixed on the terrace of a building of height 3h from the ground, as shown in the figure. A spherical ball of mass m is released on the slide from rest at a height h from the top of the terrace. The ball leaves the slide with a velocity = 00 ˆ uux and falls on the ground at a distance d from the building, making an angle θ with the horizontal. It bounces off with a velocity v and reaches a maximum height h1. The acceleration due to gravity is g and the coefficient of restitution of the ground is 1 3 Which of the following statements is(are) correct? (2023) (1) = 0 ˆ 2 ughx (2) ()=− 2 ˆˆ vghxz
by applying an impulse of π × 2 10N-s 2 at
a distance 2/3 cm from its centre. The coin spins about its diameter and moves along the +z direction. By the time the coin reaches back to its initial position, it completes n rotations. The value of n is __________. [Given: The acceleration due to gravity g = 10 ms−2] (2022)
Passage-I:
A pendulum consists of a bob of mass m = 0.1 kg and a massless inextensible string of length L= 1.0 m. It is suspended from a fixed point at height H = 0.9 m above a frictionless horizontal floor. Initially, the bob of the pendulum is lying on the floor at rest vertically below the point of suspension. A horizontal impulse P = 0.2 kg-m/s is imparted to the bob at some instant. After the bob slides for some distance,
CHAPTER TEST – JEE MAIN
Section - A
1. From a uniform circular disc of radius a , an isosceles right angled triangle with the hypotenuse as the diameter of disc is removed. The distance of the centre of mass of the remaining portion from the centre of the circle is (1) 3(π–1)a (2) () π− 1 6 a (3) ()31 a π− (4) ()31 a π+
2. The centre of mass of the letter ‘F’, which is cut from a uniform metal sheet from point
the string becomes taut and the bob lifts off the floor. The magnitude of the angular momentum of the pendulum about the point of suspension just before the bob lifts off is J kg–m2/s . The kinetic energy of the pendulum just after the lift-off is K joules. (2021)
30. The value of J is
31. The value of K is
32. A solid sphere of mass 1 kg and radius 1 m rolls without slipping on a fixed inclined plane with an angle of inclination θ = 30° from the horizontal. Two forces of magnitude 1 N each, parallel to the incline, act on the sphere, both at a distance r = 0.5 m from the centre of the sphere, as shown in the figure. The acceleration of the sphere down the plane is ____ms–2. (Take g = 10 ms–2) (2021)
(1) 1533 , 77 (2) 1523 , 77 (3) 2233 , 77 (4) 3322 , 77
3. The coordinates of centre of mass of particles of mass 10, 20, and 30 g are (1, 1, 1) cm. The position coordinates of mass 40 g, which when added to the system, the position of combined centre of mass be at (0, 0, 0) are, (1) (3/2, 3/2, 3/2) (2) (–3/2, –3/2, –3/2) (3) (3/4, 3/4, 3/4) (4) (–3/4, –3/4, –3/4)
4. A projectile of mass 3m explodes at highest point of its path. It breaks into three equal parts. One part retraces its path, and the second one comes to rest. The range of the projectile was 100 m, if no explosion would have taken place. The distance of the third part from the point of projection, when it finally lands on the ground, is (1) 100 m (2) 150 m (3) 250 m (4) 300 m
5. The object at rest suddenly explodes into three parts with the mass ratio 2 : 1 : 1. The parts of equal masses move at right angles to each other with equal speed v. The speed of the third part after explosion will be
(1) v (2) 2v
(3) 2v (4) /2 v
6. Two billiard balls of same size and mass are in contact on a billiard table. A third ball of the same size and mass strikes symmetrically and then comes to rest right after the impact. The coefficient of restitution between the balls is
(1) 1/2 (2) 3/2 (3) 2/3 (4) 9/4
7. Two particles of masses m 1 and m 2 in projectile motion have velocities 12 , v andv respectively, at time t = 0. They collide at time. Their velocities become 12 v andv at time 2t0, while still moving in air. The value of ()() '' 11221122 || mvmvmvmv +−+
is
(1) zero (2) ()120 mmgt +
(3) ()120 2 mmgt + (4) ()120 1 2 mmgt +
8. A body of mass m, moving with velocity collides head-on elastically with another body of mass 2m, which is initially at rest. The ratio of KE of the colliding body before and after the collision will be
(1) 1 : 1 (2) 2 : 1
(3) 4 : 1 (4) 9 : 1
9. A particle P moving with speed v undergoes a head-on elastic collision with another particle Q of identical mass but at rest. After the collision, (1) both P and Q move forward with speed v/2
(2) both P and Q move forward with speed 2 v
(3) P comes to rest and Q moves forward with speed v (4) P and Q move in opposite directions with speed 2 v
10. A neutron travelling with a velocity v and kinetic energy E collide perfectly elastically head on with the nucleus of an atom of mass number A at rest. The fraction of total energy retained by neutron is (1) 2 1 1 A A
(2) 2 1 1 A A +
(3) 2 1 A A
(4) 2 1 A A +
11. Mass per unit area of a circular disc of radius a depends on the distance r from its centre as σ(r) = A + Br2. The moment of inertia of the disc about the axis perpendicular to the plane and passing through its centre is
(1)
4 2 45 ABa a (2)
2 4 2 46 ABa a (3)
4 2 45 AaB a (4)
2 4 2 46 AaB a
12. From a uniform circular disc of radius R and mass 9 M , a small disc of radius R /3 is removed, as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is R 2R 3
(1) 4MR2 (2) 10MR2
(3) 2 40 9 MR (4) 2 37 9 MR
13. A uniform cylinder of mass M and radius R is to be pulled over a step of height a(a < R) by applying a force F at its centre O, perpendicular to the plane through the axes of the cylinder on the edge of the step (see figure). The minimum value of F required is
(1) 2 1 Ra MgR
(2) 2 1 R MgRa
(3) a MgR (4) 2 2 1 a MgR
14. A wheel having moment of inertia 2 kgm2 is rotating about its geometrical axis at the rate of 60 rpm about its axis. The average torque which can stop the wheel’s rotation in one minute would be
(1) 18 Nm π (2) 2 15 Nm π
(3) 12 Nm π (4) 15 Nm π
15. If a particle moving horizontally hits a stationary ring, then about point of collision,
(1) angular momentum of the ring is conserved
(2) angular momentum of the particle is conserved
(3) angular momentum of particle-ring system is conserved
(4) all of these
16. Two rotating bodies A and B of masses m, and 2m, with moments of inertia IA and IB ( I B > I A ), have equal kinetic energy of rotation. If L A and L B be their angular momenta respectively, then
(1) 2 B A L L = (2) LA = 2LB
(3) LB > LA (4) LA > LB
17. A cubical block of side 30 cm is moving with velocity 2 ms–1 on a smooth horizontal surface. The surface has a bump at a point O, as shown in figure. The angular velocity (in rad/s) of the block immediately after it hits the bump is a
(1) 13.3 (2) 5.0
(3) 9.4 (4) 6.7
18. The moment of inertia of a body about a given axis is 1.2 kgm2. Initially, the body is at rest. In order to produce a rotational kinetic energy of 1500 J, an angular acceleration of 25 rad/s2 must be applied about that axis for a duration of
(1) 4 s (2) 2 s
(3) 8 s (4) 10 s
19. As shown in the figure, a bob of mass m is tied by a massless string whose other end portion is wound on a flywheel (disc) of radius r and mass m. When released from rest, the bob starts falling vertically. When it has covered a distance of h, the speed of the bob will be
20. The angular position of a particle on the rim of a rotating wheel is given by θ = 4t – 3t2 + t3, where θ is in radians and t is in seconds.
(a) What is the angular velocity at t = 2 s?
(b) What is the average angular acceleration for the time interval that begins at t = 2 s and ends at t = 4 s?
(1) 4 rad/s, 12 rad/s2
(2) 6 rad/s, 14 rad/s2
(3) 4 rad/s, 6 rad/s2
(4) 10 rad/s, 12 rad/s2
Section - B
21. The distance between the centres of carbon and oxygen atoms in the carbon monoxide gas molecule is 1.13 × 10–10 m. The distance of centre of mass of the molecule relative to carbon atom is______ × 10 –12 m.
22. Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 ms–1 to the heavier block in the direction of the lighter block. The velocity of the centre of mass is m/s–1.
23. A rubber ball is dropped from a height of 5 m on a planet, where it rises back to a height of 1.8 m. The ball loses its velocity on bouncing by a factor of x/5. Then, x is
(1) 3 2 gh
(2) 3 4 gh
(3) 4 3 gh
(4) 2 3 gh
24. A body of mass 2 kg, moving with a speed of 4 m/s, makes an elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its initial speed. The speed of the two-body centre of mass x/10 m/s. Then, the value of x is .
25. A shaft rotating at 3000 rpm is transmitting a power of 3.14 kW. The magnitude of torque is . Nm. (Take π=3.14)
CHAPTER TEST – JEE ADVANCED
2022 P2 Model
Section-A
[Integer Value Questions]
1. Two particles A and B, initially at rest, move towards each other by mutual force of attraction. At the instant when the speed of A is n and speed of B is 3n, the speed of the centre of mass of the system is .
2. A linear rod of length l has linear mass density 0 1,
X
where λ0 is a constant and X is distance from left end of the rod. The distance of centre of mass from its left end is (5/N)l. The value of N is___.
3. Particle A experiences a perfectly elastic collision with a stationary particle B. The particles fly apart symmetrically relative to the initial direction of motion of particle A with angle of divergence θ(0 < θ < 90°). Ratio between the masses of the two bodies B A m m is 1 , 12cos+θ n where n is
4. Three objects A, B, and C are kept in a straight line on a frictionless horizontal surface. These have masses m, 2m, and m, respectively. The object A moves towards B with a speed of 9 m/s and makes an elastic collision with it. Thereafter, B makes completely inelastic collision with C. All motions occur on the same straight line. Find the final speed (in m/s) of object C.
mm B 2m C
5. Consider a uniform cubical box of side a on a rough floor that is to be moved by applying minimum possible force F at a point distance b above its centre of mass (see in figure). If the coefficient of friction is μ = 0.4, the maximum possible value of × 100 b a for box not to topple before moving is
6. A thin rod of mass 0.9 kg and length 1 m is suspended, at rest, from one end so that it can freely oscillate in the vertical plane. A particle of mass 0.1 kg moving in a straight line with velocity 80 m/s hits the rod at its bottom-most point and sticks to it (see figure). The angular speed (in rad/s) of the rod immediately after the collision will be 10x. The value of x is
7. A thin disc rotates about an axis passing through its centre and perpendicular to its plane with a constant angular velocity ω I is the moment of inertia of that disc and L is its angular momentum about the given axis. Then, rotational kinetic energy of the disc E is proportional to I–n. The value of n is
8. A spool, initially at rest, is kept on a frictionless fixed inclined plane, making an angle θ = 37° with the horizontal. The mass of the spool is 2 kg and it is pulled by a string, as shown, with a force of T = 10 N. The string connecting the spool and the pulley is initially horizontal. Find the initial acceleration of the spool on the incline. Express your answer in m/s 2 .
kinetic energy, the magnitude of the relative velocity between the two particles after collision, is
(1) 0 4 v
(2) 0 2v
(3) 0 2 v
(4) 0 2 v
11. In the figure shown, if the coefficient of restitution between A and B is 1 2 e = , then
Section – B
[Multiple Option Correct MCQs]
9. A ball of mass is thrown at an angle of 45° to the horizontal from the top of a 65 m high tower AB, as shown in fig, at t = 0. Another identical ball is thrown with velocity 20 m/s horizontally towards AB from the top of a 30 m high tower CD, after 1 s from the projection of the first ball. Both the balls move in the same vertical plane. If they collide in mid-air,
(1) the two balls will collide at time t = 5 s
(2) the two balls witll collide at time t = 2 s
(3) distance AC is 40 m
(4) velocity of combined ball just after they strike together is –1 ˆ 520s ˆ m ij
10. In a collinear collision, a particle with an initial speed v0 strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original
(1) velocity of B after collision is v/2
(2) impulse between two during collision is 3 4 mv
(3) loss of kinetic energy during the collision is 2 3 16 mv
(4) loss of kinetic energy during the collision is 2 1 4 mv
12. The potential energy of a particle of mass m at a distance r from a fixed point O is given by () 2 , 2 = Vrkr where k is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius R about the point O. If v is the speed of the particle and L is the magnitude of its angular momentum about O, which of the following statements is (are) true?
(1) 2 k vR m =
(2) k vR m =
(3) 2 LmkR = (4) 2 2 mk LR =
13. A particle of mass m is moving in a horizontal circle inside a smooth inverted fixed vertical cone above height h from apex. Angle of cone is θ. Then, θ h
(1) normal force on particle by surface of cone is mg cos θ
(2) normal force on particle by surface of cone is mg cosecθ
(3) time period of revolution of particle increases if θ increases, keeping h constant
(4) time period of revolution increases if h increases, keeping θ fixed
14. Three solid cylinders A, B, and C, each of mass m and radius R, are allowed to roll down on three inclined planes A', B', and C' respectively (each of inclination θ). For A' and A, coefficient of friction is zero. For B' and B, coefficient of friction is greater than zero but less than tan 3 θ , and for C' and C, coefficient of friction is more than θ tan . 3 On
reaching the bottom of the inclined plane, (1) C has maximum angular speed
(2) B has minimum total kinetic energy
(3) A takes minimum time to reach the bottom
(4) B has minimum translational kinetic energy
Section – C
[Single Option Correct MCQs]
15. A non-uniform rod of length L is lying along the x-axis with one end at origin O, as shown in figure. The linear mass density (i.e., mass per unit length) λ varies with x as λ = a+ bx , where a and b are constants. Find the distance of the centre of mass from origin O.
(1) L(3a+ 2bL)/4(2a+bL)
(2) L(2a + 3bL)/4(a + 2bL)
(3) L(3a + 2bL)/3(2a + bL)
(4) L(2a + 3bL)/3(a + 2bL)
16. A smooth ball is dropped from height h on a smooth fixed incline, as shown in figure. After collision, the velocity of the ball is directed horizontally. Then, the coefficient of restitution is
h θ
(1) cot2 θ
(2) sin2 θ
(3) tan2 θ
(4) cos2 θ
17. Point masses m 1and m 2 are placed at the opposite ends of a rigid rod of length L and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point P on this rod, through which the axis should pass so that the work
CHAPTER 9: System of Particles and Rotational Motion
required to set the rod rotating with angular velocity ω 0 is minimum, is given by
18. A rod PQ of mass M and length L is hinged at end P. The rod is kept horizontal by a massless string tied to point Q, as shown in figure. When string is cut, the initial angular acceleration of the rod is
(1) 2 1 m xL m = (2) 2 12 mL x mm = + (3) 1 12 mL x mm = + (4) 1 2 m xL m =
ANSWER KEY
JEE Main
7 (61) 14
(11) 1 (12) 3 (13) 3 (14) 4 (15) 4 (16) 3 (17) 3 (18) 1 (19) 20 (20) 30 (21) 4 (22) 3 (23) 4 (24) 4 (25) 1 (26) 3
Level-III
Theory Based Questions
JEE Advanced
Brain
(11) 1 (12) 14 (13) 4.25 (14) 120 (15) 34.82 (16) 4.12 (17) 1.5 (18) 1 (19) 1
| CHAPTER 9: System of Particles and Rotational Motion
Chapter Test - JEE Main
Chapter Test - JEE Advanced
GRAVITATION CHAPTER
Chapter Outline
10.1 Basic Forces in Nature
10.2 Kepler’s Laws
10.3 Universal Law of Gravitation
10.4 The Gravitational Constant
10.5 Acceleration Due to Gravity of the Earth (g)
10.6 Variation of (g) with Depth and Height and Due to Rotation of Earth
10.7 Gravitational Field
10.8 Gravitational Potential and Potential Energy
10.9 Escape Speed
10.10 Earth Satellite
10.11 Energy of an Orbiting Satellite
10.12 Geostationary and Polar Satellites
10.13 Weightlessness
From the earliest time, ‘gravity ’ means the tendency of most bodies to fall to earth. In contrast, things that leaped upwards, like flames of fire, were said to have ‘ levity ’. Aristotle was the first writer to attempt a quantitative description of falling motion. He wrote that a freely falling object attained a constant speed shortly after being released, and heavier things fall faster in proportion to their mass. Aristotle by most likely observed the slower motion of things falling through water, where buoyancy and fluid resistance dominate, and assumed that to be a slowed-
down version of falling through air, which it is not.
Galileo was the first to get it right. He realised that a falling body picked up speed at a constant rate. In other words, it had constant acceleration. He also made the crucial observation that if air resistance and buoyancy can be neglected, all bodies fall with the same acceleration; bodies of different weights dropped together reached the ground at the same time.
Gravitation is one of the fundamental forces of nature. It is a common experience that when a stone is thrown up, it returns to the earth instead of just going up and up. Newton first revealed that both terrestrial bodies, like stone, and celestial bodies, like moon, fall with certain accelerations towards the centre of the earth because of the earth’s gravitational force. After some time, the stone hits the earth but the moon does not hit the earth, why?
Aryabhatta, a famous Indian astronomer, studied the motion of the moon and the planets, and suggested that the earth is a solid sphere and it spins around itself. According to Rig veda, paths of the planets in the solar sytem were suggested to be elliptical. The earliest recorded model for planetary motions proposed by the Greek astronomer Ptolemy about 2000 years ago, was a ‘ geocentrore model, ’ in which all celestial bodies, stars, the sun, and the planets, revolved around the earth in circular orbits.
In the 16th century, the astronomer Nicolas Copernicus (1473-1543) suggested that all the planets move around the sun in circular orbits and he proposed the heliocentric theory.
It was already mentioned by Aryabhatta (5 th centrury A.D.). His theory was discredited by the Church, but notable amongst its supporters was Galileo who had to face prosecution from the state for his beliefs. Tycho Brahe (1546-1601), a Danish astronomer, made very careful and accurate measurements of the motion of the planets. His compiled data was analysed later by his assistant Kepler (1571-1640).
Kepler formulated three laws of planetary motion, which accurately described the motion of planets around the sun. These laws were known to Newton and enabled him to make a great scientific leap in proposing his universal law of gravitation. This law of universal gravitation helped to understand the universe, in general, and the solar system, in particular. Albert Einstein proposed that gravitation is a consequence of the curvature in space-time continuum produced by the presence of matter. In this chapter, we discuss gravitation at the terrestrial level by considering Newton’s law of gravitation. A very brief idea of Kepler’s planetary motion, gravitational potential energy, is discussed. We conclude this chapter with explanations about escape speed, polar and geostationary satellites, and weightlessness.
10.1 BASIC FORCES IN NATURE
We know that there exits a gravitational field around mass. Similarly, an electric field exists around a charge and magnetic field exists around a magnetic pole. It is in these fields that there develops an interaction (force) between two bodies. Thus, we have the gravitational force, electromagnetic force, etc., in nature. These forces come into play due to the exchange of field particles, such as gravitons, photons, leptons, etc. Various types of interactions in nature can be classified into the following four categories: gravitational force, electromagnetic force, strong nuclear force, and weak nuclear force
10.1.1 Gravitational Force (Fg)
Gravitational force has the following important properties:
■ This force is always attractive.
■ It is a conservative force (i.e., work done is independent of path follwed).
■ This force forms action–reaction pair (i.e., F12 = – F21).
■ This force is independent of medium.
■ This is the weakest of all the forces.
■ It has the longest range (i.e., it can act over infinite distances).
■ This force operates by the exchange of gravitons (in gravitational field).
■ This force can provide radial acceleration.
■ This force can provide tangential acceleration.
■ The force between two masses is independent of the presence of other masses.
10.1.2 Electromagnetic Force (Fe)
The electromagnetic force is dominant on the atomic level and it is responsible for holding the atoms together that make up the molecules. It also binds the electrons of an atom to the nucleus. Apart from determining the properties of atoms and molecules, electromagnetic forces are also responsible for other kinds of forces, such as friction, air resistance, elasticity, tension, etc.
It may be attractive or repulsive, depending upon the type of charge of the interacting particles. It is a long range force. It is communicated through photons.
10.1.3 Strong Nuclear Force
Strong neclear force is the strongest force in nature and it exists between different nucleons inside a nucleus. This force is responsible for keeping the protons and neutrons bound in the nucleus. It has the following characteristics:
■ It is basically attractive in nature. However, when the distance between two nucleons is of the order of 0.4 fm, then it becomes repulsive.
■ It is charge independent.
■ It is spin dependent.
■ It is a very short range force.
■ It is about 100 times stronger than electromagentic force.
■ It is communicated through π -mesons.
10.1.4 Weak Nuclear Force
Weak nuclear force is responsible for a type of radioactive decay known as beta decay and other similar decay processes involving fundamental particles. The forces acting between leptons (*) are called weak nuclear forces.
Thus force is responsible for beta decay in radio-activity. It is a short range force. It is communicated through weak bosons (***).
Table 1: Relative strengths of basic forces between protons
Basic Force Relative Range strength
Gravitational 1 Long range, infinite
Weak nuclear range <<1 fm 1031 Extremely short
Electromagnetic 1036 Long range, infinite
Strong force 1038 Short range, 1 fm
The more recent advances in particle physics suggest that the electromagnetic force, the weak nuclear force, and possibly strong nuclear force may all be the same type of interaction.
Key Insights:
■ Range of gravitational force > range of electromagnetic force > range of nuclear force
■ Strength of nuclear force > strength of electro-magnetic force > strength of weak
nuclear forces > strength of gravitational force
(*) Leptons: Electrons, positrons – mesons and neutrions
(**) Hardons: Mesons and baryons
(***) Bosons: Fundamental particles having integral spin
■ Neutron changes into a proton emitting electron and antineutrino. The basic force involved is weak nuclear force.
10.2 KEPLER’S LAWS
A heavenly body revolving around a planet in an orbit is called a natural satellite. For example, moon revolves around the planet earth. So, moon is the satellite of earth. Their motions can be studied with the help of Kepler’s laws, as stated below.
10.2.1 Law of Orbits
“All planets revolve around the sun in elliptical orbits, with the sun at one of the focii of the ellipse.”
As shown in the figure, the sun may be at F1 or F2. Here, a and b denote lengths of semi-major and semi-minor axes. The closest point A is called the perihelion and the farthest point B is called aphelion.
For circular orbits, F1 and F2 merge into one and the semi-major axes become the radius of the circle.
Key Insights:
■ The distance of closest approach with the sun at F1 is AS. This distance is called ‘Perigee’.
■ The greatest distance (BS) of the planet from the sun is called ‘apogee’.
■ Perigee (AS) = AO–OS = a(1 – e), where e = eccentricity = SO OA
■ Apogee (BS) = OB + OS = a(1 + e)
10.2.2 Law of Areas
“The line joining the planet to the sun sweeps out equal areas in equal intervels of time.” (OR)
“Areal velocity of radius vector is constant.”
This law comes from the observations that planets appear to move slower when they are farther from the sun than when they are nearer. As shown in fig, a planet takes the same time to travel from A to B as from C to D (shaded areas are equal). Naturally, the planet has to move faster from C to D. This law is identical with the law of conservation of angular momentum, which is valid for any central force.
Let the sun be at the origin and the position and momentum of the planet be denoted by r and p . Then, the area swept by the position vector of the planet of mass m in time interval dt is ‘dA’, given by 1 .() 2 dArrd=θ Hence, areal
and as angular momentum ( L) = mvr, So, 2 dAL dtm = (or) 2 dAL dtm =
rVr dt ==ω
The gravitational force is a central force. L is the angular momentum equal to () rp × .
The gravitational force, which is directed along r due to the sun, does not create any torque on the planet. In the absence of torque, angular momentum is conserved. Hence, dA dt is constant, and hence, the law of areas follows.
Key Insights:
■ Let V P and V A be the magnitudes of the velocity of the planet at the perihelion and aphelion. S denotes the position of the sun. SP = r A and SA = r B. If m is the mass of the planet, from conservation of angular momentum, S O P
Angular momentum at P = Angular momentum at A
LP = LA; mVP rP = mVArA or VPrP = VArA It means that VP > VA as rA > rP
Hence, out of aphelion and perihelion, the speed of planet is more at perihelion because the planet has to cover greater linear distance to keep the aerial velocity constant.
■ Areal velocity versus time graph:
(t)
■ The direction of areal velocity of the earth around the sun is normal to the plane containing the earth and the sun, as areal velocity
Now, speed of the planet is V = 2 R T
Substituting the value in above equation,
■ The rate of change of areal velocity is a measure of normal acceleration, which is zero.
■ The path of the orbit depends on total energy and value of e.
■ If e > 1 and total energy (KE + PE) > 0, the path of the satellite is hyperbolic and it escapes from its orbit.
■ If e < 1 and total energy is negative, it moves in an elliptical path.
■ If e = 0 and total energy is negative, it moves in circular path.
■ If e = 0 and total energy is zero, it will take parabolic path.
■ The path of the projectiles thrown to lower heights is parabolic and thrown to greater heights is elliptical.
■ Kepler’s laws may be applied to natural and artificial satellites as well.
10.2.3 Law of Periods
“The square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of the ellipse traced out by the planet or the average distance between the sun and the planet.”
If T is the time period of revolution of the planet and 2a is the length of major axis of the ellipse traced out by the planet, then 23Ta ∝
i.e., 23 11 22 Ta Ta
Key Insights:
■ As the planet revolves around the sun, Centripetal force = Gravitational force
Since 2 (4/) GM π is constant, therefore, 23TR ∝ (or) 2 3 T R = constant.
■ If gravitational force is proportional to r–n , then time period of a planet
Tr
Table 2: The semi-major axes and time periods of the planets revolving around the sun
10.2.4 Deduction of Newton’s Law of Gravitation from Kepler’s Laws
Let us consider a planet of mass ‘m’ revolving around the sun of mass ‘ M ’ in an orbit of radius ‘r’, with a constant angular velocity ω . Let T be the time period of revolution of the planet around the sun.
The centripetal force acting on the planet
Fmr=ω
According to Kepler’s third law, 23Tr ∝ (or) T2 = Kr3 (2) → where K is proportionality constant. Therefore, F = 22 32 44 .(3)mrm K krr ππ =→
This centripetal force is provided by the gravitational attraction exerted by the sun on the planet. According to Newton, the gravitational attraction between the sun and the planet is mutual. If force F is directly proportional to the mass of the planet ( m ), it should also be directly proportional to the mass of the sun (M). Hence the factor, 4422 MorGM KK ππ ∝=
Substituting this value in Eq. (3), we get 2 GMm F r = , which is Newton’s law of gravitation.
Key Insights:
■ Both earth and moon are subjected to the gravitational force of the sun. As observed from the sun, the orbit of the moon will not be strictly elliptical because the total gravitational force on it is not central.
■ In our solar system, the inter-planetary region has chunks of matter (much smaller size compared to planets) called asteroids. They move in orbits like planets and obey Kepler’s laws.
■ If the sun and the planets carried huge amounts of opposite charges, then the second law will not change and the first law will still be valid.
■ According to Newton’s third law of motion, if the earth pulls the stone, the stone must also pull the earth with the same force but due to the large mass of the earth, its acceleration towards the stone is quite negligible and we just cannot notice it. Hence, only the stone appears to fall towards the earth
1. A Saturn year is 29.5 times the Earth year. How far is saturn from the sun if the Earth is 1.50 × 108 km away from the sun?
Sol: Given that TS = 29.5 T e and R e = 1.5 × 108 km
According to Kepler’s third law, we can write 23TR ∝
R s = 1.43 × 109 km
2. A planet revolves around the sun in an elliptical orbit of eccentricity ‘e’. If ‘T’ is the time period of the planet, then show that the time spent by the planet between the end of the minor axis and close to the sun is
Sol: A B a S constant; dA dt = 1 () (Area)SAB 42 (Area)ellipse
Try yourself:
1. Let the speed of the planet at the perihelion P be VP and the sun–planet distance SP be r p . Relate ( r p , V P ) to the corresponding quantities at the aphelion (rA, VA). Will the planet take equal time to tranverse BAC and CPB?
Ans: The planet will take a longer time to traverse BAC than CPB.
TEST YOURSELF
1. The period of revolution of a surface satellite around a planet of radius R is T. The period of another satellite around the same planet in an orbit of radius 3 R is (1) T (2) 3T (3) 9T (4) 33 T
2. Suppose that the earth revolves around the sun in a circular orbit of radius 15 × 10 10 m with a time period of 1 year. The time taken by another planet, which is at a distance of 540 × 1010 m, to revolve around the sun in a circular orbit once will be (1) 216 years (2) 144 years (3) 72 years (4) 36 years
3. If the area swept by the line joining the sun and the earth from February 1 to February 7 is ‘A’, then the area swept by the radius vector from February 8 to February 28 is (1) A (2) 2A (3) 3A (4) 4A Answer Key (1) 4 (2) 1 (3) 3
10.3 UNIVERSAL LAW OF GRAVITATION
By studying the motions of the moon and planets, Newton discovered a fundamental law of gravitation that describes the gravitational attraction between any two bodies.
This law states that “every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and is inversely proportional to the square of distance between them and this force acts along the line joing the two particles.”
Consider two particles A and B of masses m1 and m2. Let r be the distance between their centres and F be the force of attraction between them.
According to universal law of gravitation,
Here, G is constant, called the universal gravitational constant.
The value of G = 6.67 × 10–11 N-m2/kg2
Unit of G: SI unit: Nm2/kg2, CGS unit: dynes cm2/g2
Dimensional formula of G: 2 12 Fr G mm =
Definition of G:
Let m1 = m2 = 1 and r = 1
From Eq. (1), F = 2 11 1 G G ×× =
Thus, universal gravitational constant is equal to the force of attraction acting between two bodies, each of unit mass, whose centres are placed unit distance apart.
Gravitational constant is a scalar quantity. Its value is same throughout the universe and it is independent of the nature of the bodies as well as the nature of the medium between the bodies.
Key Insights:
■ Newton’s law of gravitation holds good for objects lying at very large distances and also very short distances. It falls when the distance between the objects is less than 10–9 m, (i.e, of the order of intermolecular distances).
3. The gravitational force between two identical objects at a seperation of 1 m is 0.06667 mgwt. Find the mass of each object.
(G = 6.67 × 19–11 Nm2/kg2 and g = 10 m/s2)
Sol: Given that r = 1 m, F = 0.0667 × 10–6 × 10 N
Let m1 = m2 = m
The gravitational force F = 12 2 Gmm r
0.0667×10–6×10 = 6.67×10–11× 2 12 m
⇒ m = 100 kg
Try yourself:
2. Magnitude of gravitational force between two particles kept at a separation is F . If mass of one particle is quadrupled and separation is halved, find the new gravitational force between the particles.
Ans: 16F
Vector form of universal law of gravitation:
Let 12 ˆ r = unit vector from A to B
21 ˆ r = unit vector from B to A
12F = gravitational force exerted on body A by body B
21F = gravitational force exerted on body B by body A
According to Newton’s law of gravitation, 12 12 21 2 ˆ (1)
The negative sign shows that the gravitational force is attractive in nature.
From Eqs. (2) and (3), we have 2112FF =−
If 1r and 2r are the position vectors of point masses m1 and m2 at points A and B, as shown in fig., then 1221 rrr =−
. Now,
Key Insights:
■ If we have a collection of point masses, the gravitational force on any one of them is the vector sum of the gravitational forces exerted by the other point masses, as shown in fig. (principle of superposition).
Gravitational force on point mass m 1 is the vector sum of the gravitational forces exerted by m2, m3 and m4
4. Two point objects of masses 4 kg and 3 kg are placed along x and y axes, respectively, at a distance of 1 m from the origin. An object of mass 1 kg is kept at the origin of the coordinate system. Calculate the resultant gravitational force attraction on the object at the origin.
Sol: Force due to 4 kg mass on 1 kg mass at the origin is
1 2 41 4 1 G FG ×× == Force due to, 3 kg mass on 1 kg mass at the origin is
2 2 31 3 1 G FG ×× == The resultant force is found by using parallelogram law of vectors 22 1212 22 0 2cos (4)(3)2(4)(3)cos905
Try yourself:
3. Gravitational force between two point masses m and M , seperated by a distance r , is F . Now, if a point mass 3m is placed next to m, what will be the (a) force on M due to m (b) total force on M? Ans: (a) 2 GMm F r = (b) 4F
10.3.1 Characteristics of Gravitational Force
■ It is always attractive.
■ It is independent of the medium between the particles.
■ F – r 2 graph is a rectangular hyperbola, as shown. F r2 0 distance2 Force
■ It is an action–reaction pair, i.e, the force with which one body (say earth) attracts the second body (say moon) is equal to the force with which the moon attracts the earth. (However, the accelerations will not be equal). This force is mutual and internal.
■ It is a central force. This force exhibits no angular dependence. Its magnitude depends on r We say that gravitational force posseses spherical symmetry.
■ Gravitational force between two bodies does not depend upon the properties of the medium.
■ It is a conservative force, i.e, work done by it is path independent.
■ In applying Newton’s law of gravitation to the case of extended bodies of finite size, instead of point masses, care should be taken in assuming that the whole mass of a body can be regarded as concentrated at its centre of mass.
■ The law of gravitation holds true for point masses.
Key Insights:
■ The law is not directly applicable for the gravitational force between extended object (like the earth) and a point mass. Each point mass in the extended object will exert force on the given point mass. So, all these forces due to various point masses of extended object will be added vectorially to get the net force on the given point object. For spherically symmetrical object, its entire mass is assumed to be concentrated at its centre.
5. Three particles of identical masses ‘m’ are kept at the vertices of an equilateral triangle with each side having length ‘ a’. Find the gravitational force of attraction on any one of the particles.
Sol: Force due to 4 kg mass on 1 kg mass at the origin is
Let FA and FB be the forces due to the masses at A and B on the mass at C.
Here, FA = FB
Hence, the resultant force on the mass at C 2cos 2 AFF θ =
Try yourself:
4. Three equal masses of m kg each are fixed at the vertices of an equilateral triangle ABC.
a) What is the force acting on a mass 2m placed at the centroid G of the triangle?
b) What is the force if the mass at the vertex A is doubled?
Take AG = BG = 1 m
Ans: (a) 0 (b) 2 ˆ 2Gmj
10.3.2 Shell Theorem
The force of attraction between a hollow spherical shell/sphere of uniform density and a point mass situated outside is just as if the entire mass of the shell/sphere were concentrated at the centre.
1. If a point mass lies outside the shell, the gravitational force caused by the various regions of the shell have components along the line joining the point mass to the centre, as well as along an attraction perpendicular to this line. When summing the gravitational forces over all the regions of the shell, the perpendicular components cancel out, leaving only a resultant force along the line joining the point to the centre. P m M R o r 2 GMm F r = (If r >
where R is the radius of the spherical shell.
2. The force of attraction due to a spherical shell of uniform surface density on a point mass situated inside it is zero.
If a point mass lies inside the shell, various regions of the spherical shell attract the point mass inside it in various directions. These forces cancel each other completely.
Hence, F = 0 (If r < R )
3. The nature of F vs r graph:
4. If centres of two spheres are separated by a distance
6. Two metal spheres of same material, each of radius r, are in contact with each other. The gravitational force of attraction between the spheres is proportional to
Try yourself:
5. Two thin spherical shells, each of radius R, are kept in contact with each other. If F is the magnitude of gravitational force acting between them, then F ∝ Rn. Find ‘n’.
Ans: 2
10.3.3 Central Force
fixed point O
A central force is that force which is always directed towards or away from a fixed point, i.e; along the position vector of the point of application of the force, with respect to the origin or fixed point. The magnitude of central force depends on r
■ For the motion of a body under a central force, the angular momentum of the body is always conserved because torque acting on the body by the central force is zero.
0constant dL L dt τ==∴=
■ The motion of a particle under the central force is always confined in a plane.
■ The radius vector of the particle in a circular or elleptical orbit under the central force has constant areal velocity.
■ Gravitational force and electrostatic force are the examples of central force.
Expression for Areal Velocity
Let us consider the motion of a planet under the gravitational force of the sun. The gravitational force is directed towards the sun. This force also satisfies the requirement of central force.
12 2 Gmm F r = where m 1 and m 2 are the masses of the planet and the sun, respectively. Assume that, the sun is at ‘O’ and the planet moves along the trajectory Tr. At any instant of time, the planet is at P and the central force is directed along OP
. In time ∆t, the particle moves from P to P'. \ PP'= ∆ s = v ∆ t → (1)
Here, the tangent PQ gives the direction of velocity at P. The area swept by the planet in ∆t is given by 1 ''sin 2 PQPPPr=××α (from the Fig) 1 sin 2 Asr ∆=∆×α 1 sin 2 Avtr ∆=∆α (from (1))
7. Three identical bodies, each of mass m, are located at the vertices of an equilateral triangle with side a. At what speed must they move if they all revolve under the influence of one another’s gravitation in a circular orbit circumscribing the triangle, while still preserving the equilateral triangle?
Sol: The force of attraction on body C due to bodies at A and B are
2 1 2 Gm F a = along CA and 2 2 2 Gm F a = along CB
The resultant force on the body at C is 22 012122cos60FFFFF =++
This resultant force is acting along CD.
Here, OC = 3 a
When each body is describing a circular orbit with centre of orbit at O, the force F provides the required centripetal force.
The radius of the circular orbit is OC = /3, a V is the speed of the body in circular orbit. Centripetal force = Resultant gravitational force 22 2 3 /3 mVGmGm orV aaa ==
8. Three uniform spheres, each having mass ‘m’ and radius ‘r’, are kept in such a way that each touches the other. The magnitude of the gravitational force on any sphere due to the other two is
Sol: Let F1 = F2 = F
Here, F = 2 (2)422 GmmGm RR =
∴ The resultant gravitational force
FR = 2F cos /2 θ FR = 2F cos 300 22 22 333 (2)4
Try yourself:
6. Find the point at which the gravitational force acting on any mass is zero due to the earth and the moon system (such a point is called neutral point). The mass of the earth is 81 times the mass of the moon and the distance between the earth and the moon is 3,85,000 km.
m2 = 81M, (mass of the earth)
From moon here, m1 = M (mass of the moon)
Ans: 38,500 km
TEST YOURSELF
1. Two spherical balls, each of mass 1 kg, are placed 1 cm apart. The gravitaional force of attraction between them (in newtons) is (1) 6.67 × 10–9 (2) 667 × 10–7 (3) 6.67 × 10–8 (4) 6.67 × 10–5
2. Two bodies of masses m 1 and m 2 are separated by a certain distance. If 12F is the force on m1 due to m2 and 21F is the force on m2 due to m1, then
(1) F12 = F21 (2) 12F = 21F (3) 12F = – 21F (4) both 1 and 3
3. Two spheres of masses m and M are situated in air and the gravitational force between them is F. The space around the masses is now filled with a liquid of specific gravity 3. The gravitational force will now be (1) F/9 (2) 3F (3) F (4) F/3
4. Two spherical balls of mass 1 kg and 4 kg are separated by a distance of 12 cm. The distance from 1 kg, at which the gravitational force on any mass becomes zero, is (1) 4 cm (2) 12 cm (3) 6 cm (4) 8 cm
5. Two metal spheres, each of radius ‘r ’ and made of same material, are separated by a certain fixed distance d ( d > 2 r ) in a medium. The gravitational force between those spheres is proportional to (1) 2 1 r (2) 4 1 r (3) r4 (4) r6
Answer Key
(1) 2 (2) 4 (3) 3 (4) 1 (5) 4
10.4 THE GRAVITATIONAL CONSTANT (CAVENDISH EXPERIMENT)
The value of the gravitational constant (G) was first determined by English scientist Henry Cavendish in 1798. It was a torsion balance made of a wooden rod suspended from a wire, with a small lead sphere attached to each end. Two big lead spheres were located near the smaller spheres and held in space with a separate suspension system. The big spheres attract the nearby small ones by equal and opposite force.
If d is the separation between the centres of the big sphere and its neighbouring smaller sphere, and M and m are their masses, then the gravitational force between them is
2 (1) Mm FG d =→
If L is the length of the rod, then the torque due to gravitational force τ = FL
2 .(2) GMm L d τ=→
In equilibrium, this is equal to restoring torque, and hence, 2 (3) GMmLk d =θ→
By measuring the value of angle of twist (θ), and substituting the values of M, m, d, L and θ in the above equation, G is calculated and it is found to be 6.67 × 10–11 Nm2/kg2.
Key Insights:
Torsion wire
Diagram of torsion balance
There is no net force on the bar but there is a torque acting on it. This torque is equal to F times the length of the rod, where F is the gravitational force between a big sphere and its neighbouring small sphere. Due to this torque, the suspended wire gets twisted. The bar stops rotating when it reaches an angle where the restoring torque of the wire equals the gravitational torque.
If θ is the angle of twist of the suspended wire, then the restoring torque = k θ , where k is the restoring couple per unit angle of twist.
i) A thin rod of mass M and length L is bent in a semi circle. Gravitational force on a particle with mass ‘m’, placed at centre of curvature, is F= 2 2 GMm L π
ii) A thin rod of mass M and length L is bent into a complete circle. Then, the resultant force on it is zero.
iii) Cavendish is credited with weighing the earth. Cavendish determined experimentally the value of G . Then, he used the equation 2 GM gR = to determine the mass (M) of the earth.
9. In Cavendish’s experiment, let each small mass be 20 g and each large mass be 5 kg. The rod connecting the small masses is 50 cm long, while the small and the large spheres are separated by 10.0 cm. The torsion constant is 4.8 × 10–8 kg m2 s–2 and the resulting angular deflection is 0.4 0 . Calculate the value of universal gravitational constant G from this data.
Sol: Here, m = 20 g = 0.02 kg, M = 5 kg
r = 10 cm = 0.1 m, l = 50 cm = 0.5 m
θ = 0.40 = (0.40)(2 π /3600) = 0.007 rad,
k = 4.8×10–8 kg m2 s–2
Thus, from 2 kr G Mml θ =
On substitution, G = 6.72×10–11 N m2 kg–2
Try yourself:
7. Find dimension of L in the dimensional formula of G.
Ans: 3
10.5 ACCELERATION DUE TO GRAVITY OF THE EARTH ( g
)
“If the force on the body is due to gravity of earth, then acceleration in the body is called acceleration due to gravity, which is denoted by g,” i.e., a = g
Then, F = mg (or) g = F/m .
If m = 1, then g = 1 F = F.
Thus, acceleration due to gravity is defined as the force of gravity acting on a unit mass of the body placed on or near the surface of the earth.
Acceleration due to gravity is the acceleration set up in the body when it falls freely under the effect of gravity alone. The SI unit of acceleration due to gravity is ms –2 or N/kg. It is a vector. Its direction is towards the centre of the earth. At a given place, the value of g is constant. However, it differs from place to place on the surface of the earth. It also varies with altitude, depth, and rotation of the earth. On the surface of the earth, the value of g is taken to be 9.8 ms–2 for all practical purposes. The dimensional formula for g is [M0LT–2].
The value of g is independent of the shape, size, mass etc. of the body but depends upon mass and radius of the earth or planet, due to which there is a gravity pull.
10.5.1 Relation between Universal Gravitational Constant (G) and Acceleration Due to Gravity (g)
Mr P m r RE d
Fig. 10.1 The mass m is located at a depth d below the surface of the earth
Let us consider the earth to be made of a large number of concentric spherical shells. The total mass of the shells is just the mass of the earth.
The gravitational force due to any shell on a point, as situated outside, is just as if the entire mass of the shell is concentrated at the centre of the shell. However, the force of attraction due to any shell on a point mass situated inside it is zero.
Consider a point mass m situated inside the earth, at a distance r from the centre. The point p lies outside the sphere of radius r. For certain shells, whose radius is greater than r, the point p is inside point and these shells exert no gravitational force on the mass m at p. However, the shells with radius less than or equal to r are assumed to be a sphere of radius r , for which the point p lies on the surface. Hence, a gravitational force is exerted on the mass m at p by the earth of radius r
If M r is the mass of the assumed sphere, then, the force on the mass m at p is given by
GmM
If R E or R is the radius of the earth and mass of the earth is ME or M, then density ρ
Similarly, the mass of the sphere of ‘r’ is given by
As 2 GM gR = But 3 4 3 MR
The gravitational force on the point mass is
If the mass ‘m’ is situated on the surface of the earth, then r = RE; then, the gravitational force on the mass m placed on the surface of the earth is 2 E E GmM F R =
The force acting on the body due to gravitational pull of the earth is F = mgE
From (4) and (5), 2 E E E GmM mgR = 22 () E E E GMGM gorgRR ==
This relation is valid for any planet, including the earth.
a) In terms of radius (R) and mean density ( ρ ),
10. If acceleration due to gravity on the surface of earth (Mass = M, Radius = R) is g, find the acceleration due to gravity on the surface of planet of mass M/2 and radius R/4.
b) In terms of mass ( M) and density ( ρ ),
Try yourself:
8. Acceleration due to gravity on the surface of a planet of mass M and density r is g. Then, what will be the acceleration due to gravity on the surface of a planet of mass 27 M and density 8r?
Ans: 12g
TEST YOURSELF
1. A spherical planet far out in space has a mass M 0 and diameter D0. A particle of mass ‘ m ’ falling freely near the surface of this planet will experience an acceleration due to gravity, which is equal to
2. Mass of the moon is 1/81 of the earth but gravitational pull is 1/6 of the earth. It is due to the fact that
(1) radius of the moon is 81/6 of the earth
(2) radius of the earth is 81 6 of the moon
(3) radius of the moon is 81 6 of the earth
(4) moon is the satellite of the earth
3. Weight of a body on the surfaces of two planets is the same. If their densities are d1 and d2, then the ratio of their radius is
10.6 VARIATION OF ( g ) WITH DEPTH AND HEIGHT AND DUE TO ROTATION OF EARTH
Now, let us study the variation of ‘ g ’ with altitude, depth, latitude, and the shape of the earth. We know that g = F/m. The force (F) exerted by the earth on the body of mass ( m) is affected by a number of factors, like height, depth, latitude, and hence, g also depends upon these factors.
10.6.1 Variation of ‘g’ above the Surface of the Earth or with Altitude
Altitude refers to the height measured from the surface of the earth.
4. A person can jump safely from a height of 2 m on the earth. On a planet where acceleration due to gravity is 2.45 ms –2, the maximum height from which he can jump safely is
(1) 8 m (2) 4 m
(3) 16 m (4) 64 m
5. The value of ‘g’ at a particular point on the surface of the earth is 9.8 ms–2. Suppose the earth suddenly shrinks uniformly to half of its present size without losing any mass. The value of g at the same point (assuming that the distance of the point from the centre of the earth does not shrink) will now be
(1) 4.9 ms–2
(2) 3.1 ms–2
(3) 9.8 ms–2
(4) 19.6 ms–2
Answer Key
(1) 3 (2) 2 (3) 2 (4) 1 (5) 3
Consider a body of mass m at a height h from the surface of the earth of mass M and radius R. By Newton’s law of gravitation, the gravitational force of attraction of the earth on the body is
()2 GMm F Rh = + ............(1)
But the gravitational force of attraction of the earth on the body is equal to its weight mgh where gh is the acceleration due to gravity at that height.
()2 h GMm mgRh∴= + ............(2)
()2 h GM gRh = + ............(3)
To express it in terms of the acceleration due to gravity ‘g’ on the surface of the earth: We know, 2 2 GM gGMgR R =⇒=
2 2 2 () ........(4) h h gR gRh R ggRh ∴= +
From this expression, we can find that acceleration due to gravity g h decreases as height increases,
For smaller heights, h << R 2 h Rh ggR +
Using binomial expansion up to first two terms, (1)1if1 n n +=+<< xxx
2 1 h h ggR
The equation (5) is only approximate and is valid when the height of the body above the ground is very small compared to the radius of the earth.
11. If acceleration due to gravity on the surface of earth (Radius = R) is 10 m/s2, what is the value at an altitude R above the surface of earth?
Sol: ' 22 and (2) GMGM ggRR == 2 ' '2 2 10 m/s 24 2.5m/s gR
Try yourself:
9. Acceleration due to gravity at a height R/2 from the surface of earth (radius R ) is g Then, what is the acceleration due to gravity at an altitude R from the surface of earth?
Ans: 12g 9g/16
10.6.2 Variation of g with Depth
Let us assume that the earth to be a homogeneous uniform sphere of radius ‘ R’, density ‘ ρ ’ and mass ‘M’. m d R-d M R
Variation of g with depth o
Let g be the acceleratio n due to gravity at a place on the surface of the earth. When a body is on the surface of the earth, its entire mass M attracts that body towards its centre. We know that 2 GM gR = -----------(1)
Now, consider a point mass m at a depth d, from the surface of the earth. The outer shell of thickness d, which is exterior to the body, exerts no resultant gravitational force on the body. The force on the body will only be due to the mass of the earth confined in the inner solid sphere of radius (R-d). The mass of inner solid sphere is Ms; then, () () () () 33 33 or s s MRdMRd M MRR ==
The gravitational force on the point mass is () () () () 3 2323 s GMmRdmGMmRd GM F RRRdRd ===
The acceleration due to gravity is the force for the unit mass; and let it be gd at the depth d. () 3 d FGM gRd mR ∴==− ---------- (2)
From equations 1 and 2, we get 1 ........(3) d d gRdd gg gRR
From the above equation, it is clear that the value of ‘g’ decreases with the depth.
Equation 3 gives the variation of ‘ g’ with depth.
■ ‘ g’ decreases as depth d increases.
■ ‘ g’ becomes zero at the centre of the earth.
■ ‘ g’ is maximum on the surface of the earth and it decreases with height and depth.
■ Decrease in ‘g’ at small heights is more than the decrease in ‘g’ at small depths.
■ At large heights, decrease in ‘g’ is less than the decrease in g at large depths.
12. If ‘g’ on the surface of the earth is 9.8 ms –2 , find its value at a depth of 3200 km. (Radius of the earth = 6400 km)
Sol: With d = 3200 km
Try yourself:
10. Find the depth at which the value of g becomes 25% of that at the surface of the earth. (Radius of the earth = 6400 km)
Ans: 4800 km
The variation of g with the distance r of a place from the centre of the earth is shown below.
Acceleration due to gravit y 0 g = 9.8 ms–2
Inside the Earth
Distance from centre of Ear th (r) R above the sur face g ∝ 1 r2 g ∝ r x
Inside the earth: () d g gRd R =−
()() dd g grRdrgr R =−=⇒∝
⇒ g d versus r graph is a straight line passing through the origin, as shown in fig.
Above the earth: 2 ()2 h gR gRh = + 2 2 () h gR gRhr r =+= 2 1 gh r
⇒ gh versus r graph is a curve, as shown in the figure.
10.6.3 Variation of g with Latitude
The earth rotates about the axis passing through its north and south poles, a frame attached to the earth is a non-inertial frame of reference. In such a frame, so as to apply Newton’s laws, we have to assume a centrifugal force on the objects at rest.
Latitude at a point on the surface of the earth is defined as the angle, which the line joining the point to the centre of earth makes with the equatorial plane. It is measured in degrees.
Latitude at equator is 0° and at poles is 90°. It is denoted by φ. Let us find out the effect of latitude on acceleration due to gravity.
The earth rotates about its axis from west to east. Consider that an object on its surface at a latitude φ moves in a circular path of radius r.
Then, r = Rcos φ
Centrifugal force acting on the particle is F c = 22 (cos)mrmRω=φω as the component of centrifugal force which opposes gravitational force is 22 coscos C FmRφ=ωφ
The reslutant force acting on the particle towards the centre of earth is
13. What is the time period of rotation of the earth around its axis so that the objects at the equator become weightless? (g = 9.8 m/ s2, Radius of earth = 6400 km)
Sol: g at the equator is g0 = g –2 g' gR =−ω
If bodies are to become weightless at the equator, g0 = 0.
22 0,gRRggR =−ω⇒ω=ω=
Time period of rotation, 2 2 R T g π ==π ω R = 6400 × 103 m; g = 9.8 m/s2 6.4106 2 5078 s84 minutes 38 s. 9.8 T × =π==
Try yourself:
11. If acceleration due to gravity at pole of earth is ‘a’ and that at the equator is ‘b’, then what is the acceleration due to gravity on the surface of earth at lattitude 60°?
Ans: 1 2 ab
a) At equator, 02 0 0 ggR φ=⇒=−ω ; which is minimum.
b) At poles, 0 90 90 ggφ=⇒= ; which is maximum. (As latitude φ increases, g φ increases.)
The value of acceleration due to gravity at poles is independent of angular velocity of earth.
c) If earth stops spinning, then g at equator increases by R ω 2 and remains unchanged at poles. The value a = ω 2R
Key Insights:
■ Variation of ‘g’ with shape of the Earth: It is known that the earth is not a perfect sphere. It appears bulged at the equator and flattened at the poles. Hence, its equatorial radius is greater than the polar radius, i.e., R e > R p
From the relation, 2 GM gR =
From the above relation, 2 1 gR ∝ Rp Re P E ⇒ ge < gp
As the equatorial radius is greater than polar radius, the value of g at the equator is minimum and it is maximum at the poles.
Variation in the value of ( g) h 22 ,;g, GMGM gforrRforrR Rr ===>
3 ,for d GM grrR R =<
TEST YOURSELF
1. A man weights 80 kg on the earth’s surface. The height above ground where he will weigh 40 kg is (radius of earth is r)
(1) 0.31 times r (2) 0.414 times r
(3) 0.51 times r (4) 0.61 times r
2. If R is the radius of the earth, the height above the surface of the earth where the weight of a body is 36% less than its weight on the surface of the earth is
(1) 4R/5 (2) R/5
(3) R/6 (4) R/4
3. A seconds pendulum is mounted in a rocket. Its period of oscillation decreases when the rocket
(1) comes down with uniform acceleration
(2) moves around the earth in a geostationary orbit
(3) moves up with uniform velocity
(4) moves up with uniform acceleration
4. If g is the acceleration due to gravity on the surface of the earth, its value at a height equal to double the radius of the earth is (1) g (2) 2 g (3) 3 g (4) 9 g
5. A body weighs 500 N on the surface of the earth. How much would it weight half way below the surface of the earth?
(1) 100 N (2) 900 N (3) 250 N (4) 125 N
6. Let ω be the angular velocity of the earth’s rotation about its axis. An object weighed by a spring balance gives the same reading at the equator as it does at a height h above the poles (h<<R). The value of h is
Answer Key (1) 2 (2) 4 (3) 4 (4) 4 (5) 3 (6) 2
10.7 GRAVITATIONAL FIELD
In the field of physics, a gravitaional field, also known as a gravitational acceleration field, is a vector field employed to elucidate the effects exerted by a celestial body throughout the space surrounding it.
Limitations of Newton’s Third Law: Newton’s third law has certain limitations and it is not strictly valid in all situations. The limitations are as follows:
1) The law does not apply strictly when the objects move with speed close to that of light.
2) It does not hold true where the gravitational fields are very strong.
3) It also fails to explain how one body exerts gravitational force on the other though they are not in physical contact.
4) The law is not strictly applicable when the gravitational interaction between two bodies involves vast distances.
5) The transmission of gravitational force of one body on another cannot be instantaneous, as stated by this law
Here, the que stion arises as to how two non-contacting bodies, separated by astronomical distances, influence one another instantaneously without any intervening signal that can move faster than light or that can move with an infinite velocity. This indicates a limitation of Newton’s third law when applied to transmission of gravitational force.
Action at a distance concept: Newton’s law of universal gravitation assumes that gravitational force acts directly and instantaneously even when the bodies are not in physical contact or when they are separated by a vast distance. This point of view is known as ‘action at a distance’ or ‘force at a distance’ concept.
However, this theory cannot explain how the gravitational force can be transmitted instantaneously over large distances without any intervening medium between them.
Gravitational field: It is the space around a mass in which its influence is felt.
The field concept assumes that the presence of a body (mass) modifies the space around it in some way and, thus, a gravitational field is developed.
A body placed in this field in turn exerts a gravitational force of attraction on the other. This gravitational field has both momentum and energy. Thus, the field is the agent through which two bodies exert forces on each other
Propagation of gravitational field:
■ According to Einstein’s general theory of relativity, whenever a body with mass is accelerated, the gravitational field around it undergoes rapid changes.
■ Just as photon in electromagnetic field, a quantum of energy is associated with gravitational field called ‘graviton’.
■ Gravitons, like photons, are massless, electrically uncharged particles, assumed to travel at the speed of light, and they are emitted by highly accelerating and extremely massive objects, such as stars.
10.7.1 Gravitational Intensity or Strength of Gravitational Field
Gravitational intensity, at any point in a gravitational field, is defined as the gravitational force experienced by a unit mass placed at that point.
If F is the force experienced by a body of mass m, gravitational intensity E is given by
I t is a vector quantity and it is always directed towards the source of field.
SI Unit:- N/kg CGS Unit:- dyne/ g
10.7.2 Representation of Gravitational Field
Gravitational field intensity is a vector quantity. The magnitude of the field at any point is the force acting on a unit mass kept at that point. The direction of the field is the direction in which a freely placed unit mass moves on its own.
In order to represent the gravitational field, we make use of the fact that the gravitational field intensity is a vector quantity. A vector can be asso-ciated with every point in the gravitational field. The magnitude of a vector at any point in the field is equal to the magnitude of the force acting on a unit mass placed at that point. The direction of this vector will be the direction in which the test mass placed at that point will move, if free to do so. So, a gravitational field can be represented completely with the help of vectors. The gravitational field is a vector field.
An important way to represent gravitational field is by gravitational field lines. A gravitational field line is the path traced out by a unit mass when allowed to move freely in the gravitational field. The direction of the gravitational field, at a point, is given by the tangent to the field line at that point.
In the figure, the gravitational field of an isolated spherical body is represented with the help of gravitational field lines. The field lines are directed towards the centre of the body.
It is not necessary that the gravitational field lines should be straight. In the case of two or more masses, the gravitational field lines will be curved.
Try yourself:
12. A and B are two points from a particle at distances r and 2 r, respectively. If gravitational field at A is E, what is the difference between magnitude of gravitational fields at A and B?
Ans: 3 4 E
Gravitational field due to a uniform solid sphere:
i) Field at an external point: A uniform sphere may be treated as a single particle of same mass placed at its centre for calculating the gravitational field at an external point. Thus, 2 () GM Er r = for rR ≥ or 2 1 ()Er r ∝
Lines of force
Field due to a point mass: Suppose, a point mass M is placed at point O. We want to find the intensity of gravitational field E at a point P, at a distance r from O. Magnitude of force F acting on a particle of mass m placed at P is,
2 GMm F r =
The direction of the force F, and hence of E, is from P to O, as shown in fig. O M r P E
14. Find the intensity of gravitational field due to a 1 kg point mass at a distance of 1 m from it.
Sol: () 22 4 ,'. 9 /2
Here, r is the distance of the point from the centre of the sphere and R is the radius of sphere.
ii) Field at an internal point: The gravitational field due to a uniform sphere at an internal point is proportional to the distance of the point from the centre of the sphere. At the centre itself, it is zero and at surface, it is 2 , GM R where R is the radius of the sphere.
Thus, 3 () GM ErrforrR R =≤ or () Err ∝
Hence, E versus r graph is as shown in fig. Er 2 1 E r E r R 2 GM R
15. If gravitational field at the surface of a sphere of radius R is E, what is the field at a distance R/2 from its surface?
GM E r =
Sol: -11 2 22 -112 6.673×10×1 = m/s 1 = 6.673×10 m/s.
Try yourself:
13. If gravitational field due to a solid sphere of radius R at a distance R from its surface is E, then what is the field at a distance R/2 from its centre?
Ans: 2E
Field due to a uniform spherical shell
i) At an external point: For an exteranl point, the shell may be treated as a single particle of same mass placed at its centre. Thus, at an external point, the gravitational field is given by
This is directed towards the centre of the ring. It is zero at the centre of the ring and maximum at 2 R r = (can be obtained by putting 0 dE dr = ). Thus, E – r graph is as shown in fig.
The, maximum value is E max = 2 2 . 33 GM R
16. If gravitational field at an axial point of a thin ring of radius R at a distance R from its centre is E, what is the maximum possible value at an axial point?
Sol:
2
2 () GM ErForrR r =≥ at r = R (the surface of shell)
2 GM E R = and otherwise, 2 1 E r ∝
ii) At an internal point: The field inside a uniform spherical shell is zero. Thus, E versus r graph is as shown in the above fig.
Field due to a uniform circular ring at a point on its axis: Field strength at a point P on the axis of a circular ring of radius R and mass M is given by 223/2 () () GMr Er Rr = + R r
Try yourself:
14. Find the gravitational field at an axial point of a thin ring of radius R at a distance 2R from its centre. Mass of the ring is M
Ans: 2 2 55 GH R
Field due to circular disc: Gravitational field intensity due to a circular disc at any point on the axial line is 2 22 2 1 g GMx E RxR
+
or θ x 2 2 (1) g GM Ecos R =−θ (in terms of ‘ θ ’)
where θ is angle of elevation of edge of the disc with respect to axial point.
Field due to concentric shells: Two concentric shells of masses M 1 and M 2, are situated as shown in fig. We then find out force on a particle of mass m, when it is at a distance ‘ r’ from centre of the two shells.
At c E = 0, due to M1 and M2, 12 , 0 mm E =
At 12 2 2 0, MM GM bEE b ⇒==
At A 12 12 22 , MM GMGM EE aa ==
Total field at point A, 2 12 ()GMM E a + =
17. Two concentric thin shells of masses m and 2m have radii R and 2R respectively. What is the gravitational field at a distance 3 R from their common centre?
Sol: () ()22 2 . 3 3 GmmGm E RR + ==
Try yourself:
15. Two concentric shells of masses m and 3 m have radii R and 3 R, respectively. If gravitational field at the surface of inner shell is E, what is the gravitational field at the surface of outer shell ?
Ans: 4 9 E
10.7.3 Gravitational Intensity on the Surface of Earth
To determine the gravitational intensity on the surface of earth, let us imagine a test mass on the surface of the earth. Therefore, gravitational intensity ‘E s ’ on the surface of earth is obtained by putting r = R (radius of earth) in equation (1).
where M = mass of earth.
We know, 2 GM Rg = where ‘g’ is the acceleration due to gravity. \ E s = g
Thus, gravitational intensity at any point on the surface of earth is equal to the value of acceleration due to gravity at that point. 4 3 s ERG =πρ
10.7.4 Null Point
The point at which the resultant gravitational field is zero is known as null point.
Let two objects of masses m 1 and m 2 be separated by a distance r and the null point be at a distance ‘x’ from mass m1, as shown in fig. 12 22 () GmGm xrx ∴= m1 m2 x r 12 22 2 1 () 1 mmr x xrxm m ⇒=⇒= +
18. Two particles of masses m and 4 m are separated by a distance d. Find the distance of the point from m where the resultant gravitational field is zero.
dd x m m
Sol: 3 4 1 == +
Try yourself:
16. Separation between two particles A and B is d and distance of null point from A is d/3. What is the mass of B?
Ans: 4m
10.7.5 Properties of Gravitational Field
1) The source of gravitational field is the mass or energies of matter.
2) Gravitational field of an object at a point is indicated by i) vector function of position called gravitational intensity, (or) ii) scalar function of position called gravitational potential.
3) Gravitational fields are added according to rules of vector addition.
4) The direction of gravitational field at a point is indicated by the tangent drawn to the line of force at that point.
5) Gravitational fields are supposed to propagate through particles known as gravitons.
6) Gravitational fields are supposed to propagate with velocity of light.
TEST YOURSELF
1. The following figure shows two shells of masses m1 and m2. The shells are concentric. At which point will a particle of mass ‘m ’ experience zero force? m1 m2
(1) A (2) B (3) C (4) D
2. If a spherical shell is cut into two pieces along a chord, as shown in figure, and point P and Q have gravitational intensities IP and IQ, respectively, then
Q
(1) IP > IQ
(2) IP = IQ = 0
(3) IP = IQ ≠ 0
(4) IP < IQ
3. A body of mass 60 g experiences a gravitational force of 3.0 N, when placed at a particular point. The magnitude of the gravitational field intensity at that point is (1) 50 N/kg (2) 20 N/kg
(3) 180 N/kg (4) 0.05 N/kg
4. The gravitational field at some point in space is () g ˆˆ g3i4j N/k =+ . The force exerted on a 2 kg mass placed at that point is
(1) 5 N, 53° with x-axis
(2) 5 N, 37° with x-axis
(3) 10 N, 53° with x-axis
(4) 10 N, 37° with x-axis
5. A spherical hole of radius R/2 is made in a solid sphere of radius R from its edge. The mass of the sphere before hollowing was M. The gravitational field at the centre of the hole due to the remaining mass is R
(1) zero (2) 2 4 GM R
(3) 2 8 GM R (4) 2 2 GM R
Answer Key
(1) 4 (2) 3 (3) 1 (4) 3
(5) 4
10.8 GRAVITATIONAL POTENTIAL AND POTENTIAL ENERGY
When a body moves in a gravitational field from one point to another, work is done by the external agent on the body to move it opposite to the field direction, which increases its potential energy. On the other hand, if the body moves in the direction of field, work is done by the field, and as a result, PE of the body decreases.
“The amount of work done in bringing a unit mass from infinity to a point in the gravitational field is called gravitational potential.” (or)
The gravitational potential energy of a particle of unit mass at a point is called gravitational potential at that point.
Its unit, in CGS system, is erg g –1 and its SI unit is J kg–1
It is always negative since gravitational force is always attractive in nature.
10.8.1 Potential Due to Different Objects
Potential due to a point mass: Suppose a point mass M is situated at a point O. We want to find the gravitational potential due to this mass at a point P, at a distance r from O. For this, let us find work done in taking the unit mass from P to infinity. This will be,
■ With an increase in ‘ r ’, it becomes less negative, i.e., it increases as we move away from the source of gravitational field.
■ Maximum value of gravitational potential is zero and this happens only at infinite distance from the source
Potential due to a uniform solid sphere:
(i) Potential at an external point: To calculate gravitational potential of a solid sphere at an external point, total mass of the sphere is to be taken at its centre and distance of the external point is to be measured from the centre of the sphere.
() GM VrrR r =−≥ M r At the surface, r = R and GM V R =− P r R
Hence, the work done in bringing unit mass from infinity to P will be GM r
Thus, the gravitational potential at P will be,
V r =−
Conclusions:
■ Gravitational potential at any point is a scalar quantity.
(ii) Potential at internal point: At some internal point P, potential at a distance r from the centre is given by, R V 1.5 GM R r GM R
() 22 3 3 2 r GMRr V R =− for r < R, R for a point on the surface of sphere, r = R hence, potential on the surface is () r GM V R =−
At the centre of sphere, r = 0. Hence, potential at the centre of sphere is V = 3 2 GM R , i.e., at the centre of the sphere, the potential is 1.5 times the potential at the surface. The variation of V versus r graph is as shown in figure.
19. If gravitational potential at the surface of a solid sphere of radius R is v , then find the gravitational potential at a distance R/2 from its centre.
Try yourself:
17. If the gravitational potential at the surface of a solid sphere of radius R is v, then what is the gravitational potential at a distance R from its surface?
Ans: v/2
Potential due to a uniform thin spherical shell:
(i) Potential at an external point: To calculate gravitational potential of a spherical shell at an external point, total mass of the shell is to be taken at its centre and distance of the external point is to be measured from the centre of the shell.
at r = R,
(ii) Potential at an internal point: The potential due to a uniform spherical
shell is constant throughout any point inside the s hell and this is equal to . GM R . Thus, V–r graph for a spherical shell is as shown in fig.
GM R R V r O R r
20. A point mass m is kept at the centre of a thin spherical shell of mass m and radius R. What is the gravitational potential at a radial distance r = R/2?
Sol: 3 . /2 GmGmGm v RRR =−+−=−
Try yourself:
18. If gravitational potential at the centre of a thin spherical shell of radius R is v, what is the gravitational potential at a distance R/2 from its surface at an external point?
Ans: 2v/3
Potential due to a uniform ring at a point on its axis: The gravitational potential at a distance r from the centre on the axis of a ring of mass M and radius R is given by 22 () 0 GM Vrr Rr =−≤≤∞ +
At r = 0, , GM V R =− o R r P
i.e., at the centre of the ring, gravitational potential is GM R .
R
The V–r graph is as shown in fig.
10.8.2 Relation between Gravitational Field and Potential
Consider gravitational field in which E is intensity at a point p having position vector r Let unit mass be moved from r to rdr + . The work done by the field = . Edr . By definition, it is equal to negative change of in potential energy per unit mass or decrease in potential.
. EdrdVorEdV dr =−=−
The gravitational field intensity is equal to negative gravitational potential gradient.
i.e., E =− gradient V
Here, V∂ = ∂x Partial derivative of potential function V w.r.t. x, i.e., we will differentiate V w.r.t. x, assuming y and z to be constant.
Eq. (i) can be written in the following different forms.
(1) dV E d =−
x , if gravitational field is along x-direction only.
(2) dVEdr =−
Key Insights:
■ If E is given, V can be calculated by the formula rr VdVEdr ∞∞ ==−∫∫
■ The negative of the slope of V – r curve gives E.
21. In a region, gravitational potential varies with x according to the equation v = (2x2 + 8) J/kg. What is the gravitational field at x = 1 m?
Sol: 2 2 4 m/s At 1 m, 4 m/s. dv Eixi dx xEi =−=− ==−
Try yourself:
19. The gravitational field in a region is given by ˆ ˆ (43) N/kg. Eij =+
Find the gravitational potential at the points (3 m, 0) and (0, 4 m), if the potential at the origin is taken to be zero.
Ans: 12 J/kg
10.8.3
Gravitational Potential Energy of a System
When two or more bodies interact with each other due to gravitational forces, some work has to be done in assembling them together in their respective places.
Total work done against the conservative force in assembling the bodies together, in their respective places, is called potential energy of the system.
Case-I
A system of two masses: Potential energy of a system of two masses is defined as the amount of work done in bringing these two masses from infinity to their respective places.
yz dVEdEdyEdz ⇒=−++ x x Here, ˆ ˆ ˆ drdidyjdzk =++
x and
yz EEiEjEk =++
x
Consider a space where there is no gravitational field. In order to bring a body of mass m1 from infinity to a point A in the space, no work is done because there is no gravitational force on m1. Now, m1 produces a gravitational field in the space. To bring another body of mass m2 from infinity to a point B at a distance r from m1, some work is required. This work is done by the gravitational force of attraction of m1 acting on m2, and this work is equal to negative change in potential energy.
Let P be the position of ‘ m 2’ (distance x from m1) at any instance. Gravitational force F acting upon it is
This work is equal to gravitational potential energy of the system of two point masses.
Conclusions:
■ Negative sign of the potential energy is due to the fact that the gravitational forces are attractive in nature.
■ The above equation indicates that with an increase in r , the potential energy of the system becomes less negative, i.e., it increases.
■ When distance between two bodies is infinite, gravitational potential energy is zero. This is the maximum value of gravitational potential energy of the system.
Let dW be the small work done against gravity to displace it through a distance dx from x to x + dx
The total work done against the field in moving the mass from x = r1 to x = r2 can be obtained by integrating the above expression between the limits r1 and r2.
■ Gravitational potential energy of an object of mass ‘m’ present on the surface of the earth of mass M and radius R is GMm U R =
■ Two bodies of masses m 1 and m 2 are separated by a distance r1. If the distance between them is increased to r2 , the change in PE of the system,
UGmm rr ∆=−
We know that work done against conservative force is equal to change in potential energy.
The work done in moving the mass m 2 from infinity to a point p can be obtained by substituting
and
= . Now,
In the above case, if distance between the two bodies increases, the PE of the system also increases.
Key Insights:
In the above cases:
a) Work done by gravitational force = negative of change in potential energy.
Wgravitational force = ifUUU−∆=−
b) Work done by external agent = Change in potential energy Wexternal agent = fi UUU∆=−
c) The gravitational potential due to the gravitational force of the earth is defined as the potential energy of a particle of unit mass at that point.
Case-II
A system of three masses: Consider a system comprising of three particles of masses m1, m2 and m3, located at A, B and C, respectively.
i) When mass m1 is brought from infinity to A, in free space, no work is done.
ii) When particle of mass m2 is brought from infinity to B in the gravitational field of m1, work done is
iv) For an n-particle system, there are () 1 2 nn pairs and PE is calculated for each pair and added to get the total PE of the system.
22. Find the potential energy of a system of four particles placed at the vertices of a square of side l. Also obtain the potential at the centre of the square.
Sol: Consider four masses, each of mass m, at the corners of a square of side l, (see fig). We have four mass pairs at distance l and two diagonal pairs at distance 2l
iii) When particle of mass m3 is brought from infinity to C in the combined gravitational field of masses m1 and m2, work done is W13 + W23, where
Hence, 22 ()42 2 GmGm Ur ll =−− 2122 25.41 2 GmGm ll =−+=−
The gravitational potential at the centre of the square () 2/2rl = is ()42 Gm Vr l =−
Try yourself:
20. Three point masses m, 2m, and 3m are placed at the vertices of an equilateral triangle of side length ℓ. Find the gravitational potential energy of the system.
Ans: 11Gm
Key Insights:
■ Relation among U, V, E: a) 22 11 . rr rr dv EdvEdr dr =−=−∫∫
2 1 21 . r r VVEdr −=− ∫ b) dU FdUFdr dr =−⇒=− 2 1 21 r r UUmEdr −=− ∫
c) 21 21 field UUW VV mm −==−
2 1 . r extagent r W Edr m =−= ∫
■ If a body is moving only under the influence of gravitational force, law of conservation of mechanical energy is applicable.
1122 UKUK∴+=+
■ To find the velocity required to send an object to height ‘h’ above the earth’s surface, we apply conservation of mechanical energy. M R m u=v M R m v=0 h
When the body is on the surface of the earth, 2 11 1 , 2 GMm PEKEmV R ==
When the body is at its maximum height, 22,0 GMm PEKE Rh == +
Using the law of conservation of energy, TE1 = TE2
1122 PEKEPEKE⇒+=+
22 11 2 . rr rr GMh V RRh dvEdr
23. The gravitational field in a region is given by ˆ ˆ (512) N/kg. Eij =+ Find the change in gravitational potential energy, if a particle of mass 1 kg is taken from the origin to the point (12 m, 5 m). Sol:
Try yourself:
21. The gravitational field in a region is given by ˆ ˆ (23) N/kg. Eij =+
Find the work done by the field when the particle is moved on the line 3y + 2x = 5.
Ans: zero
10.8.4 Self Energy of Earth
The energy possessed due to interactions between particles of the body is called self energy. Consider an element of thickness dr at a radial distance r
Let m1 and m2 are the masses of sphere of radius r and element respectively.
If M and R are the mass and radius of earth,
The potential energy of system of masses m1 and m2 is given by 12Gmm dU r =
The total potential energy (self energy) of the earth,
TEST YOURSELF
1. The amount of work done in lifting a body of mass ‘m’ from the surface of the earth to a height equal to twice the radius of the earth is
(1) 2 3 GMm R (2) 3 2R GMm
(3) 5 3 GMm R (4) 3 5 GMm R
24. The self gravitational potential energy of a spherical shell of mass M and radius R is Sol. When mass of the shell is m, the potential is . GM R
To add an elementary mass dm, the work done is GMdm R
\ Net work done = self potential energy 2 0 2
Try yourself:
22. What is gravitational potential energy of system consisting of two concentric spherical shells of mass, m, and 4 m and radii R and 2 R.
2. A point P lies on the axis of a ring of mass M and radius a, at a distance a from its centre C. A small particle starts from P and reaches C under gravitational attraction only. Its speed at C will be
(1) 21 1 2 GM a (2) 21 1 2 GM a
(3) () 2 21 GM a (4) zero
3. Four particles, each of mass 1 kg, are at the four corners of a square of side 1 m. The work done to move one of the particles to infinity is
(1) 22G (2) 221 2 G +
(3) () 221 G + (4) 3G
4. The gravitational field due to a mass distribution is 2 A = E x in x-direction. Here, A is a constant. Taking the gravitational potential to be zero at infinity, potential at x is
(1) 2A x (2) 3 2A x
(3) A x (4) 2 A 2 x
Answer Key
(1) 1 (2) 2 (3) 2 (4) 3
10.9 ESCAPE SPEED
The minimum velocity with which a body must be projected from the surface of a planet so as to escape from the gravitational influence of the planet is called escape velocity of that planet.
Expression for escape speed from a height: Consider an object of mass m at a height ‘h’ above the surface of the planet. The distance of the object from the centre of the planet is (h + R). Let it be thrown with an initial velocity vi. The total energy of the object at that point is
2 1 1 (1) 2()hi GMm EmvW hR =−+→ +
where W1 is the gravitational potential energy at infinity (practically, W1 is zero).
If vf is the speed of the object at infinity, then the total energy of the object at infinity,
2 1 1 (2) 2 f EmvW ∞ =+→
From equations (1) and (2), according to law of conservation of energy, 1122 2()2if GmM mvmv hR −= +
Here, the RHS is a positive quantity and its minimum value is zero.
2 1 0 2() i GMm mv hR −≥ +
Hence, for minimum value of vi,
() 2 min 2 min 1 0 2() 1 2() i i GMm mv hR GMm mv hR −= + = +
The minimum speed required for an object to reach infinity is the escape velocity.
mine 2 () v () i GM vor hR = +
If the object is thrown from the surface of the planet, h = 0, and we get 2 e GM v R =
In terms of g, the expression for v e is
On the surface of earth, v e =11.2 km/s
*Escape velocity of an object on earth will be 2 e VgR = Where g = 9.8 m/s2 R= 6.4 × 106 m
29.86.4106 e V ⇒=××× = 11.2 km/s
Key Insights:
■ The escape velocity does not depend on the mass of the body. Hence, from a minute atom to a mighty rocket, every object will have the same escape velocity, irrespective of its mass, for a particular planet.
■ It is also independent of angle of projection.
■ It depends on the mass of the planet, radius of the planet, and height of projection.
Alternative method: Consider a body of mass ‘ m ’ placed at rest at a height ‘ h ’ above the surface of a planet of mass M and radius R. Then, gravitational potential energy of the plant body system is PE = GMm Rh + Minimum kinetic energy imparted to the body for it to escape from the planet is
2 1 2 2 2 e eh GMm KEPE Rh GMm mV Rh GM VgRh Rh =−= + ⇒= + ⇒==+ + where gh = acceleration due to gravity at height h
10.9.1 Escape Velocity of a Body from the Surface of a Planet
Consider an object of mass ‘ m’ at rest on the surface of a planet of mass M and radius R
The gravitational potential energy of the body is GMm PE R = Negative sign indicates that the body is attracted by the planet and the body is bound to the planet. Now, the body can be sent out of gravitational field by making its total energy either zero or positive.
Suppose the body is projected with a velocity Ve, such that kinetic energy imparted to the body must be equal and opposite to the potential energy. Then, total energy of the body becomes zero and body escapes.
2 PE + KE = 0(or)PE = KE. 1 2 e GMm mv R
2 1 2 e GMm mv R =
2 e GM v R = (or) 2 e vgR =
Earth has atmosphere, but on the surface of moon, there is no atmosphere: Gas molecules present in the earth’s atmosphere move with certain mean velocities depending upon the nature of the gas and the temperature of surroundings. At ordinary temperatures, the root mean square (r.m.s.) velocity of hydrogen molecules is around 2 km/s and for oxygen, nitrogen, carbon dioxide, and water vapour, it is around 0.5 to 1 km/s. The escape velocity of earth is 11.2 km/s –1. These velocities are much below the escape velocity of the earth, and hence, we find oxygen, nitrogen, CO2, and water vapour in the earth’s atmosphere.
On the surface of moon, there is no atmosphere: The moon’s escape velocity is 2.38 km/s. In the early period of moon’s
formation, because its temperature was so high and due to thermal agitations, oxygen and nitrogen molecules acquired speeds greater than 2.38 km/s, resulting in their escape from the moon’s surroundings.
Key Insights:
ρ = mean density of the planet
25. A spaceship is launched into a circular orbit, close to the earth’s surface. What additional velocity should now be imparted to the spaceship in the orbit to overcome the gravitational pull? [Radius of earth = 6400 km and g = 9.8 m/s2]
Sol: Additional velocity to be imparted to the orbiting satellite for escaping = Ve –V0 = 11.2 – 8 = 3.2 km/s
26. Find the ratio of escape velocities of two planets if values of g on the two planets are 9.8 m/s2 and 3.3 m/s2, and their radii are 6400 km and 3400 km, respectively.
Sol: The escape velocity in terms of g and R is 2 gR The ratio of escape velocities is:
Try yourself:
23. Find the ratio of escape velocities from the earth and the moon. (Mass of the earth = 6.0 × 1024 kg, Radius of the earth = 6400 km, Mass of the moon = 7.4 × 1022 kg, Radius of the moon = 1740 km)
Ans: 4.695 : 1
10.9.2 Behaviour of a Body Projected Vertically Up with Different Velocities from the Surface of a Planet
C onsider a body of mass ‘m’ projected with a velocity ‘V’ from the surface of a planet of mass ‘M’ and radius ‘R’. Then,
Case-III: If the velocity of projection V > Ve, then total energy of the body becomes positive. Hence, the body escapes from the planet and goes to infinity. Let V ∞ be the velocity of the body at interstellar space; then, using law of conservation of energy, T.Esurface of planet = T.Einterstellar space 1122 22 GMmGMm mVmV R ∝ +=−+ ∝
(Numerically)
Case-I: If the velocity of projection V < V e, then the body fails to escape. It goes to a certain maximum height and then falls back using law of conservation of energy;
27. A projectile is fired vertically upward from the surface of earth with a velocity , e K υ where ueis the escape velocity and K < 1. Neglecting air resistance, show that the maximum height to which it will rise, measured from the centre of earth, is R/(1 –K2), where R is the radius of the earth.
Sol: If a body is projected from the surface of earth with a velocity υ and reaches a height h , by conservation of energy, 2 2 1 2 1 GMmGMm mv RRh
If e V V n = then 22 22 1 1 gRghR h h nn R =⇒= +
Case-II: If the velocity of projection V = Ve, then total energy of the body just becomes zero so that the body just escapes from the planet and goes to infinity.
28. A rocket is fired with a speed 2 gR υ= near the earth’s surface and directed upwards.
(a) Show that it will escape from the earth.
(b) Show that in interstellar space, its speed is 2 gR υ= .
Sol: (a) As PE of the rocket at the surface of earth is (–GMm/R) and at ∞ , it is zero, energy required for escaping from earth 2 0 GMmGM RmgRasgR
As total energy of the body is zero at infinity,
i.e., body possesses zero velocity at infinity.
As initial KE of the rocket 2 1 2 2 mvmgR = is greater than the energy required for escaping (= mgR), the rocket will escape.
(b) If υ is the velocity of the rocket in interstellar space (free from gravitational effects), then by conservation of energy,
Try yourself:
24. The masses and radii of the earth and moon are M1, R1, and M2, R2, respectively. Their centres are at a distance d apart. What is the minimum speed with which a particle of mass m should be projected from a point midway between the two centres, so as to escape to infinity? Ans: )()( 12 12 2 or 2 GMM GmMM d + +
TEST YOURSELF
1. The escape velocity of a body from the surface of the earth is nearly (1) 7.98 km s–1 (2) 9.82 km s–1 (3) 10.48 km s–1 (4) 11.2 km s–1
2. The relation between the escape velocity V e and orbital velocity V0 is given by (1) 0 2 e VV = (2) 0 2
VV
(3) 0 2 e VV = (4) V e = V0
3. The KE required to project a body of mass ‘ m’ from the earth’s surface to infinity is (1) mgR/4 (2) 2 mgR (3) mgR (4) 2mgR
4. The escape velocity for a body projected vertically upwards from the surface of the earth is 11 km s –1. If the same body is projected at an angle of 45° with the vertical, the escape velocity will be (1) 1 112 km s (2) 22 km s-1 (3) 11 km s-1 (4) 1 11/2 km s
Answer Key (1) 4 (2) 1 (3) 3 (4) 3
10.10 EARTH SATELLITE
An artificial satellite moves in a circular orbit around the earth. Different parameters of earth’s satellite are discussed below.
10.10.1 Orbital Speed
The horizontal velocity required for an object to revolve around a planet in a circular orbit is called orbital velocity.
Derivation: Consider a satellite of mass m revolving around a planet of mass M and radius R in a circular orbit of radius r . If h is the height of the orbit, then orbital radius r = R + h. Let V0 be the orbital velocity.
Centripetal force on the satellite = ()() 2 0 ...1 mv Rh +
Gravitational force of attraction of the planet on the satellite = ()() 2 ...2 GMm Rh +
The centripetal force required for the object to revolve in a circular orbit with a constant speed is supplied by the gravitational force of attraction of the planet on the obje ct. () 2 0 2 2 0 ...3 () () () mvGMm RhRh GMGM vrRh Rhr ⇒= ++ ⇒===+ + () 0 ...4 GMGM v Rhr == +
Orbital velocity does not depend upon the mass of the satellite. It depends on the mass of the planet, radius of the planet, and height of the orbit from the surface.
When h << R, i.e., the satellite is revolving very close to the surface of the planet, R + h >> R. Then, the above expression becomes 2 2 .....(5) o o GMgRGM vgRRR vgR
Near the surface of earth, 3 9.86400107.92km/s o vgR==××=
To find the relation between escape velocity and orbital velocity for a satellite orbiting very close to the earth’s surface:
From Eq. (5) o vgR =
We also know that 2(2)() eeo vgRvv =⇒=
Escape velocity = (2) (Orbital velocity)
29. An artificial satellite is revolving in a circular orbit at a height of 1200 km above the surface of the earth. If the radius of the earth is 6400 km, mass is 6×1024 kg. Find the orbital velocity (G = 6.67 × 10–11Nm2/kg2)
the earth and the other around Mars, close to their surface. (Mass of the earth = 6 × 1024 kg, Mass of Mars = 6.4 × 10 23 kg, Radius of the earth = 6400 km and Radius of Mars = 3400 km)
Sol: The orbital speed of a satellite close to the surface of a planet,
0 GM V R =
The ratio of orbital speeds,
Try yourself:
25. Find the ratio of the orbital speeds to two satellites of the earth if the satellites are at a height of 6400 km and 19200 km. (Radius of the earth = 6400 km.)
Ans: 2 1
GM V Rh = + ×× = ×+× ××× = × () () ()() 0 11 24 33 11 24 3 6.6710610
Sol: () () ()() 0 11 24 33 11 24 3 6.6710610 640010120010
640010120010
640010120010
GM V Rh = + ×× = ×+× ××× = × () () ()() 0 11 24 33 11 24 3 6.6710610
GM V Rh = + ×× = ×+× ××× = ×
6.6710610 = 7.26 km/s 760010
6.6710610 = 7.26 km/s
760010
6.6710610 = 7.26 km/s
760010
30. Find the ratio of the orbital speeds of two satellites, one of which is rotating around
ii) For surface satellite (i.e., h < < R) ⇒ r = R
a) In case of earth, the time period of surface satellite is T = 84.6 min
b) If two planets have same mean density, then satellites revolving close to their surfaces will have same time periods.
c) Angular momentum of a satellite:
LGMmRh
=+ for the surface satelite ( h << R)
31. A satellite is at a height of 25, 600 km from the surface of the earth. If its orbital speed is 3.536 km/s, find its time period. (Radius of the earth = 6400 km)
Sol: R = 6400 km = 6.4 × 106 m, h = 25, 600 km = 25.6 × 106 m, v0 = 3.536 × 103 m/s
The time period
Try yourself:
26. The radius of a planet is R1 and a satellite revolves round it in a circle of radius R 2. The time period of revolution is T . The acceleration due to the gravitation of the planet at its surface i s
TEST YOURSELF
1. A satellite is revolving in a circular orbit around the earth at a height h . If the acceleration due to gravity at that height g, the orbital speed of the satellite is (1) gh
(2) 2 gh
(3) ()gRh +
(4) gR
2. Two satellites of earth S1 and S2 are moving in the same orbit. The mass of S1 is four times the mass of S2. Which one of the following statements is true?
(1) The potential energies of earth satellites in the two cases are equal.
(2) S1 and S2 are moving with the same speed.
(3) The kinetic energies of the two satellites are equal.
(4) The time period of S1 is four times that of S2.
3. The radius of circular orbits of two satellites of the earth, A and B, are 4 R and R, respectively. If the speed of satellite A is 3V, then the speed of satellite B will be
(1) 3V/4
(2) 6V
(3) 12V
(4) 3V/2
Answer Key
(1) 3 (2) 2 (3) 2
10.11 ENERGY OF AN ORBITING SATELLITE
Consider a satellite of mass m revolving in an orbit of radius r, around the earth of mass M. Then,
(a) PE of satellite = GMm r
(b) Its kinetic energy 2 1 KE 2 mv = But GM v r =
(c) Therefore, total energy of satellite
E = PE + KE 2 GMm E r ⇒=−
Key Insights:
■ Total energy, TEKE 2 PE =−=
So, PE : KE : TE = –2 : 1 : –1 KE PE TE r E
■ When a satellite falls from a higher orbit to a lower orbit, its
• PE decreases
• TE decreases
• KE increases
• Speed increases
• Linear momentum increases
• Angular momentum decreases
■ When a satellite rises from lower orbit to higher orbit, its
• PE increases
• TE increases
• KE decreases
• Speed decreases
• Linear momentum decreases
• Angular momentum increases
■ The apparent weight of an object in a satellite that revolves around the earth is zero, since normal reaction is zero.
■ The moon does not hit the earth. However, the moon is also influenced by earth’s gravity. Indeed, the moon falls towards the earth’s surface as its surface is curving constantly away from it, so that they never get any closer.
Trajectories of a body projected with different velocities:
1) A body revolves around a planet only when it is projected with sufficient velocity in a direction pependicular to the gravitational force of attraction of planet on the body. point of projection part of ellipse speed v earth circle
hyperbola parabola ellipse v(2gr) > 1 v(2gr)11kms =≈ (2gr)v(gr) >> 1 v(gr)8kms =≈ v(gr) <
If vgr < , body falls on the surface of earth.
If vgr = , body revolves in a circular orbit.
If 2 grvgr << , body revolves in an elliptical orbit.
If 2 vgr = , body escapes along parabolic path.
If 2 vgr > , body escapes along hyperbolic path.
Special Cases:
Case-I: Work done to lift a body at rest from the surface of a planet to a height h is
TEi = TEsurface = 0 GMmGMm RR −+=
TEf = TEheight = 0 GMmGMm RhRh −+= ++
Work done, W = TEf –TEi = () 1 GMmGMm RRh WGMmhmgh RRhh R + ⇒== + +
Case-II: Work done to shift a body at rest from the surface of a planet into an orbit in which the body revolves around the planet is
TEi = TEsurface I m v M II m v r 0 GMmGMm RR =+=
TEf = TEorbit = () 2 0 1 22 GMmGMm mv RhRh += ++
Case-III:
Work done, W= TEf –TEi () () 2 2 2 GMmGMm RRh Rh WGMm RRh =− + + ⇒= +
Work done to shift a body revolving around the planet from one orbit to another orbit is, ih1TE = (TE) = () 2 1 11 1 22 GMmGMm mv RhRh −+= ++
TEf = () 2h TE () 2 2 22 1 22 GMmGMm mv RhRh =−+= ++
Work done, W = TEf –TEi = ()() 2212 GMmGMm RhRh ++ () () 21 12 2 GMmhh W RhRh
Case-IV: Work done or additional energy required for a body to escape, which is initially revolving around the planet, close to the surface, is
TEi = 22 GMmGMmGMm RRR +=−
TE f = O (body escapes only when its TE becomes zero or positive)
Work done or additional energy imparted to the body is fi TE-TE 2 GMm WE R =∆== = KE body
Hence, a body (satellite) revolving around the planet escapes when:
i) its KE is doubled (increases by 100%)
ii) its velocity is increased to 2 times that of the present value (increases by 41.4%)
Additional velocity imparted to the body = V e –V0 = 00 2 VV () 0 213.2 km/s V −= (nearly)
Key Insights:
■ In the above case, if the body initially revolves around the planet at a height h from the surface, then its
TE 2 GMm Rh = +
∴ Additional energy required for the body to escape is () 2 GMm Rh +
32. A sky lab of mass 2 × 103 kg is first launched from the surface of earth in a circular orbit of radius 2R (from the centre of earth) and then, it is shifted from this circular orbit to another circular orbit of radius 3R. Calculate the minimum energy required (a) to place the lab in the first orbit (b) to shift the lab from first orbit to the second orbit.
Given, R = 6400 km and g = 10 m/s2.
Sol: The energy of the sky lab on the surface of earth
(b) As for II orbit r = 3 R,
(a) So, the energy required to place the lab from the surface of earth to the orbit of radius
But as 2 g=(GM/R), i.e., GM = gR2 or () 10 10 11 12.8101.110J
33. A satellite of mass 400 kg is in a circular orbit of radius 2R about the earth, where R is radius of the earth. How much energy is required to transfer it to a circular orbit of radius 4R ? Find the changes in the kinetic and potential energies. 6 (6.3710m) R =×
Sol: Initial total energy is 2 24 GMmGMm E rR −=−= Final total energy is 2 2 28 GMmGMm E rR −=−=
The change in total energy is 21 2 6 9 8 8 9.84006.3710 88 J 3.1310 GMm E R GMmR R gmR EE E E
Change in kinetic energy = K2 – K1 = –3.13 ×109J
Change in potential energy = U2 – U1 = –6.25 ×109J
Try yourself:
27. A satellite is revolving around the earth in geostationary orbit. If R be the radius of earth, T be its time period of rotaion around its own axis, and g be the acceleration due to gravity on its surface, then find the acceleration of the satellite. Ans: 1/3 2 2 4 4 gR
TEST YOURSELF
1. The energy required to shift a satellite from orbital radius r to radius 2r is E. What energy will be required to shift the satellite from orbital radius 2r to 3r?
(1) E (2) E/2
(3) 3E/4 (4) E/3
2. By what percent does the energy of a satellite need to be increased to shift it from an orbit of radius r to 3r?
(1) 66.7% (2) 33.3% (3) 15% (4) 20.3%
Answer Key
(1) 4 (2) 3
10.12 GEOSTATIONARY AND POLAR SATELLITES
Geost ationary satellites are used as communication satellites and polar satellites are used as spy satellites.
10.12.1 Geostationary Satellite
The satellite which revolves round the earth and appears to be stationary, with respect to an observer on earth, is called geostationary satellite.
If the period of revolution of an artificial satellite is equal to the period of rotation of earth, then such a satellite is called geostationary satellite.
When an earth satellite satisfies the following conditions, then it becomes a geostationary satellite.
1. H eight of the orbit above the earth’s surface is about 35,800 km.
2. The orbit must lie in the equatorial plane of the earth.
3. The sense of revolution must be same as that of the rotation of the earth about its own axis (from west to east, along the equatorial plane).
4. Time period of revolution must be same as that of the earth’s rotation (24 hours).
5. The orbital angular velocity of geostationary satellite is same as that of the earth.
6. Speed of revolution is 3.1 km/s.
7. Relative angular velocity of geostationary satellite w.r.t earth about the centre of the earth is zero.
8. The orbit of geostationary satellite is called parking orbit.
9. Orbital radius of geostationary satellite is 42,250 km (approximately 7 R, here, R is the radius of the earth).
10.12.2 Uses of Geostationary Satellites
Geostationary sattellits are used for the following purposes:
■ Study the upper layers of the atmosphere
■ Forecast the changes in the atmosphere
■ Know the shape and size of the earth
■ Identify the minerals and natural resources present inside and on the surface of the earth
■ Transmit the TV programmes to distant places
■ Study the properties of radio waves in the upper layers of the atmosphere
■ Undertake space research, i.e., know about the planets, satellites, comets, etc.
■ Widely used for telecomunication purpose
■ The EM waves having frequences from 2 MHz to 10 MHz are reflected from the ionosphere. These are used for radio broadcasting.
■ The EM waves having frequences above 30 MHz are not reflected from ionosphere and they cannot be received by the line of sight. These signals are received and broadcasted back to a wide area of earth by geostationary satellites.INSAT group of satellites sent by the India are one such group of geostationary satellites.
10.12.3 Polar Satellite
A satillite whose orbital plane is perpendicular to equitorial plane is called a polar satellite. It is a low altitude ( h = 500 to 800 km) satellite, but it goes around the poles of the earth in a north–south direction, whereas the earth rotates around its axis in an east–west direction. Since its time period is around 100 minutes, it crosses any altitude many times a day. The satellite can view polar and equatorial regions at close distances with good resolution.
T he first polar satellite was launched on February 24, 1996, by NASA. Polar satellites are also called low earth orbits (LEO). They can view only the poles or a limited area on the earth at the same time. The earth rotates on its axis as the satellite goes around the earth. It can view only a small strip of the earth during one rotation. Adjacent strips are viewed in next successive revolutions, so that, in effect, the whole earth can be viewed strip by strip during the entire day. Information gathered from polar satellites is extremely useful for remote sensing metereology as well as for environmental studies of the earth.
Key Insights:
■ Satellites orbiting the earth have finite life and sometimes, debris of satellites falls to the earth. This is because viscous forces cause the speed of the satellite to decrease, and hence, its height grad ually decreases.
■ The angle between the equatorial plane and the orbital plane of a geostationary satellite is 00 and of polar satellite is 900
■ I.E.R.S. is India’s polar satellite.
34. A geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to earth’s gravity at the site of this satellite? (Take the potential energy at infinity to be zero). Mass of the earth = 6.0 × 1024 kg, radius = 6400 km, G = 6.67 × 10–11 Nm2/kg2
Sol: Gravitational potential at height h from the surface of earth is 11 24 6 66 GM V (Rh) 6.6710(610) 9.410Jkg (6.4103610)
Try yourself:
28. If g be the acceleration due to gravity on the surface of earth, R is the radius of earth, and T is its time period of rotation about its axis, then find the orbital speed of geostationary satellite.
Ans: (2 π 1/3/T)2gr
TEST YOURSELF
1. A geostationary satellite is orbiting the earth at height 6R from the surface of the earth. The time period of another satellite revolving around the earth at a height 2.5 R from the earth’s surface is
(1) 122hr (2) 12hr
(3) 62hr (4) 6hr
2. To have an earth satellite synchronous with the rotation of the earth, it must be launched at a proper height
(1) from west to east in equatorial plane
(2) from south to north in equatorial plane
(3) from east to west in equatorial plane
(4) from north to south in equatorial plane
Answer Key
(1) 3 (2) 1
10.13 WEIGHTLESSNESS
The force with which the earth attracts a body towards its centre is known as weight of that object in general.
An object feels weightlessness, whenever normal reaction on it is zero. This is possible when the net acceleration of the object is equal to acceleration due to gravity in magnitude and direction at that point.
Case–I: When an object is in a freely falling lift, its apparent weight is zero. This phenomenon is known as weightlessness.
Using free body diagram, a=g
mg – N = ma ()() 0 0 app Nmgamgg N W ⇒=−=− ⇒= ∴=
Case–II: Consider a person in a satellite orbiting around the earth, close to the surface of the earth. Let V0 be the orbital velocity of the satellite. Then,
0 GM V R = N
Centripetal acceleration, 2 0 2 c VGM ag RR ===
Using free body diagram, mg – N = ma c cNmgma⇒=− () 0 c NmgaN ⇒=−⇒= 0 app W ∴=
It means that everything inside the satellite is in a state of free fall. This is just like a body or a person falling towards the centre of the earth from a height. So, a person inside the satellite experiences weightlessness.
10.13.1 Condition for Weightlessness of an Object at the Equator of the Earth
We know that at equator, 22 0 0 2 where rad/s 246060 isthepresentvalue ggRCos φ=−ωφ π ω= ××
As, 0 φ= ()222 00 0.034 m/s e RggR ω=⇒=−ω
Where, 2 0 aR=ω is the centripetal acceleration.
If a = g, ge = g – a = g – g = 0, Hence, the object appears weightless.
This is possible for an angular velocity ω such that 2 gRgR =ω⇒ω=
If the present value of angular velocity is ω,
Hence, present angular velocity of earth has to increase to 17 times, for which the object becomes weightless at equator.
Key Insights:
■ If 2 Rgω=→ Object appears weightless.
■ 2 Rgω>→ Object flies off from the surface of the planet.
■ 2 Rgω<→ Object will remain stuck to with the surface of the planet.
CHAPTER REVIEW
Universal Law of Gravitation, The Gravitational Constant
■ Gravitational force is attractive.
■ Inverse square law acts along the line joining the particles.
■ Magnitude of this force is independent of the medium between the particles.
■ Magnitude of this force is independent of the material of the objects.
■ Gravitational force is central force.
■ Gravitational force is conservative force.
■ Universal law of gravitation holds true for points masses and spherically symmetrical bodies.
Acceleration Due to Gravity of the Earth, Gravitational Field
■ Intensity of gravitational field due to a uniform circular thin ring a point on its axis is given by 223/2 , () GMx E Rx = + where m = mass of ring, x = distance of the point from the centre of ring, R = radius of ring.
■ Intensity of gravitational field due to a thin uniform circular ring at a point on its axis is maximum when /2 xR =±
■ If two particles of masses m 1 and m 2 are separated by a distance r and p be a point on the line joining the particles, at a distance x from m1, the intensity of gravitational field at p is zero when 2 4 , 3 GM gGPR R ==π where p is the average density of earth.
■ Acceleration due to gravity at a distance r (r < R) from the centre of earth is given by s r ggR =
■ Acceleration due to gravity at a radial distace r (r > R) is given by 2 22 s GMR gg rr ==
■ Acceleration due to gravity at an altitude h above the surgace of earth is given by ()() 2 22 s GMR gg RhRh == ++
■ Acceleration due to gravity at an altitude h ( h << R ) above the surface of earth is given by 2 1 s h ggR
=−
■ Acceleration due to gravity at a depth d below the surface of earth is 1 s d ggR
=−
■ Acceleration due to gravity at a lititude φ when the earth is spining with angular
| CHAPTER 10: Gravitation
velocity ω is gien by 2 2 1cos, s s R gg g φ
where gs is the acceleration due to gravity on the surface of earth when the earth is not spinning.
Earth Satellite
■ Orbital speed 0 Gm V r =
■ Time period of rotation
2 23 4 . Tr GM
■ KE; 2 GMm K r = where M = mass of earth m = mass of satellite r = Orbit radius
■ PEU. GMm r ==−
■ Total energy, . 2 GMm E r =−
■ Angular momentum of a satellite about the centre of earth 2 LGMmr =
Exercises
JEE MAIN LEVEL
Level-I
Kepler’s Laws
Single Option Correct MCQs
1. A and B, two satellites revolving around the earth in circular orbits, have time periods 8 hrs and 1 hr, respectively. The ratio of their radius of orbits is
(1) 83/2 (2) 8 : 1
(3) 4 : 1 (4) 2 : 1
Numerical Value Questions
2. The period of a planet around the sun is 27 times that of earth. The ratio of the radius of planet’s orbit to the radius of earth’s orbit is ________.
Universal Law of Gravitation
Single Option Correct MCQs
3. The gravitational force between two bodies is F. If the mass of each body is doubled and the distance between them is halved, then the force between them will be
(1) 4F/9 (2) 9F/4
(3) 16F (4) F/16
4. Three equal masses m are placed at the three vertices of an equilateral triangle of side a. The gravitational force exerted by this system on another particle of mass m placed at the midpoint of a side is
5. Kepler’s third law states that the square of period of revolution (T) of a planet around the sun is proportional to third power of
average distance of r between sun and planet, i.e., T2 = kr2, where k is constant. If the mass of sun and planet are M and m, respectively, then, as per Newton’s law of gravitation, force of attraction 2 GMm F r = , the relation between G and K is described as
(1) GK= 4 π 2
(2) GMK = 4 π 2
(3) K = G
(4) 1 K G =
6. If mass M is split into two parts m and (M–m ), which are then separated by a certain distance, the ratio m/M that maximises the gravitational force between the two parts is
(1) 1 : 2 (2) 1 : 1
(3) 2 : 1 (4) 1 : 4
7. Three spherical balls of masses 1 kg, 2 kg and 3 kg are placed at the corners of an equilateral triangle of side 1 m. The magnitude of the gravitational force exerted by 2 kg and 3 kg masses on 1 kg mass is
(1) 19 G (2) 17G
(3) 19G (4) G 17
8. A thin spherical shell of mass ‘M’ and radius ‘R’ has a small hole. A particle of mass m is released at the mouth of them. Then, (1) the particle will execute S . H . M inside the shell
(2) the particle will oscillate inside the shell, but the oscillations are not simple harmonic
(3) the particle will not oscillate, but the speed of the particle will go on increasing. (4) none of these
Acceleration Due to Gravity and Its Variation
Single Option Correct MCQs
9. If the change in the value of g at a height h above the surface of the earth is same as that at depth ‘x’ below the surface of the earth, then (h<<R)
(1) x = h
(2) x = 2h
(3) x = h/2
(4) x = h2
10. The ratio of the accelerations due to gravity at the bottom of a deep mine and that on the surface of the earth is 978/980. Find the depth of the mine if the density of the earth is uniform throughout and the radius of the earth is 6300 km.
(1) 12.86 km (2) 13.0 km
(3) 25.38 km (4) 90.9 km
11. If ‘ g ’ is acceleration due to gravity on the earth, R is the radius of the earth, and angular speed of rotation of the earth about its axis is made to be 2g 5R , then the weight of a body at the equator decreases by (1) 60% (2) 50% (3) 40% (4) 75%
12. If the radius of the earth were to shrink by 1% ,its mass remaining the same, the acceleration due to gravity on the earth’s surface would (1) decrease by 2% (2) remain unchanged (3) increase by 2% (4) increase by 1%
13. Two planets of radii r 1 and r 2 are made from the same material. The ratio of the accelerations due to gravity at the surface of the two planets is
(1) 1600 m (2) 16 m (3) 160 km (4) 1600 km
15. The variation of acceleration due to gravity (g) with distance (r) from the centre of the earth is correctly represented by (Given: R = radius of earth)
14 The altitude at which the weight of a body is only 64% of its weight on the surface of earth is (Radius of the earth is 6400 km)
Numerical Value Questions
16. If an object weights 270 N on the earth’s surface, the weight of the object at an altitude equal to twice the radius of the earth is ________N.
17. Find the percentage decrease in weight of a body, when taken 16 km below the surface of the earth. Take radius of the earth as 6400 km.
Gravitational Field
Single Option Correct MCQs
18. A solid sphere of mass ‘2M’ and radius ‘2a’ is surrounded by a uniform concentric spherical shell of thickness 3a and mass 3M. The gravitational field at a distance ‘5a’ from the centre will be (1)
23. Two heavy spheres, each of mass 100 kg and radius 0.1 m, are placed 1 m apart on a horizontal table. What is the potential at the midpoint of the line joining their centres?
(1) –3.7× 10–8 J/kg (2) –1.7× 10–6 J/kg
(3) –2.7× 10–8 J/kg (4) –4.7× 10–8 J/kg
24. The change in potential energy when a body of mass m is raised to a height nR from the earth’s surface is (R = Radius of the earth)
(1) () 1 n mgR n
(2) mgR n
19. Two bodies of masses 2 kg and 8 kg are separated by a distance of 9 m. Then, the point where the resultant gravitational field is zero is at a distance of
(1) 6 m from 8 kg
(2) 3 m from 8 kg
(3) 6 m from 2 kg (4) 4.5 from each mass
Numerical Value Questions
20. The radius and acceleration due to gravity of the moon are 1/4th and 1/5th that of the earth. The ratio of the mass of the earth to mass of the moon is _______.
Gravitational Potential and Potential Energy
Single Option Correct MCQs
21. At what height from the surface of the earth will the gravitational potential and the value of g be –5.4 × 10 7 J kg –2 and 6.0 ms–2, respectively? Take the radius of the earth as 6400 km.
(1) 2600 km (2) 1600 km
(3) 1400 km (4) 2000 km
22. The potential energy of a body of mass ‘ m’ is given by U = px+qy+rz. The magnitude of the acceleration of the body will be (1) pqr m ++ (2)
(3)
+
(3) () 1 n mgR n
Earth Satellite
+
(4) () 2 2 1 n mgR n
Single Option Correct MCQs
25. A satellite is in a circular orbit around a planet. Its period of revolution is T, radius of the orbit is R, orbital velocity is V, and acceleration is ‘a’. Then,
(1) 2 and == V Vata R
(2) 2 and π == R VVaT T
(3) 2 2 and π == RV Va TR
(4) 2 1 2 VaT =
26. The satellite of mass m revolving in a circular orbit of radius r around the earth has kinetic energy E. Then, its angular momentum will be
(1) 2 E mr (2) 2 2 E m
(3) 2 2 Emr (4) 2 Emr
27. A satellite is revolving near the earth’s surface. Its orbital velocity depends on its mass ‘m’ as
(1) m0 (2) m
(3) m2 (4) m3
28. A satellite orbits the earth at a height of R/6. What is its orbital speed?
(1) 2 R GM
(2) R GM (3) 7R GM (4) 6 7R GM
29. A satellite is launched into a circular orbit of radius R around the earth, while a second satellite is launched into a circular orbit of radius 1.02 R. The percentage difference in the time periods of the two satellites is
(1) 1.5
(2) 2.0
(3) 0.7
(4) 3.0
30. A satellite of mass m is moving in a circular orbit of radius r = (R e + h) around the earth of radius R e and mass M e, and density of the earth is ρ. Match the f ollowing.
Column I
Column II
(A) Orbital velocity of the satellite (I) 3 2 r GM π
(B) Kinetic energy of the satellite (II) 2 e GMm r
(C) Potential energy of the satellite (III) 2 e GMm r
(D) Total energy of the satellite (IV) e GMm r
(E) Time period of the satellite (V) e GM r
Choose the correct answer from the options given below:
(A) (B) (C) (D) (E)
(1) V II IV III I
(2) IV I III II V
(3) III II I V IV
(4) IV III II I V
Escape Speed
Single Option Correct MCQs
31. The mass of a planet is half that of the earth and the radius of the planet is one-fourth that of the earth. If we plan to send an artificial satellite from the planet, the escape velocity will be nearly (V e of earth is 11 km s–1)
(1) 11 km s–1
(2) 5.5 km s–1
(3) 15.5 km s–1
(4) 7.7 km s–1
32. The ratio of escape velocity on earth to the escape velocity on a planet whose radius and mean density are twice that of the earth is
(1) 1 : 2
(2) 1:22
(3) 1 : 4
(4) 1:2
33. A body is projected up with a velocity equal to 3/4 th of the escape velocity from the surface of the earth. The height it reaches is (Radius of the earth is R)
(1) 10 R/9 (2) 9 R/7
(3) 9 R/8 (4) 10 R/3
Weightlessness
Single Option Correct MCQs
34. A man inside an artificial satellite feels weightlessness, because the force of attraction due to earth is
(1) zero at that place
(2) balanced by the force of attraction due to moon
(3) balanced by centrifugal force
(4) non-effective due to particular design of satellite.
Level-II
Kepler’s Laws
Single Option Correct MCQs
1. A planet revolves around the Sun in elliptical orbit. The area ‘CSA’ is 1/4th of total area of ellipse. If t1 is the time taken by the planet to move along ABC and t2 is the time taken to move along CDA, then (Here, CA is the minor axis of the ellipse )
circular orbits of radii rA and rB respectively. Which out of the following would be the correct relationship of their orbits ?
(1) 23 2 ABrr =
(2) 33 r2rAB =
(3) 33 4 ABrr =
(4) π()−=− 2 2233 4 ABBA TTrr GM
Numerical Value Questions
(1) t1 = t2
(2) t1 = 4t2
(3) 2t1 = t2
(4) t1 = 3t2
2. Two planets A and B of equal mass are having their period of revolutions TA and T such that TA=2TB. These planets are revolving in the
3. Assume the planetary orbits to be circular, and let T be the time period of revolution of the Earth around the sun of mass M. If Jupiter’s period of revolution around the sun is K times that of the Earth, the velocity of Jupiter in heliocentric reference frame is 1/x 1/y 2 GM TK π
. The product of x and y is ____.
Universal Law of Gravitation
Single Option Correct MCQs
4. In a double star system, two stars of masses m1 and m2, separated by a distance d, rotate about their centre of mass. Then, their common angular velocity would be
(1) 1 3 d Gm
(2) 2 3 d Gm
(3) () 12 3 Gmm d +
(4) () 3 12 d Gmm +
Numerical Value Questions
5. Two particles, each of mass ‘ m’, move in a circle of radius ‘r’ under the action of their mutual gravitational force. The speed of each particle is Gm xr . The value of x is _____.
Gravitational Field
Single Option Correct MCQs
6. Two identical point masses are fixed on y -axis at equal distances from the origin. A small particle starts on the x-axis, from x = ∞ towards the origin. Its acceleration a is taken positive along its direction of motion. Out of the following graphs, the one that best represents the variation of a with x-coordinate of particle i s
(1) x
(2) x a
(3) x a
(4) x a
7. A solid sphere of radius R gravitationally attracts a particle placed at 3R from its centre with a force F 1. Now a spherical cavity of radius 2 R
is made in the sphere (as shown in figure) and the force becomes F 2. The value of F1 : F2 is
(1) 50 : 41 (2) 25 : 36
(3) 41 : 50 (4) 36 : 25
8. A person whose mass is 100 kg travels from Earth to Mars in a spaceship. Neglect all other objects in sky and take acceleration due to gravity on the surface of the Earth and Mars as 10 m/s2 and 4 m/s2, respectively. Identify from the below figures, the curve that fits best for the weight of the passenger as a function of time.
(1) III (2) I
(3) IV (4) II
Numerical Value Questions
9. The gravitational field in a region is given by ()=− E3i4 ˆˆ jN/kg . Find out the work done (in joules) in displacing a particle by 1 m along the line 4y = 3x + 9.
Gravitational Potential and Potential Energy
Single Option Correct MCQs
10. A body of mass m is projected with velocity λv e in vertically upward direction from the surface of the earth into space. It is given that v e is escape velocity and λ<1. If air resistance is considered to the negligible, then the maximum height from the centre of earth, to which the body can go, will be (R: radius of earth) (2022)
11. If the gravitational field intensity at a point is given by 2.5 GM g r = , then the potential at distance r is (Take gravitational potential at infinity to be zero)
12. A point mass m and a thin uniform rod of mass M and length L are located along a straight line, as shown in figure. What is the potential energy of gravitational interaction of this system?
(3) a 1 GMm lL
(4) () a/2 GMm L +
Numerical Value Questions
13. The initial velocity vi required to project a body vertically upward from the surface of the earth to reach a height of 10 R, where R is the radius of the earth, may be described in terms of escape velo city v e such that ie x vv y =× The value of x will be_____
Earth Satellite
Single Option Correct MCQs
14. A satellite is revolving in a circular orbit at a height h from the earth’s surface (radius of earth R ; h<<R). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth’s gravitational field, is close to : (Neglect the effect of atmosphere.)
(1) gR (2) /2 gR
(3) () 21 gR (4) 2 gR
15. A satellite is orbiting just above the surface of a planet of average density D, with period T. If G is the universal gravitational constant, the quantity 3 G π is equal to
(1) D2T (2) 3πT2D (3) 3πD2T (4) T2D
16. A satellite is projected with velocity 5 6 times the escape speed from the earth’s surface. The initial velocity of the satellite is parallel to the surface of the earth. The maximum distance of the satellite from the centre of the earth will be pR, where R is the radius of earth. Find p. (1) 5 (2) 4 (3) 3 (4) 6
Numerical Value Questions
17. Two satellites of masses 50 kg and 100 kg revolve around the earth in circular orbits of radii 9R and 16R, respectively, where R is the radius of the earth. The speeds of the two satellites will be in the ratio x/y, where x and y are co-prime. Find the value of x + y.
18. A satellite of mass m is moving in a circular orbit of radius r around the earth (mass M). Its angular momentum with respect to the centre of the orbit is L = (GMmxr)1/y. Find the value of x/y.
19. The minimum energy required to take a satellite to a height ‘ h ’ above the earth’s surface (radius of earth = 6.4 x 103 km) is E1, and kinetic energy required for the satellite to be in a circular orbit at this height is E2. The value of h for which E1 and E2 are equal is ______ (in km).
Escape Speed
Single Option Correct MCQs
20. The masses and radii of the earth and moon are (M1, R1) and (M2, R2) respectively. Their centers are at a distance ‘ r’ apart. Find the minimum escape velocity for a particle of mass ‘m’ to be projected from the middle of these two masses:
(1) () + = 12 4 1 2 GMM V r
(2) () + = 12 4GMM V r
(3) () + = 12 2 1 2 GMM V r
(4) () + = 12 2GMM V r
Numerical Value Questions
21. The radius in kilometer to which the present radius of earth ( R = 6400 km) to be compressed so that the escape velocity is increased 10 times is
Geostationary and Polar Satellites
Single Option Correct MCQs
22. A satellite is seen every 6 hours over the equator. It is known that it rotates opposite to that of the earth’s direction. Then, the angular velocity (in radians per hour) of satellite about the centre of the earth will be
(1) π/2 (2) π/3
(3) π/4 (4) π/8
Level-III
Single Option Correct MCQs
1. The value of acceleration due to gravity on the surface of the earth is x. At an altitude of ‘h’ from surface of earth, its value is y. If ‘R’ is the radius of earth, then the value of h is
(1) 1 x R y
(2) 1 yR x
(3) yR x (4) x R y
2. Time periods of oscillation of a simple pendulum on the surface of the earth at depth d below the surface, and at a height d above the surface are T1, T2, and T3,respectively. Then, (d<<Rearth)
(1) T1>T2>T3 (2) T1<T2>T3
(3) T2>T1<T3 (4) T1<T2<T3
3. In older times, people used to think that the earth was flat. Imagine that the earth is indeed not a sphere of radius R, but an infinite plate of thickness H. What value of H is needed to allow the same gravitational acceleration to be experienced as on the surface of the actual earth? (Assume that the earth’s density is uniform and equal in the two models.)
(1) 2R/3 (2) 4R/3
(3) 8R (4) R/3
4. Variation of intensity of grav itational field (E) of earth with distance (r) from centre of earth is correctly represented by
(1) E r o r (2) E r o r (3) E r o r (4) E r o r
5. A thin rod of mass M and length L is bent into a semicircle. What is the gravitational force on a particle with mass m at the centre of curvature?
6. From a solid sphere of mass M and radius R, a spherical portion of radius 2 R
removed as shown in figure. Taking gravitational potential V = 0 at r = ∞, the potential at the centre of the cavity thus formed is____. (G = universal gravitational constant) (1) GM R (2)
R (4)
7. Two bodies of masses M1and M2 are placed at a distance ‘r’. The gravitational potential at the point, where the gravitational field is zero, is
(1) ()−++1122 2 G MMMM r
(2) () 12 2 G MM r −+
(3) ()2 12 G MM r −+
(4) () 12 G MM r −+
8. If three particles, each of mass M, are placed at the corners of an equilateral triangle of side a, the potential energy of the system and the work done, if the side of the triangle is changed from a to 2a are
(1) 2 33 , 2 GMGM aa
(2) 3322 , 2 GMGM aa
(3) 22 22 33 , 4 GMGM aa
(4) 2 33 , 2 GMGM aa
9. A body starts from rest from a point distance R0 from the centre of the earth. The velocity acquired by the body when it reaches the surface of the earth will be (R is the radius of the earth) (R0 > R)
(1) 0 11 2 RR GM
(2) 0 11 2 RR GM
(3) 0 11 GM RR
(4) 0 11 2 RR GM
10. A satellite is moving in a circular orbit around the earth with a speed equal to half the escape speed from the earth. If ‘R’ is the radius of the earth, then the height of the satellite above the surface of the earth is
(1) R/2 (2) 2R/3
(3) R (4) 2R
11. Imagine a light planet revolving around a very massive star in a circular orbit of radius R with a period of revolution T. If the gravitational force of attraction between the planet and the star is proportional to R–5/2 , then T2 is proportional to
(1) R3 (2) R7/2
(3) R3/2 (4) R9/2
12. The correct graph representing the variation of total energy (Et), kinetic energy (Ek)and potential energy (U ) of a satellite with its distance from the centre of the earth is
(1) ENERGY R U t e
(2) o ENERGY U R t e
(3) o ENERGY U K t e
(4) o ENERGY U R t E
13. An object of mass m, lying at rest, at a depth of R/2 from the earth’s surface, is to made an earth’s satellite at a height of R/2 from the earth’s surface. The work required to do this is (M–mass of the Earth, R–radius of the earth and neglect the rotation of earth)
(1) GMm R (2) 25 24 GMm R
(3) 17 24 GMm R (4) 15 2 GMm R
14. The escape velocity of a body on the earth’s surface is v e. A body is thrown up with a speed of kv e where k > 1. Assuming that the sun and other planets do not influence the motion of the body, the velocity of the body at the infinite distance is
(1) 2 1 e v k (2) 2 1 e vk
(3) () 2 1 e vk (4) 1 e v k
15. Two bodies, each of mass ‘M’, are separated by a distance 2 l . A particle of mass m is projected from the midpoint of the line joining their centres perpendicular to the line. The minimum initial speed of the mass m to escape from the gravitational field of the two bodies is
(1) 4 GM l (2) 2 GM l
(3) 2 4 GM l (4) GM l
16. In planetary motion, the areal velocity of a position vector of a planet depends on angular velocity ω and the distance of the planet from the sun ( r ). If so, choose the correct relations.
Numerical Value Questions
18. In a certain region of space, gravitational field is given by k E r =− . Taking the reference point to be at r = r0 with V = V0, the potential as a function of r is
Find the value of x.
19. The proper potential energy of gravitation interaction of matter (i.e., self energy), forming a uniform solid sphere of mass M and radius R, is given by 2 10 nGM R . Then, the value of n is ______. [Given: G = universal gravitational constant and gravitational potential at infinity = 0]
20. The angular velocity of the earth about its polar axis, so that the weight of the body at the equator will be zero, is ____10 –3 rad s–1
(1) B and C are true.
(2) B and D are true.
(3) A and C are true.
(4) A and D are true.
17. In the picture below, the planet orbits around the sun with a period of 40 months, and it takes 12 months to translate from point D to point E and 1 month from point B to C. The area of the ellipse is A. Mass of the satellite is m. The area of the shaded region is
21. A system consists of a thin ring of mass m and radius R and a very long uniform wire of linear mass density λ, oriented along the axis of the ring with one of its ends at the centre of the ring. Then the interaction force between the ring and the wire is KGm F R λ = Then, find the value of ‘ K’.
22. The mass density of a planet of radius R varies with the distance r from its centre as
2 2 ()1 o r r R . Then, the gravitational field is maximum at 5 rR x = . The value of x is _____. (Round off to nearest integer)
(1) A/12
(2) A/16
(3) A/20
(4) A/8
23. V 1 is gravitational potential at the centre of earth (in magnitude). V2 is magnitude of gravitational potential at a distance half the radius of the earth from the centre of the earth. Then, 2 1 5 2 Vn Vn + = , where n = ______.
24. Distance between the centres of two stars is 10 a . The masses of these two stars are M and 16 M , and their radii are a and 2 a, respectively. A body of mass m is to be projected from the surface of the larger star towards the smaller star. If the minimum initial speed of the body is , GM x a then find x () 52.24 = .
25. Two satellites S 1 and S 2 revolve around a planet in coplanar circular orbits in the same sense. Their periods of revolution are T and 8T, respectively. The angular speed of S2, as observed by an astronaut in S1, at the closest approach and at the farthest distance is ω 1 and ω2, respectively. Find the value of ω ω 1 2 9 .
26. A particle is fired vertically upwards with a speed of 15 km s–1. The speed of the particle (in km s–1) when it goes out of the earth’s gravitational pull is 5 a. Find the value of a [Given: ()2 2 -1 125kms e v = ]
27. An artificial satellite of mass m is moving in a circular orbit at a height equal to the radius R of the earth. Suddenly, due to internal explosion, the satellite breaks into two parts
THEORY-BASED QUESTIONS
Single Option Correct MCQs
1. Choose the correct statements about gravitational force.
A) It forms an action-reaction pair.
B) It is a non-central force.
C) It is a conservative force.
D) It is independent of medium between the masses.
(1) A, B, C, and D are correct.
(2) A, C, and D are correct.
(3) A, B, and D are correct.
of equal masses. One part of the satellite stops just after the explosion and then falls to the surface of the earth. The increase in the mechanical energy of the system (satellite + earth) due to explosion will be mgR x (g is the acceleration due to gravity on the surface of earth). Then, find x.
28. The time period of a satellite revolving above the earth’s surface at a height equal to R will be nR , where n is _______. (Given: g = π2 m/s2, R = radius of earth)
29. A large solid sphere has variable but radially symmetrical mass density changing with distance from the centre ( r ) as ρ = ρ 0 r n . For what absolute value of n , field will be independent of r?
30. At a height of 10 n km, above the pole, value of gravitational acceleration is same as that on the earth's surface at equator, assuming earth to be spherical. What is the approximate value of n? 2 11 2 [6.6710, Nm G kg =× mass of earth = 6 × 10 24 kg, Radius of earth = 6400 km] (Round off to nearest integer)
(4) B, C, and D are correct.
2. Two bodies of masses m1 and m2 are separated by a certain distance. If 12F is the force on m1 due to m2, and 21F is the force on m2 due to m1, then
(1) F12= F21
(2) 1221FF =
(3) 1221FF =−
(4) Both 1 and 3
3. Planets move round the sun due to (1) mutual attraction and repulsion between sun and planet (2) frictional force
(3) centrifugal force
(4) gravitational attraction between the sun and planets
4. The torque acting on a planet due to gravitational force of attraction is (1) ∞ (2) 0
(3) moderate (4) very high
5. The weight of an object in the coal mine, at sea level and at the top of the mountain ,are W1, W2, and W3, respectively. Then, (1) W1<W2>W3 (2) W1 =W2 =W3 (3) W1<W2<W3 (4) W1>W2>W3
6. Consider the earth to be a homogeneous sphere. Scientist A goes deep down in a mine and scientist B goes high up in a balloon. The gravitational field measured by (1) A goes on decreasing and that by B goes on increasing (2) B goes on decreasing and that by A goes on increasing (3) each decreases at the same rate (4) each decreases at different rates
7. An artificial satellite of the earth releases a packet. If air resistance is neglected, the point where the packet will hit will be (1) a head
(2) exactly below (3) behind (4) it will never reach the earth
8. An earth satellite is moved from one stable circular orbit to a farther stable circular orbit. Which one of the following quantities will increase?
(1) Gravitational force
(2) Gravitational PE
(3) Linear orbital speed
(4) Centripetal acceleration
9. The PE of a sattellite is – U. Its KE is (1) U
(2) U/2
(3) –U
(4) 2U
10. Choose the wrong statement. The escape velocity of a body from a planet depends upon (1) the mass of the body
(2) the mass of the planet
(3) the average radius of the planet
(4) the average density of the planet
11. If planets have same density but different radii, then the relation between escape velocity V and radius of planet R is
(1) ∝ VR
(2) ∝ 3/2 VR
(3) ∝ VR
(4) ∝ V1/R
12. A planet revolves in elliptical orbit around the sun. The linear speed of the planet will be maximum at
(1) A (2) B
(3) C (4) D
13. The period of a geostationary artificial satellite of the earth is
(1) 24 hr (2) 48 hr
(3) 12 hr (4) zero
14. Time period of polar satellite (for the earth) is around
(1) 24 hours (2) 100 minutes
(3) 84 minutes (4) 6 hours
15. A person sitting in a chair in a satellite feels weightless because
(1) the earth does not attract the objects in a satellite
(2) the normal force by the chair on the person balances the earth’s attraction
(3) the normal force is zero
(4) the person in satellite is not accelerated
Statement Type Questions
Each question has two statements: statement I (S-I) and statement II (S-II). Mark the correct answer as
(1) if both statement I and statement II are correct,
(2) if both statement I and statement II are incorrect
(3) if statement I is correct but statement II is incorrect,
(4) if statement I is incorrect but statement II is correct
16. S-I : The law of gravitation holds good for any pair of bodies in the universe.
S-II : The weight of any person becomes zero when the person is at the centre of the earth.
17. S-I : There are points outside the earth where the value of g is equal to its value at the equator. (Take the effect of bulging of earth and its rotation into account)
S-II : There are points outside the earth where the value of g is equal to its value at the poles. (Take the effect of bulging of earth and its rotation into account)
18. S-I : If E is the total energy of a satellite moving around the earth, then its potential energy will be E/2.
S-II : The kinetic energy of a satellite revolving in an orbit is equal to half the magnitude of the total energy E
Asser tion and Reason Questions
In each of the following questions, a statement of Assertion (A) is given, followed by a corresponding statement of Reason (R). Mark the correct answer as
(1) if both (A) and (R) are true and (R) is the correct explanation of (A),
(2) if both (A) and (R) are true but (R) is not the correct explanation of (A), (3) if (A) is true but (R) is false, (4) if both (A) and (R) are false.
19. (A) : The escape velocities of planet A and B are same. But A and B can be of unequal mass.
(R) : The product of their mass and radius must be same, M1R1 =M2R2
20. (A) : Angular momentum of the planet orbiting around the sun remains constant.
(R) : Torque acting on the revolving planet is zero.
21. (A) : An astronaut in an orbiting space station above the earth experiences weightlessness.
(R) : An object moving around the earth under the influence of earth’s gravitational force is in a state of ‘free-fall’.
22. (A) : A particle placed inside a spherical shell away from its centre will remain stationary.
(R) : Gravitational field is zero inside a spherical shell.
23. (A) : Gravitational potential is equal to the work done in bringing a unit mass from infinity to a point inside gravitational field.
(R) : Gravitational potential is zero inside a shell.
24. (A) : Acceleration due to gravity at a depth ‘d’ from the surface of earth is 1 d gR
(R) : A body becomes massless at the centre of earth.
25. (A) : The speed of revolution of an artificial satellite revolving very near the earth’s surface is 8 kms–1.
(R) : Orbital velocity of a satellite becomes independent of height of near earth.
26. (A) : Generally, the path of a projectile from the earth is parabolic but it is elliptical for projectiles going to a very large height.
(R) : The path of a projectile is independent of the gravitational force of earth.
JEE ADVANCED LEVEL
Multiple Option Correct MCQs
1. If the gravitational force between sun and earth is inversely proportional to r3, where r is distance between earth and sun, then which of the following statements are correct?
(1) Its orbit around sun can’t be a stable elliptical orbit.
(2) Its orbit around sun can’t be a stable circular orbit.
(3) Motion of earth around sun will be in one plane.
(4) Tidal forces acting on earth will be inversely proportional to r4
2. Two objects of mass m and 4m are at rest at an infinite separation. They move towards each other under mutual attraction, If G is the universal gravitational constant, then at separation r,
(1) the sum of energy of the two objects is negative
(2) their relative velocities of approach is
1/2 10Gm r in magnitude.
27. (A) : The planet appears to move slower when they are farther from the sun than when they are nearer.
(R) : Planetary motion does not obey the law of conservation of angular momentum.
28. (A) : An astronaut inside a massive spaceship orbiting around the earth will experience a finite but small gravitational force.
(R) : The centripetal force necessary to keep the spaceship in orbit around the earth is provided by gravitational force between the earth and the spaceship.
(3) the sum of kinetic energy of the objects is 2 4Gm r
(4) the sum of angular moment of both the objects is zero about any point
3. A particle is dropped from a height equal to the radius of the earth above the tunnel dug through the earth, as shown in the figure.
R: Radius of earth; M: Mass of earth C o R
(1) Particle will oscillate through the earth to a height R on both sides.
(2) Particle will execute simple harmonic motion.
(3) Motion of the particle is periodic.
(4) Particle passes the centre of earth with a speed 2 R GM =
4. Inside a uniform spher e of mass M and radius R, a cavity of radius R/3 is made in the sphere, as shown.
(1) Gravitational field inside the cavity is uniform.
(2) Gravitational field inside the cavity is non-uniform.
(3) The escape velocity of a particle projected from point A is 88 45 GM R
(4) Escape velocity is defined for earth and particle system only.
Numerical Value Questions
5. Three particles of equal mass M are situated at the vertices of an equilateral triangle of side l. The speed of each particle, so that they move on a circular path, is . GM xl Find the value of x.
6. Consider the arrangement shown in figure. A thin rod (mass M and length 4R) and a sphere of equal mass M and radius R are located as shown in figure. The centre of the sphere is at a distance 2 R from the centre of the rod on its perpendicular bisector. Assume that the system of rod and sphere is in gravity-free space. If the force acting on the rod due to sphere is given by 2 2 , GM nR find the value of n
7. Angular speed of rotation of earth is ω 0. A train is running along the equator at a speed v from west to east, with respect to surface of earth. A very sensitive balance inside the train shows the weight of the object as W 1. During the return journey, when the train is running at same speed from east to west (with respect to surface of earth), the balance shows the weight of the object to be W 2. Weight of the object shown by the balance at the poles is W0 The value of W2 –W1 was found out to be 00 pWv g ω (g is the acceleration due to gravity on the surface of the earth). Find the value of ‘ p’.
8. A spherical planet of radius R has a uniform density ρ and does not rotate. If the planet is made up of some liquid, the pressure at any point at distance r from the centre is ηπGρ2 (R2 –r2). Find η.
9. A satellite is launched in the equatorial plane in such a way that it can transmit signals up to 60° latitude on the earth. The angular velocity of the satellite is 3 GM nR . Find n
10. A satellite is transferred from one circular orbit to another circular orbit by Hohmann Transfer orbit maneuver, by temporarily entering in an intermediate elliptical orbit by increasing speed instantly by firing propeller. To transfer a satellite in circular orbit of radius r to another orbit of radius 3r, additional energy to be imparted at point
P (in orbit of radius r ) is GM n r . Find n. (M = Mass of planet)
Passage-based Questions
Passage-I:
The gravitational force is a central force. Acceleration due to gravity has different values for different points on the earth’s surface. g varies with altitude or depth. Gravitational acceleration has different values for different planets and depends on mass and radius of planets. For the given density of planets, gravitational acceleration is directly proportional to radius of planets.
11. The value of g at the equator is less than that of pole due to
(1) shape of earth only (2) rotation of earth only
(3) both shape and rotation of earth
(4) mass of the sun
12. The object on equator will have zero weight, when
(1) Earth rotates 17 times faster than the present rotational speed
(2) duration of day becomes 4 hrs
(3) Earth rotates 8 times faster than the present rotational speed
(4) duration of day becomes 8 hrs
Passage-II:
The value of acceleration due to gravity is g1 at a height h (R = radius of the earth) from the surface of the earth. It is again equal to g1 and a depth d below the surface of the earth.
13. If 2 hR = , then the ratio d R
Find 9k.
is equal to k.
14. If 4 hR = , then the ratio d R
Find 25p.
Passage-III:
is equal to p
An artificial satellite is moving in a circular orbit with a speed equal to 1/4th of magnitude of escape velocity from earth.
15. Height of satellite above earth’s surface is ______×102 km.
16. Kinetic energy with which it is to be projected from the surface of earth so as to orbit is . 16 x mgR Find x. (R is the radius of earth)
Passage-VI:
An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of escape velocity from the earth. (Radius of earth = 6400 km, g at earth’s surface = 10 ms–2)
17. The height of the satellite above the earth’s surface is n × 104 km. The value of n is ____.
18. If the satellite is stopped suddenly in its orbit and allowed to fall freely onto the earth, then the speed with which it hits the surface of the earth is ______ km/s.
Matrix Matching Questions
19. A planet of mass M has two natural satellites with masses m1 and m2. The radii of their circular orbits are R 1 and R 2, respectively. Ignore the gravitational force between the satellites. Define v 1 , L 1 , K 1 , and T 1 to be, respectively, the orbital speed, angular momentum, kinetic energy, and time period of revolution of satellite1; and v 2 , L 2 , K 2 and T 2 to be the corresponding quantities of satellite 2. Given
and
Match the ratios in List-I to the numbers in List-II.
1 2 L L (II) 1
1 2 K K (III) 2
BRAIN TEASERS
1. A body stretches a spring by a certain amount at the equator. Assuming the earth to be a perfect sphere of radius R, g, the acceleration due to gravity on the surface of the earth, and ω , the angular speed of the earth, the height at which, at poles, it will stretch the spring by the same amount at the equator is
(1) R
(2) () 2 11/ RRg
(3)
(4)
2. Given below are two statements. One is labelled Assetion (A) and the other is labelled Reason (R).
Assertion (A) : For a mass M kept at the centre of a cube of side ‘a’, the flux of gravitational field passing through its sides is 4πGM.
Reason (R) : If the direction of a field due to a point source is radial and its dependence on the distance ‘ r ’ for the source is given as 1/ r 2 , its flux through a
(D) 1 2 T T (IV) 8
(A) (B) (C) (D)
(1) IV II I III
(2) III II IV I
(3) II III I IV
(4) II III IV I
closed surface depends only on the strength of the source enclosed by the surface and not on the size or shape of the surface.
In light of the above statements, choose the correct answer from the options given below.
(1) Both (A) and (R) are true and (R) is the correct explanation of (A).
(2) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(3) (A) is true but (R) is false.
(4) Both (A) and (R) are false.
3. A spherical body of radius R consists of a fluid of constant density and is an equilibrium under its own gravity. If P ( r ) is the pressure at r ( r < R ), then the correct option(s) is (are)
(1) P(r = 0) = 0
(2) () () 3/4 63 2/380 PrR PrR = = =
(3) () () 3/5 16 2/521 PrR PrR = = =
(4) () () /2 20 /327 PrR PrR = = =
4. Find the gravitational force of attraction between the ring and sphere, as shown in the diagram, where the plane of the ring is perpendicular to the line joining the centres. If 8 R is the distance between the centres of a ring (of mass ‘m’) and a sphere (mass M’)where both have equal radius R. m R x √8 R M R y
(1) 2 8 . 27 GmM R
(2) 2 1 . 38 GMm R
(3) 2 22 . 3 GMm R
(4) 2 8 9 GmM R
5. Two earth’s satellites (each of mass m and considered as point masses) move in a common plane along circular orbits in the same direction. The orbital radius of one satellite is r while that of the other satellite is r– ∆ r (Here, ∆ r<<r). Then,
(1) time interval that separates the periodic approaches of the satellites to each other over the minimum distan ce is nearly () 5/2 1/2 4 3 r GMr π ∆
(2) time interval that separates the periodic approaches of the satellites to each other over the minimum distance is nearly () π ∆ 5/2 1/2 2 3 r GMr
(3) at the instant of closest approach, the pseudo force on outer satellite, as seen by inner one, is m(Δr)( ω 2 rel) outward, where ω rel is relative angular velocity of approach between two satellites
(4) difference of angular velocities of two satellites is ()1/2 5/2 3 2 GMr r ∆
6. An artificial satellite of a planet revolves in a circular orbit, whose radius exceeds the radius of the planet by 7 times. In the process of motion, the satellite experiences slight resistance due to cosmic dust. Assuming the resistive force depends on the speed of the satellite as F = kv2 (where k is a constant), the time taken by the satellite to stay in orbit until it falls on to the planet’s surface is () 1 m x kgR .
(Given m = mass of the satellite, g = acceleration due to gravity at the surface of planet, and R = radius of the planet) Find the value of x.
7. A satellite of mass m is orbiting around the earth (mass M, radius R) in a circular orbit of radius 4R. It starts losing energy slowly at a constant rate dE dt −=η due to friction.
Find the time (t) in which the satellite will spiral down to the surface of the earth.
(1) 3 8 GMm Rη (2) 3 48 GMm Rη
(3) GMm Rη (4) 8 GMm Rη
8. A particle is projected from the surface of the earth with velocity equal to its escape velocity at 45° with horizontal. The angle of its velocity with horizontal at height h = R, where horizontal at some point means a line parallel to tangent on earth just below that point, is ____ (in degrees).
9. The minimum and maximum distance of a satellite from the centre of the earth are 2 R and 4 R , respectively, where R is radius of earth, M is mass of earth and m is mass of satellite. Then, choose the correct statement.
(1) Angular momentum of satellite about centre of earth is 2 8 3 GMRm .
FLASHBACK (Previous JEE Questions)
JEE Main
1. A light planet is revolving around a massive star in a circular orbit of radius R with a period of revolution T. If the force of attraction between planet and star is proportional to R−3/2 then choose the correct option: (2024)
(1) T2 ∝ R5/2 (2) T2 ∝ R7/2
(3) T2 ∝ R3/2 (4) T2 ∝ R3
2. The acceleration due to gravity on the surface of earth is g. If the diameter of earth reduces to half of its original value and mass remains constant, then acceleration due to gravity on the surface of earth would be: (2024)
(1) g/4 (2) 2g (3) g/2 (4) 4g
3. Given below are two statements. One is labelled Assertion (A) and the other is labelled Reason (R).
Assertion (A) : The angular speed of the moon in its orbit about the earth is more than the angular speed of the earth in its orbit about the sun.
Reason (R) : The moon takes less time to move around the earth than the time taken by the earth to move around the sun.
In light of the above statements, choose the correct answer from the options given below. (2024)
(2) Radius of curvature at the point when it is at minimum distance from the centre of earth is 8R/3.
(3) Total energy of satellite is 6 GMm R .
(4) Minimum speed of satellite is 6 GMm R
(1) Both (A) and (R) are true and (R) is the correct explanation of (A).
(2) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(3) (A) is true but (R) is false.
(4) (A) is false but (R) is true.
4. At what distance above and below the surface of the earth a body will have same weight, (take radius of earth as R) (2024)
(1) 5RR (2) 3 2 RR
(3) 2 R (4) 5 2 RR
5. A planet takes 200 days to complete one revolution around the Sun. If the distance of the planet from Sun is reduced to one fourth of the original distance, how many days will it take to complete one revolution? (2024)
(1) 25 (2) 50
(3) 100 (4) 20
6. The gravitational potential at a point above the surface of earth is –5.12 × 10 7 J/kg and the acceleration due to gravity at that point is 6.4 m/s2. Assume that the mean radius of earth to be 6400 km. The height of this point above the earth’s surface is: (2024)
(1) 1600 km
(2) 540 km
(3) 1200 km
(4) 1000 km
7. Escape velocity of a body from earth is 11.2 km/s. If the radius of a planet be one-third the radius of earth and mass be one-sixth that of earth, the escape velocity from the planet is: (2024)
(1) 11.2 km/s (2) 8.4 km/s (3) 4.2 km/s (4) 7.9 km/s
8. Four identical particles of mass m are kept at the four corners of a square. If the gravitational force exerted on one of the masses b y the other masses is 2 2 221 32 Gm L
+
, the length of the sides of the square is (2024)
(1) 2 L (2) 4 L (3) 3 L (4) 2 L
9. The mass of the moon is 1/144 times the mass of a planet and its diameter 1/16 times the diameter of a planet. If the escape velocity on the planet is v, the escape velocity on the moon will be: (2024)
(1) 3 v (2) 4 v
(3) 12 v (4) 6 v
10. Assuming the earth to be a sphere of uniform mass density, the weight of a body at depth = 2 dR from the surface of earth, if its weight on the surface of earth is 200 N, will be
(Given: R = radius of earth) (2023)
(1) 500 N (2) 400 N
(3) 300 N (4) 100 N
11. Given below are two statements.
Statement I : Rotation of the earth shows effect on the value of acceleration due to gravity (g).
Statement II : The effect of rotation of the earth on the value of ‘g’ at the equator is minimum and that at the pole is maximum.
In light of the above statements, choose the correct answer from the options given below. (2023)
(1) Statement I is false but statement II is true.
(2) Both statement I and statement II are false.
(3) Both statement I and statement II are true.
(4) Statement I is true but statement II is false.
12. A planet has double the mass of the earth. Its average density is equal to that of the earth. An object weighing W on earth will weigh on that planet. (2023)
(1) 22/3W (2) 21/3W (3) 2W (4) W
13. A body of mass m is taken from earth surface to the height h equal to twice the radius of earth (Re). The increase in potential energy will be (g = acceleration due to gravity on the surface of Earth) (2023)
(1) e 2 3 mgR (2) 3mgRe
(3) e 1 3 mgR (4) e 1 2 mgR
14. Two satellites A and B move round the earth in the same orbit. The mass of A is twice the mass of B. The quantity which is same for the two satellites will be (2023)
(1) speed (2) kinetic energy
(3) total energy (4) potential energy
15. If the earth suddenly shrinks to 1/64th of its original volume, with its mass remaining the same, the period of rotation of earth becomes 24/x h the value of x is. (2023)
16. Two objects of equal mass, placed at a certain distance from each other, attract each other with a force F. If one-third mass of one object is transferred to the other object, then the new force will be (2022)
(1) 2 9 F (2) 16 9 F (3) 8 9 F (4) F
17. Given below are two statements.
Statement I : The law of gravitation holds good for any pair of bodies in the universe.
Statement II : The weight of any person becomes zero when the person is at the centre of the earth. In the light of the above statements, choose the correct answer from the options given below.
In light of the above statements, choose the correct answer from the options given below. (2022)
(1) Both statement I and statement II are true.
(2) Both statement I and statement II are false.
(3) Statement I is true but statement II is false.
(4) Statement I is false but statement II is true.
18. Two satellites A and B having masses in the ratio 4 : 3 are revolving in circular orbits of radii 3r and 4r, respectively, around the earth. The ratio of total mechanical energy of A to B is (2022)
(1) 9 : 16
(2) 16 : 9
(3) 1 : 1
(4) 4 : 3
19. If the acceleration due to gravity experienced by a point mass at a height h above the surface of earth is same as that of the acceleration due to gravity at a depth α h (h << R e ) from the earth surface, the value of α will be _______. (use R e = 6400 km) (2022)
20. Inside a uniform spherical shell
I. the gravitational field is zero.
II. the gravitational potential is zero.
III. the gravitational field is same everywhere. IV. the gravitation potential is same everywhere.
V. All of the above
Choose the most appropriate answer from the options given below. (2021)
(1) I, III, and IV (2) Only V (3) I, II, and III (4) II, III, and IV
21. A body of mass 2M splits into four masses {m, M–m, m, M–m},which are rearranged to form a square, as shown in the figure. The ratio of M/m, for which the gravitational potential energy of the system becomes maximum, is x : 1. The value of x is________ (2021)
JEE Advanced
22. Two satellites P and Q are moving in different circular orbits around the Earth (radius R). The heights of P and Q from the Earth surface are h P and h Q, respectively, where = . 3 P hR The accelerations of P and Q due to Earth’s gravity are gP and gQ, respectively. If = 36 25 P Q g g , what is the value of hQ?(2023)
(1) 3 5 R (2) 6 R (3) 6 5 R (4) 5 6 R
23. Two spherical stars A and B have densities ρ A and ρ B, respectively. A and B have the same radius, and their masses MA and MB are related by MB = 2MA. Due to an interaction process, star A loses some of its mass, so that its radius is halved, while its spherical shape is retained, and its density remains ρ A. The entire mass lost by A is deposited as a thick spherical shell on B with the density of the shell being ρ A. If v A and v B are the escape velocities from A and B after the interaction
process, The ratio 1/3 10 15 B A vn v = The value of n is (2022)
24. The distance between two stars of masses 3MS and 6MS is 9R. Here, R is the mean distance between the centres of the Earth and the Sun, and MS is the mass of the Sun. The two stars orbit around their common centre of mass in circular orbits with period nT, where T is the period of Earth’s revolution around the Sun. The value of n is ____. (2021)
CHAPTER TEST – JEE MAIN
Section – A
1. The weight of a body on earth is 400 N. Then, the weight of the body when taken to a depth half of the radius of the earth will be (1) 300 N (2) 100 N (3) zero (4) 200 N
2. If the angular velocity of earth’s spin is increased such that the bodies at the equator start floating, the duration of the day would be approximately [Take g = 10 ms –2 , the radius of earth, R = 6400×103 m, π = 3.14] (1) 60 minutes (2) does not change (3) 1200 minutes (4) 84 minutes
3. A uniform rod of mass M and length L is considered. Find the gravitational field intensity at point P on perpendicular bisector.
25. Consider a spherical gaseous cloud of mass density ρ(r) in a free space, where r is the radial distance from its centre. The gaseous cloud is made of particles of equal mass m moving in circular orbits about their common centre with the same kinetic energy K. The force acting on the particles is their mutual gravitational force. If ρ(r) is constant with time, the particle number density n(r) = ρ(r)/m is (2019) [G is universal gravitational constant]
(1) + 22
(3) + 22 22 4 GM dLd (4) none
4. A solid sphere of mass ‘ M ’ and radius ‘ a ’ is surrounded by a uniform concentric spherical shell of thickness 2 a and mass 2 M . The gravitational field at distance 3 a from the centre will be (1) 2 2 9 GM a (2) 2 9 GM a (3) 2 3 GM a (4) 2 2 3 GM a
5. The diagram showing the variation of gravitational potential of earth with distance from the centre of earth is
(1) O V R r (2) O V R r
(3) O V R r (4) O V R r
6. Two identical spherical masses are kept at some distance, as shown. Potential energy, when a mass m is taken from surface of one sphere to the other, m
(1) increases continuously (2) decreases continuously (3) first increases, then decreases (4) first decreases, then increases
7. The time period of a satellite, revolving above the earth’s surface at a height equal to R will be (Given: g = π2 m/s2, R = Radius of earth)
(1) 32 R (2) 2 R
(3) 4 R (4) 8 R
8. The energy required to shift a satellite from orbital radius ‘ r ’ to radius ‘2 r ’ is E. What energy will be required to shift satellite from orbital radius ‘2r’ to ‘3r’?
(1) E (2) E/2
(3) 3E/4 (4) E/3
9. A planet having mass 16 M e and radius 4Re, where M e and R e are mass and radius of earth, respectively, has escape velocity (in km/s) (Given: escape velocity on earth v e = 11.2 × 103 m/s)
(1) 67.2 (2) 16.8
(3) 33.6 (4) 22.4
10. Speed of a planet in an elliptical orbit with semi major axis a about sun of mass M at a distance r from sun is
(1) 21 GM ra
(3) 12 GM ra
(2) 11 GM ra
(4) 2 2 GMr a
11. A body is fired with a velocity of magnitude
2 gRvgR << at an angle of 30° with the radius vector of the earth. If, at the highest point, the speed of the body is v/4, the maximum height attained by the body is equal to ( R = radius of earth; use the appropriate assumptions)
(1) 2 8 v g (2) R (3) 2 R (4) 2R
12. A satellite is orbiting around earth in orbit, very close to earth surface, in equatorial orbit, with a rotation period of 1.5 hrs. If, at an instant, it is vertically above a point P on earth surface, find the time after which it will again come above P (if it is moving from east to west).
(1) 1.1 hrs (2) 1.2 hrs
(3) 1.3 hrs (4) 1.4 hrs
13. A black hole is an object whose gravitational field is so strong that even light cannot escape from it. To what approximate radius would earth (mass = 5.98 × 1024 kg) have to be compressed to be a black hole?
(1) 10–9 m (2) 10–6 m (3) 10–2 m (4) 102 m
14. The minimum and maximum distances of a satellite from the centre of the earth are 2R and 4R, respectively, where R is radius of earth and m is mass of the earth. The maximum speed of the satellite is
(1) 6 Gm R (2) 2 3 Gm R (3) 3 Gm R (4) 2 5 Gm R
15. A ssuming that a tunnel is dug across the earth, passing through its centre, find the time that a particle takes to reach the centre of the earth, if it is projected into the tunnel from the surface of earth with speed needed for it to escape the gravitational field of earth. (1)
(1) 2 8 GMm R (2) 2 4 GMm R
(3) 2 3 8 GMm R (4) 2 83 GMm R
16. A point mass m and a uniform thin rod (mass M and length L) are shown in figure. Find the magnitude of gravitational force of attraction between them m M,L a
(1) GMm ah
(2) () + GMm aaL
(3) () + 2 GMm aL
(4) () + 2 4 2 GMm aL
17. A uniform ring of mass ‘m’ and radius ‘R’ is placed directly above a uniform sphere of mass M and of equal radius. The centre of the ring is directly above the centre of the sphere at a distance 3R , as shown in the figure. The gravitational force exerted by the sphere on the ring will be
18. A straight rod of length L extends from x = a to x = L + a. The linear mass density of the rod varies with x -coordinate as λ = A + Bx 2 The gravitational force experienced by a point mass m at x = 0 is
(1)
+ A GmBL a
(2)
+ + A GmBL aL
(3)
−+ + 11 GmABL aaL
(4) + 2 BLA Gm aa
19 A tunnel is dug along a diameter of the earth of mass M e and radius R e . The force on a particle of mass m, placed in the tunnel at a distance r from the centre is
(1) 3 e e GMm r R (2) 3 e e GMm Rr
(3) 3 ee GMmR r (4) 2 e e GMm r r
20. Three identical particles, each of mass m, are at the vertices of an equilateral triangle of side L. If they are to be in equilibrium, the speed with which they must revolve under the influence of one another’s gravity in a circular orbit circumscribing the triangle is
22. The radius of the earth is R. The height of a point vertically above the earth’s surface at which the accelearation due to gravity becomes 1% of its value at the surface is R.
23. The density of a solid sphere of mass M and radius R varies with distance r from its centre as k r ρ= , where k is a constant. The gravitational field due to this sphere at a distance 2R from its centre is given by kG N π Then, the value of N is______.
24. Two satellites of masses m and 3m revolve around the earth in a circular orbit of radii r and 3r, respectively. The ratio of orbital speeds of the satellites, respectively, is k Then, the value of k is_____.
Section-B
21. A body weighs 64 N on the surface of the earth. What is the gravitational force (in N) on it due to the earth at a height equal to one-third of the radius of the earth?
CHAPTER TEST – JEE ADVANCED
2022-P2 Model
Section – A [Integer Value Questions]
1. An infinite number of particles, each of mass 1 kg, are placed on the positive x-axis at 1 m, 2 m, 4 m, 8 m…….from the origin. The magnitude of the resultant gravitational force on 1 kg mass kept at the origin is aG F b = Then a + b =______.
2. Four identical particles of equal mass 1 kg are made to move along the circumference of radius 1 m under the action of their own mutual gravitational attraction.
25. A satellite is launched into a circular orbit of radius ‘R’ around the earth while a second satellite is launched into an orbit of radius 1.02 R. The percentage change in the time periods of the two satellites is______.
The speed of each particle will be_________
(122)
3. A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F 1 on a particle placed at P , at a distance 2 R from the centre O of the sphere. A spherical cavity of radius R /2 is now made in the sphere, as shown in figure. The sphere with cavity now applies a gravitati onal force F 2 on the same particle placed at P . The 2 1 F F is 9 x . The value of x is______.
4. A particle is projected from a point () 3,0PR along the axis of an annular uniform disc of mass M and inner and outer radii R and 6 R , respectively, as shown in the figure. If the escape velocity of the particle from this point is 2 GM NR × , find N ( 3 R,0)
7. A ring of mass m and radius 3R is rotating with a constant angular speed of ω. + 2 BLA Gm aa around a planet of mass M and radius R. The centres of ring and planet coincide with each other. Tension in the ring is given as = π 2 . 3 GMm T nR Find the value of n. (Assume elements of the ring do not gravitationally interact with each other) ω
5. Distance between the centres of two stars is 10 a. Masses of these stars are M and 16M, and their radii are a and 2a, respectively. A body of mass ‘m’ is fired straight from the surface of the larger star towards the surface of the smaller star. The minimum initial speed of the body to reach the surface of the smaller star is 3x 2a GM
. Find ‘x’.
6. A satellite revolves around a planet of radius R in an orbit of radius nR. Due to change in weather, atmospheric density increases and the satellite experiences a resistance force, which depends on velocity v of satellite as 2 ˆ fvv =−α , where α is a positive constant. The time after which the satellite hits the surface of the planet is found to be
0 1 xm tn ygR = α The value of (x + y), when x and y are the least possible integers, is________.
8. Gravitational potential at the centre of the earth is V, assuming the earth as a uniform sphere of radius R with uniform density. If the central portion of earth of radius R/2 is removed, leaving behind a cavity (assume), gravitational potential on the surface of the earth will change by () 2 31 NV N + , where N =_____.
Section – B
[Multiple Option Correct MCQs ]
9. A tunnel is dug along a chord of the earth at a perpendicular distance R/ 2 from the earth’s centre. The wall of the tunnel may be assumed to be frictionless. A particle is released from one end of the tunnel. The pressing force by the particle on the wall and the acceleration of the particle varies with x (distance of the particle from the centr e of the earth) according to
(1) Pressing Force
x = R 2 x=R x
(2) Pressing Force
(3) Acceleration
x = R 2 x=R x
x = R 2 x=R x
(4) Acceleration
x = R 2 x=R x
10. There are two concentric spherical shells of masses m1 and m2 and radii R1 and R2(R2 > R1). The force on a particle of mass m0, kept at a distance ‘r’ from the common centre, is F. Then,
(1) () + => 120 2 2 ,when Gmmm FrR r
(2) =<< 10 12 2 ,when Gmm FRrR r
(3) F = 0, when r < R1
(4) 21 2 Gmm F r =
sectional diameter is d(d<<r). C is a small ball of mass m ( m << M ); which is moving freely inside the tunnel, without friction, in a uniform circular motion. Take acceleration due to gravity on surface of A as g.
A B R r C d
(1) If contact force on C is double the gravitational force, its time period is
2 3 R g π .
(2) I f contact force on C is double the gravitational force, its time period is 2 R g π
(3) If contact force is zero, time period is T, radius of tunnel is r, then T ∝ r.
(4) If contact force is zero, time period is T, radius of tunnel is r, then T would be independent of r
12. P is a point on the axis of a ring centred at O. The gravitational field at P does not change its magnitude but changes direction by 90° when a spherical shell of same mass and same radius as the ring is placed in the region with centre at O'. Then,
(1) O'P ≥ R
(2) O'P>R
(3) O'PO is an acute angled triangle
(4) O'PO is obtuse
11. ‘A’ is a huge planet with uniform mass distribution of mass M and radius R. B is a circular tunnel made in it, concentric with planet. Radius of tunnel is 2 R r =
; cross-
13. Consider a planet of mass M and radius R, as shown in the figure. A particle is projected from the surface of the planet at an angle θ from radial direction and with velocity equal to the orbital velocity of circular orbit at that point. Then, choose the correct statement(s).
(1) Maximum distance of particle from centre of the planet is ( R + Rcos θ).
(2) Path of the particle is parabola.
(3) Time taken by the particle to return t o the surface of planet is 3 R GM .
(4) Time taken by the particle to reach the surface of planet is equal to 3 ð R GM .
14. Three planets of same density and with radii R1, R2 and R3, such that R1=2R2=3R3, have gravitational fields on the surface g1, g2, and g3, and escape velocity v1, v2, and v3, respectively. Then,
(1) = 1 2 1 2 g g (2) = 1 3 3 g g
(3) = 1 2 2 v v (4) = 1 2 1 3 v v
Section – C [Single Option Correct MCQs ]
15. An object is weighed at the North Pole by a beam balance and a spring balance, giving readings of W B and W S, respectively. It is again weighed in the same manner at the equator, giving reading of W B ' and W S ', respectively. Assume that the acceleration due to gravity is the same everywhere and that the balances are quite sensitive. Then,
a) WB = WS
b) WB' = WS '
c) WB' = WB
d) Ws' < WS
(1) only a and b are true
(2) only a, b, and d are true (3) only a, c, and d are true (4) all are true
16. A solid sphere of mass M and radius R has a spherical cavity of radius R/2 such that the centre of cavity is at a distance R/2 from the centre of the sphere. A point mass m is placed inside the cavity at a distance R/4 from the centre of the sphere. The gravitational pull between the sphere and the point mass m is
(1) 2 11 R GMm (2) 2 14 R GMm
(3) 2 2R GMm (4) 2 R GMm
17. The magnitudes of the gravitational fields at distances r 1 and r 2 from the centre of a uniform sphere of radius R and M are F 1 and F2, respectively. Then,
(a) =<<11 12 22 if and Fr rRrR Fr
(b) =>> 2 12 12 21 if and Fr rRrR Fr
(c) =>>11 12 22 if and Fr rRrR Fr
(d) =<< 2 11 12 22 if and Fr rRrR Fr
(1) only a and b are true (2) only b and c are true (3) only a,b and d are true (4) all are true
18. A satellite rotating around the earth is orbiting with a speed V 0. If it is suddenly stopped, then it reaches the earth’s surface with a velocity [Ve→ escape velocity on surface)
(1) 22 eo vv (2) 22 /2 eo vv
(3) 22 2 eo vv (4) 22 2 eo vv
ANSWER KEY
JEE Main Level
Theory-based Questions (1) 2 (2) 4 (3) 4 (4) 2 (5) 1 (6) 4 (7) 4 (8)
JEE Advanced Level
Brain Teasers
Flashback (1) 1 (2) 4 (3) 1 (4) 4 (5)
(11) 4 (12) 2 (13) 1 (14) 1 (15) 16 (16) 3 (17) 1 (18) 2 (19) 2 (20) 1 (21) 2 (22) 1 (23) 2.3 (24) 9 (25) 4
Chapter Test-JEE-Mains (1) 4 (2) 4 (3) 2 (4) 3 (5) 3 (6) 3 (7) 1 (8) 4 (9) 4 (10) 1 (11) 2 (12) 4 (13) 3 (14) 2 (15) 3 (16) 2 (17) 3 (18) 3 (19) 1 (20) 2 (21) 36 (22) 9 (23) 2 (24) 3 (25) 3
Chapter Test-JEE-Advanced (1) 4 (2) 2 (3) 7 (4) 5 (5) 5 (6)
(9) 2,3 (10) 1,2,3 (11) 1,4 (12) 1,4 (13) 1,4 (14) 2,3 (15) 3 (16) 3 (17) 1 (18) 4
