2025 FINAL EXAM PHYS 204 Mechanics notes + examples 2025 Concordia University

2025 FINAL EXAM PHYS 204 Mechanics notes + examples 2025 Concordia University

Notes On

MECHANICS (Physics 204)

2

Table of Contents . ......................................................................................................................................................... 3 Chapter 1 Review of Vectors ...................................................................................................... 4 1.1 Mechanics ............................................................................................................................ 4 1.2 Physical Quantities .............................................................................................................. 4 1.3 Addition of Vectors ............................................................................................................. 4 1.4 Multiplication of vectors ..................................................................................................... 6 1.5 Component of a Vector ....................................................................................................... 6 1.6 Relationships Among Unit Vectors ˆi, ˆj, kˆ ........................................................................ 7 1.7 Solved Examples .................................................................................................................. 9 Chapter 2 Units and Significant Figures ................................................................................. 12 2.1 Introduction ........................................................................................................................ 12 2.2 Length ................................................................................................................................ 12 2.3 Mass ................................................................................................................................... 12 2.4 Time ................................................................................................................................... 13 2.5 Volume, Density, and Angle............................................................................................. 13 2.6 Significant Figures ............................................................................................................ 14 Chapter 3 Kinematics .............................................................................................................. 16 3.1 Position Vector and Displacement.................................................................................... 16 3.2 Newton’s Laws of Motion ................................................................................................. 19 3.3 Friction ............................................................................................................................... 21 3.4 Weight ................................................................................................................................ 22 3.5 Equilibrium of Forces ....................................................................................................... 22 3.6 Conditions of Equilibrium of a Particle ........................................................................... 22 3.7 Solved Examples ................................................................................................................ 23 Chapter 4 Rectilinear Motion ................................................................................................. 27 4.1 Rectilinear Motion with Constant Acceleration .............................................................. 27 4.2 Solved Examples on Rectilinear Motion ......................................................................... 28 5.1 Newton’s Law of Gravitation: It states .......................................................................... 33 5.2 Free-Fall under Gravitational Force............................................................................... 33 5.3 Gravitational Field ............................................................................................................ 34 5.4 Solved Examples ................................................................................................................ 34 Chapter 6 Motion in a plane................................................................................................... 39 6.1 Motion of a projectile ........................................................................................................ 39 6.2 Solved Examples on Projectiles ........................................................................................ 40 6.3 Solved Examples on Motion in a Circle .......................................................................... 46 Chapter 7 Work and Energy .................................................................................................. 51 7.1 Work .................................................................................................................................. 51 7.2 Power ................................................................................................................................. 51 7.3 Energy and its Conservation ............................................................................................. 52 7.4 Potential and Kinetic Energy ........................................................................................... 53 7.5 Transformation of Potential and Kinetic Energy ........................................................... 54 7.7 Gravitational Potential Energy ........................................................................................ 55 7.8 Elastic Potential Energy.................................................................................................... 55 7.9 Solved Examples on Work and Energy ........................................................................... 55 Chapter 8 Momentum and Impulse ........................................................................................ 61 8.1 8.2

Momentum ......................................................................................................................... 61 Conservation of Momentum ............................................................................................. 61

3 8.3 Center of Mass .................................................................................................................. 63 8.4 Collision Phenomena ......................................................................................................... 65 Chapter 9 Equilibrium............................................................................................................. 73 9.1 Translational Equilibrium................................................................................................. 73 9.2 Rotational Equilibrium ...................................................................................................... 73 9.3 Rotational Motion ............................................................................................................. 74 9.4 Motion of a Point on a Rotating Body............................................................................. 75 9.5 Moment of Inertia .............................................................................................................. 76 9.6 Kinetic Energy of Rotation ............................................................................................... 78 9.7 Dynamics of Pure Rotation ............................................................................................... 78 9.8 Solved Examples ................................................................................................................ 79 Chapter 10 Dynamical Systems ................................................................................................. 87 10.1 Momentum of a Dynamical System ................................................................................. 87 10.2 Conservation of Momentum ............................................................................................. 88 10.3 Angular Momentum of System of particles ..................................................................... 88 10.4 Rolling Motion of a Rigid Body ....................................................................................... 88 10.5 Solved Examples ................................................................................................................ 89 Chapter 11 Elasticity and Periodic Motion ................................................................................ 95 11.1 Elastic Properties ............................................................................................................... 95 11.2 Spring ................................................................................................................................. 96 11.3 Solved Examples on Elasticity .......................................................................................... 97 11.4 Periodic Motion ................................................................................................................. 98 11.5 The total energy of a particle executing simple harmonic motion ................................ 98 11.6 Simple Pendulum ............................................................................................................... 99 11.8 Examples on SHM .......................................................................................................... 100

4

Chapter 1 1.1

Review of Vectors

Mechanics It is an area of physics which is concerned with the study of motions of material particles. Mechanics is highly successful in describing the motions of material bodies, such as planets, rockets and baseballs.

Motion implies change in position. A change in position of a point body is called ‘displacement’. Position of a particle is specified by coordinates in a reference frame. A simple reference frame is the Cartesian frame shown in Fig. 1.1, where the three axes OX, OY, OZ are at right angles to each other.

1.2

Physical Quantities The characteristics of a moving body are described in terms of some parameters called “physical quantities”, like mass, length, speed, momentum, velocity, energy, etc.

Generally, we come across two types of physical quantities in physics.

Type 1: A physical quantity in which only ‘magnitude’ (a number with unit) is required to specify it. Examples are: mass, length, speed, work, energy, power, etc. Such a physical quantity is called ‘scalar’. A scalar is denoted by an algebraic quantity (number) like, A, B, C, D etc.

Type 2: A physical quantity in which both ‘magnitude ’and the ‘direction’ are required to specify it. Examples are: displacement, velocity, momentum, acceleration, etc. Such a physical quantity is called ‘vector’.→ A vector is denoted by putting an arrow on top of an → C→. algebraic number (magnitude) as in A, B,

- Geometrically, a vector is represented by drawing a line whose length is proportional to the magnitude of the vector and the direction is indicated by drawing the line in the direction of the vector with an arrow. The direction of the vector can be specified by an angle  it makes with the reference line OX. (Fig. 1.2) →

- A vector A at a point P can be moved to any point Q by drawing a line parallel to the line at P and of the same length as the line at P. (Fig. 1.3).

1.3

Addition of Vectors →

→

→

Vector→ A + Vector→ B = Vector→ R (resultant) A+B=R This can be written as, →

(1.1)

→

- When vector A and vector B are parallel to each other, the resultant is simply, →

→

→

R = A + B, and the direction of Vector R is the same as that of vector A or vector B .

5 →

→

But, when A and B have different directions, the addition is done by the, Law of parallelogram of addition If the two adjacent sides OS and OP of a parallelogram OPQS (Fig. → 1.4), → represent in magnitude and →direction the two vectors A and B , then the resultant vector R of addition of → → A and B

is given by the diagonal OQ of the parallelogram in direction and magnitude. If θ is the angle between the two → → → vectors, A and B , the magnitude of resultant vector R is given by: 2 2 R 2 = R 2 =(A + B cos ) + (B sin ) = A2 + B2 cos2θ + 2 A B cos θ + B2 sin2θ = A2 + B2 (cos2 θ + sin2 θ) + 2 A B cos θ = A2 + B2 + A B cos θ (1.2)  R = A2 + B2 + 2ABcos → → → - The direction of the resultant of A and B is given by an angle φ, which R makes with → vector A , and is given by, tan  =

Bsin A + B cos 

or

 B sin     A + B cos  

 = tan -1

(1.3)

Law of triangle of addition Law of triangle of addition of two vectors is only a re-statement of the law of parallelogram in a different form. In Fig. 1.4, if we focus our attention on OSQ, the law of parallelogram can be restated in the following way: ‘If two sides (OS, and SQ) of a triangle taken in cyclic order, represent in magnitude and → → direction two vectors ( A and B ), then the third side (OQ) of the triangle taken in the → opposite order represents the resultant→ ( R ) of the two vectors in magnitude and direction’ In OSQ, magnitude of the resultant R is given by, R 2= A2 + B2 + 2ABcos (180 -  ) = R 2= A2 + B2 + 2ABcos ( )

Subtraction of vectors →

→

- A is a vector→A in the opposite direction (Fig. 1.5) → → ⇀ ⇀  A - B = A +(- B ) = R

→ ⇀ → R  is a vector which gives A - B .

Again we can use the law of parallelogram of addition of two vectors as shown in Fig. 1.6. →

The magnitude of R is given by, R2 = A 2 + B2 − 2AB cos  →

The direction (   ) of R is given by,

tan   =

(1.5) B sin  A - B cos 

(1.6)

(1.4)

6

1.4

Multiplication of vectors There are two kinds of products of multiplication of two vectors: 1. Scalar Product or Dot Product: in this kind of product, a vector multiplies with another vector to give a scalar quantity as a product of multiplication. We represent scalar product by a (dot) between the two vectors. → → Scalar product→ A  B = C (1.7) → Eq. (1.7) represents the scalar product of a vector A with another vector B resulting in a product which is a scalar and is given by: A  B = C = A B cos θ , where, (1.8)  is the angle between vectors A and B (Fig. 1.7) Example: (work done →W) =

→ → Force → F  Displacement → D

(→

)(

)

→ → → = (  F  ) (  D  )  (cosine of  θ between F and D )

Or,

W = F D cos θ (Fig. 1.8)

(1.9)

2. Vector Product or Cross Product: In this kind of multiplication a → → another vector B → to →result→ in a vector quantity A multiplies with → product which is also a vector C . It is written as, A  B = C , the symbol () identifies a vector product. The magnitude of the product → vector is given by,  C  = C = AB sin θ, where θ is the angle between → → vectors A and B . The →direction of resultant product vector is→at right → → angles to the plane of A and B , coming out of the plane of A and B as shown in Fig. 1.9.

We can write this product as, → → → → → A  B = C = AB sin θ (in direction ⊥ to the plane of A and B , coming out)

Note that,

1.5

A  B = - BA

Component of a Vector

Another way of interpreting the equation,

→

⇀ →

(1.10) (1.11)

, is that a

A + B= R → vector R→ can be broken up into two other vectors A and B→ , which → when added will result in a vector R . This process of breaking a

vector into two parts (components) is called ‘resolution of a vector into two components’. Generally, it is convenient to resolve a vector into two components → which are at right angles to each other.

For example, a vector R in (Fig.1.10) can be resolved into two components along (OC) and (CB) at right angles to each other and their magnitudes are R cos θ and R sin θ, respectively. If point O in the figure is taken as origin and x-axis along → (OC) and y-axis parallel to (CB), then R cos θ is called the ‘x-component’ of R along x→ axis, and R sin θ the ‘y-component’ of R. →  X-component of R → Rx = R cos θ, y-component→ Ry = R sin θ

7 - Let ˆi represents a unit vector along x-axis, and ˆj a unit vector along y-axis, then we can → ˆi say that: R has a component Rx = R cos  along the direction and the component →

Ry = R sin  along ˆj . Therefore, we can express any vector A in terms of its components along x- and y-axes as, → ˆ ˆ ˆ A = (A cos θ) + (A sin θ) j = Ax i + Ay j → - Similarly, in three dimensions any vector B can be written in terms of three components Bx, By, and Bz as, B = Bx ˆi + B y ˆj + B z kˆ

(1.12)

ˆi, ˆj, kˆ are three unit vectors along x-, y-, z-axes, respectively. → The magnitude of vector B in terms of three components is given by, Magnitude→| B | = B =

B2x + B2y + B2z

(1.13)

- Thus, the position of a point P(x,y) in two dimensions can be specified by a vector r ( OP ) in (Fig. 1.11).is called the ‘position vector or displacement vector’ of point P. Obviously, the coordinates (x, y) of point P are x- and y-components of the position vector r . We can then write, r = x ˆi + y ˆj = (r cos θ) ˆi + (r sin θ) ˆj (1.14) 2 2 2 Clearly, r = x + y

1.6

Relationships Among Unit Vectors ˆi, ˆj, kˆ ˆi, ˆj, kˆ are unit vectors along x-, y-, z-axes which are mutually at right angles to each other.(Fig. 1.12) Scalar Product:

Vector Product:

ˆi . ˆi = ˆj.ˆj = kˆ .k̂ =1

(1.15)

ˆi.ˆj = ˆj. kˆ = kˆ .ˆi = 0

(1.16)

ˆi ˆi = ˆj ˆj = kˆ kˆ = 0

(1.17)

ˆiˆj = k̂,

ˆjkˆ = ˆi, kˆ ˆi = ˆj

ˆjˆi = - k̂, kˆ  ˆj = - ˆi, ˆikˆ = - ˆj

(1.18) (1.19)

→ → Example 1: If A = AX ˆi + AY ˆj + AZ kˆ and B = BX ˆi + BY ˆj + BZ kˆ are two vectors, then the scalar product of these two vectors is:

8 → →

ˆ ˆ ˆ ˆ ˆ ˆ A  B = (AX i + AY j + AZ k)  (BX i + BY j + BZk) = (AXBXˆi  ˆi + AXBYˆi  ˆj + AXBZˆi  k) + (AYBX ˆj  ˆi+ AYBY ˆj ˆj+ AYBZˆj  k̂ ) + + (AZBX kˆ  ˆi + AZBY kˆ ˆj + A Z B Z k̂  k̂ ) Using relations in Eqs. (1.15) and (1.16), we get, A  B = AX BX + AY BY + AZ BZ

(1.20)

→ → Example 2: If A = AX ˆi + AY ˆj + AZ kˆ and B = BX ˆi + BY ˆj + BZ kˆ are two vectors, then The (cross) product of these vectors is: → vector → ˆ ˆ ˆ ˆ ˆ ˆ A  B = ( AX i + AY j + AZ k ) ( BX i + BY j + BZ k ) = (AXBXˆi  ˆi + AXBYˆi  ˆj + AXBZˆi  k̂ ) + (AYBXˆj ˆi + AYBYˆj ˆj + AYBZˆj k̂ ) + + (AZBXkˆ  ˆi + AZBYkˆ  ˆj + A ZBZk̂  k̂ ) → → Eqs. (!.17), (1.18) and (1.19), we get Using ˆ ˆ ˆ ˆ ˆ ˆ A  B = (0 + AXBYk - AXBZ j ) + (−AYBX k + 0 + AYBZi) + (AZBX j - AZBY i + 0) A  B = (A Y BZ − A Z B Y )iˆ + (A ZB X − A X B Z ) ˆj + (A XB Y − A Y B X )kˆ Or (1.21) Vector product in terms of a determinant A cross product of two vectors can be neatly represented in terms of a determinant as follows: ˆi → → AX AB= BX

ˆj AY

kˆ AZ

BY

BZ

ˆ AY =i BY

AZ BZ

ˆ AX - j BX

AZ BZ

ˆ AX +k BX

AY BY

→ → ˆ ˆ ˆ AB = i (AYBZ – AZBY) - j (AZBX – AXBZ) + k (AXBY – AYBX) Equations (1.21) and (1.22) are identical.

(1.22)

Determinants A 2  2 determinant is defined as, A B Determinant→ D → D

C

Similarly,

A D→ D

B E

G

H

= AD – BC,

C E F=A H I

F I

-B

D

F

G

I

+C

D

E

G

H

= A (EI – FH) – B (DI – FG) + C (DH – EG)

9

Example:

2 4 −6 1 3 5 2 1 4

Find the value of the determinant→

2 4 −6 5 1 3 3 +(-6) =2(12-5)-4(4-10) - 6(1-6)=14+24+30= 68 Solution: 1 3 5 =2 5 - 4 1 1 4 2 4 2 1 2 1 4

1.7

Solved Examples Example 1: The rectangular coordinates of a point are, x = 4 m, y = 3 m. Find the position vector of this point. → Solution: If rP is the position vector of the point P whose coordinates are (4 m, 3 → m), the position vector → is, rP = (42 ˆi +2 3 ˆj) m.

The magnitude of → r = (x + y ) = =5m rP P → Direction of rP →θ = tan-1 3 = 36.9∘ 4 ( rP , θ) are also the polar coordinates of the point P.

( )

Example 2: A car drives 5 km east, then 4 km south, then 2 km west. Find the magnitude and direction of the resultant displacement.

Solution: In Fig. 1.14, Displacement AB = 5 km (east), displacement BC = 4 km (south), and Displacement CD =2 km (west), then AD gives the resultant displacement.

AB = 5 ˆi + 0 ˆj , BC = 0 ˆi - 4 ˆj , CD = -2 ˆi + 0 ˆj  AD = AB + BC + CD = (5 ˆi ) + (- 4 ˆj ) + (- 2 ˆi ) = 3 ˆi - 4 ˆj )

( 3) = 53.1 (south easterly direction).

Direction of (AD)→ = tan−1 4

∘

→

Example 3: Find the angle between two vectors, Â = 3ˆi + 4ˆj + 5kˆ , and B = 3iˆ + 4jˆ - 5kˆ → → Solution: If  is the angle between vectors A and B , then the scalar product of the two → → vectors is, A  B = A → B → cos   cos  = cos  =

A B AB 0 AB

→ →

→

, A  B = ( Â = 3ˆi + 4ˆj + 5kˆ)  ( B = 3ˆi + 4ˆj - 5kˆ ) = (3 3 + 4  4 + (−5)  5) = 0 = 0,

  = 90∘ , or

→ → A is ⊥ to B

→

Example 4: Given two vectors  = 2ˆi + 3ˆj + 4kˆ and B = ˆi - 2jˆ + 3kˆ . Obtain the following: (a) Find the magnitude of each vector. (b) Write an expression for the vector sum using unit → → vectors. (c) Find the magnitude of vector sum. (d) Write an expression for A - B . (e) Find → →

the magnitude of A - B

10 →

Solution: (a) Magnitude of A :A = A2x + A2y + A2z = 5.39 km

22 + 32 + 42 = 29

→

Magnitude of B :B = B2x + B2y + B2z = 12 + (−2)2 + 32 = 14 = 3.74 km (b)

→ → → → Vector sum→ A + B = ( A = 2ˆi + 3ˆj +4kˆ ) + ( B = ˆi - 2ˆj + 3kˆ ) = {(2 + 1) ˆi + (3 – 2) ˆj + (4 + 3) kˆ } = 3 ˆi + ˆj + 7 kˆ →

→

(c) Magnitude of vector sum→ A + B  = 32 + 12 + 72 = 59 = 7.68 km →

→

→

→

(d) A - B = ( B = iˆ - 2jˆ + 3kˆ ) – ( B = 3iˆ + 4jˆ- 5kˆ ) = {(2 - 1) iˆ + (3 +2) ˆj + (4 – 3) kˆ } = ˆi + 5 ˆj + kˆ →

→

(e) Magnitude of vector difference→  A - B  = 12 + 52 +12 = 27 = 5.2 km →

→

→

ˆ ˆ Example 5: Given two vectors  = 2ˆi + 3 ˆj and→B =→ i - 2 j . (a) Find the vector product, A B. (b) Find the magnitude of the product vector A  B .

Solution:

(a)

→ → ˆ - 4 ( ˆi  ˆj ) + 3( ˆj iˆ ) – 6( j ˆj ) A  B = ( 2iˆ + 3 ˆj )  ( iˆ - 2 ˆj ) = 2 (iˆ i)

Using the identities,

{ ˆi ˆi = ˆj ˆj = kˆ kˆ = 0 }, { ˆi ˆj = kˆ, ˆjkˆ = ˆi, kˆ ˆi = ˆj } { ˆjˆi = - k̂, kˆ  ˆj = - ˆi, ˆikˆ = - ˆj }, we get,

→ → ˆ ˆ ˆ A  B = (2  0) – (4  k ) - (3  k ) - (6  0) = -7 k

(b)

→

→

 A B  = -7 kˆ  = 7

11

12

Chapter 2 2.1

Units and Significant Figures

Introduction: In Mechanics, the three basic quantities are length (L), mass (M), and time (T). All other physical quantities in mechanics can be expressed in terms of these three quantities.

To report the results of measurements and to verify the results, a standard must be defined.

In 1960, an international committee established a set of standards for these fundamental quantities. The system is called the ‘International System’ (SI) of units. SI is the abbreviation of French name ‘Syste’me International’.

In (SI), the units of length, mass and time are meter, kilogram, and second, respectively. The units of other quantities which can be expressed in terms of fundamental quantities are called ‘derived units’. Definitions of units are under constant review and are changed from time to time.

2.2

Length Until recently, the meter was defined as 1650763.73 wavelengths of orange-red light emitted from a Krypton-86 lamp. However, in October 1983, the meter was redefined as the distance traveled by light in vacuum during the time (1/299792458) second.

In metric system the following prefixes are used: kilo→ 1000 = 103, mega→ 1000,000 = 106, giga→ 109, On the lower scale side, deci→ 10-1, milli→ 10-3, -2 centi→10 , micro(µ)→10-6,

tera→ 1012

nano→ 10-9, pico→10-12,

famto→ 10--15, atto→ 10-18

Conversion to other units 1 meter (m) = 39.37 inch (in) = 3.281 foot (ft), 1 inch (in) = 2.54 centimeter (cm) = 0.0254 meter (m) 1 kilometer (km) = 3281 foot (ft) = 0.6214 mi, 1 mile (mi) = 5280 foot (ft) = 1.609 km 1 cm = 0.3937 in,

2.3

∘

1 Angstrom ( A ) = 10 -10 m = 10-8 cm = 10-1 nm

Mass The basic SI unit of mass is kilogram (kg). It is defined as the mass of a specific Platinum-Iridium alloy cylinder kept at the International Bureau of Weights and Measures at Sevres, France.

1 kilogram (kg) = 1000 gram (g) = 103 g Atomic Mass: Matter consists of atoms, and each atom is made of electrons and a nucleus.

Practically all of the mass of an atom is contained in the nucleus which consists of protons and neutrons. Different elements contain indifferent number of neutrons and protons.

The mass of C12 atom is defined to be exactly 12 atomic mass units (u), where,

13 1 u = 1.6605402 x 10-27 kg Mass of proton = 1.0073 u, Mass of neutron = 1.0087 u, Mass of electron = 9.11x10-31kg - One mole (mol) of a substance consists of Avogadro’s number (NA) of molecules.

Avogadro’s number is so defined that one mole of C12 atom has a mass of exactly 12g. Its value is found to be, NA = 6.02 x 1023 molecules/mol -Therefore, mass per atom of a given element is given by, mA =

2.4

atomic mass of the element NA

Time In 1967, the SI unit of time, second (s), was redefined using the characteristic frequency of a particular kind of Cesium atom as the reference clock. The second is defined as 192631770 periods of radiation from Cesium-133 atoms.

Before 1967, second was defined as, 1 second =  131,556,925.9747  of solar year 1900.   1 hour (hr) = 60 minutes (min) = 3600 seconds (s) 1 day (d) = 86,400s 1 year (yr) = 3.156 x 107 s

2.5

Volume, Density, and Angle Volume: The unit of volume is liter (L).

1 m3 = 106 cm3 = 6.102 x 104 in3 1 liter = 1000 cm3 = 1.0576 quart (qt) = 0.0353 ft3, 1 gallon (gal) =3.786 L = 231 in3 Density: Density () of a substance is defined as the mass per unit volume. m Density→  = (2.1) V Unit of density is (kg/m3) or (g/cm3).

Density of gold = 19.3 103 kg m3 Density of iron = 7.86kg m3 Density of air = 0.0012 103 kg m3 Angle: Unit of angle is degree (). One degree is the angle subtended at the center of a circle by (1/360) of the circumference.

1 minute () = 1/60 of a degree (),1 second () = 1/60 of a minute () In physics, generally, angle is measured in terms of ‘radian’, which is dimensionless. One radian (rad) is the angle subtended at the center of a circle by a portion of the circumference equal to the length of

14 the radius of the circle (Fig. 2.1).

- Any angle  in radian is given by (Fig 2.2), =

s

= length of the arc radius R

(2.2)

It is clear that, 2 (radian) = 360, or 1 = (/ 180) rad.

2.6

Significant Figures When certain quantities are measured, the measured values are known only within the limits of experimental uncertainty. Suppose we want to find the area of a rectangular plate by measuring its dimensions. Assume that we can measure a particular dimension of the plate within an accuracy of ± 0.1 cm. If the length of the plate is measured to be 16.3 cm, we can only claim that its length lies between 16.2 cm and 16.4 cm. In this case we say that the measured value has only three significant figures. Likewise, if its width is measured to be 4.5 cm, the actual value lies between 4.4 cm and 4.6 cm. The measured value has only two significant figures. Thus we can write the measured values as, 16.3 ± 0.1cm, and 4.5 ± 0.1 cm.

If we were to claim that the area is (16.3 cm)  (4.5 cm) = 73.35 cm2, our answer would be unjustified because it contains four significant figures in either of the measured lengths. A good rule is: ‘When multiplying several quantities, the number of significant figures in the final answer is the same as the number of significant figures in the least accurate of the quantities being multiplied. Least accurate means ‘having lowest number of significant digits’ Thus we can claim that the area is 73 cm2.

- Zeros used to position the decimal point are not significant. Thus, in 0.03 and 0.0075, there are one and two significant figures, respectively. When the zeros come after other digits (e.g 1500g-mass) we use scientific notation to indicate the number of significant figures. We would express mass of 1500 g as 1.5×103 g if there are two significant figures in the measured value, 1.50×103 g if there are three significant figures. The same rule holds for numbers less than one, so that 2.3×10-4 has two significant figures - For addition and subtraction, the number of decimal places must be considered when one is determining how many significant figures to report. When numbers are added or subtracted, the number of decimal places in the result should equal the smallest number of decimal places of any term in the sum.

For example, to compute 123 + 5.35, the answer would be 128, and not 128.35. If we compute the sum, 1.0001 + 0.0003 = 1.0004, the result has five significant figures even though one of the terms in the sum is 0.0003 has only one significant figure.

15 - Similarly, if we perform 1.002 -. 998 = 0.004, the result has only one significant figure, even though one term has four significant figures, and the other has three.

Note: The rule for addition and subtraction is different from that for multiplication and division. For addition and subtraction, the important consideration is the number of decimal places, not the number of significant figures.

Example: A rectangular plate has a length of (21.3±0.2) cm and a width of(9.80 ±0.10) cm.

Find the area of the plate and the uncertainty in the calculated area.

Solution: Area = Length  Width = (21.3 ± 0.2)  (21.3 ± 0.2)  (21.3  9.80) + 21.3  (± 0.1) + 9.80  (± 0.2)  (208.74) ± 2.13 ± 1.96  (209 ± 4) cm2

16

Chapter 3

Kinematics

Kinematics is a branch of mechanics concerned with description of motion. Motion implies a change of position. In motion, the position of a point changes with time. The change in position is called ‘displacement’.

3.1

Position Vector and Displacement Position Vector:

The position of any point P can be specified → → by a vector OP = rP as shown in Fig. 3.1. rP is called the ‘position vector’ of point P. If (x,y) are the coordinates of the point P, we can write, → → ˆ ˆ rP = | rP | = rP = OP = x2 + y2 , and direction → x i + y j , and of → y tan  = , where θ is the angle r rP is given by, P makes with x the x-axis.

Displacement: Let a body in Fig. 3.2 move from a point P at any time ti to a point Q at time tf, along a curved path of length s. Then → → the displacement of the body is given by the vector, PQ = D . D is called the displacement of Q →from P. It is clear from Fig. 3.2, → → =  = {x ˆi + y ˆj } – {x ˆi + y ˆj} Displacement→ → D = rQ − rP r f f i i ˆ ˆ = (xf – xi) i + (yf – yi) j → → →  Change in position vector→ r = r − r . The direction of the displacement → Q

D is along

P

PQ. - Change in time → t = tf - ti Average Velocity: of a body is defined as, Average velocity→ v =

t

=

f

(tf − ti )

(in the direction of PQ)

(3.1)

Note: The average velocity is independent of the path taken by the body → between two points P and Q (Fig. 3.3). The displacement D depends only on the position of the points P and Q, not on how it moved between P and Q i.e. independent of the path taken.

If a body initially at P (Fig. 3.4), after a curved path comes back to the point P after an interval of time t, then the displacement of the body is zero, and the average velocity is also zero. However, the distance traversed by the body in coming back to the point P after a time t is s, which is not zero. We then define the average speed as,

17

Average speed→ v =

s Total distance traveled = t Total time to cover the distance

(3.2)

Instantaneous Velocity and Speed: It is very important to define the velocity of a particle at a particular instant of time, rather than just over a finite interval of time. The velocity of particle at any instant of time at some point is called the ‘instantaneous velocity’. In Fig. 3.5, as Q is brought closer to P along the path taken, the time needed to travel the distance gets smaller. As Q approaches P, the time interval approaches zero and the direction of the displacement approaches the direction of tangent at the point P. We define the instantaneous velocity at a point as, →

Instantaneous velocity→ v = Lim

D

→ t - → v is simply the rate of change of displacement D . t→0

= dD

(3.3)

dt

Similarly, instantaneous speed can be defined as, Instantaneous speed→ v = Lim

s

 t→0 t

=

ds

(3.4)

dt

Relative Velocity: The relative velocity of a point A, with respect to a point B (Fig. 3.6) is obtained by finding the resultant of the velocity of A and an equal and opposite velocity of B, superimposed on the velocity of A. In Fig. 3.6, the diagonal AC of the parallelogram v →

gives the relative velocity, AB , of the point A with → → respect to the point B. If vA is the velocity of particle A, and vB the velocity of → the particle B, then vAB ( AC in Fig. 3.6) gives the relative velocity of A with respect to B. (3.5)

vAB = vA - vB

It then follows: →

If the relative velocity of a particle A relative to B is vAB , and the relative velocity of → particle B relative to a particle C is vBC , then the relative velocity of A with respect to C is given by, (3.6) vAC = vAB + vBC

18 Example 1: An aeroplane flies at airspeed of 200 km/hr. At the altitude at which it is flying, there is a wind of 100 km/hr to the west. The plane desires to proceed to north relative to the ground. In what direction should it head and what will be its ground speed?

Solution: Speed of airoplane with respect to air→ vPA = 200 km/hr Velocity of wind with respect to ground→ v→AG = 100 km/hr to west → If vPG is the relative velocity of the plane relative to ground, → → → From Eq. 3.6, vPG = vPA + vAG (3.7) We now take the axes as shown in Fig. 3.7. Since the plane wants to go to north, we know → the direction of vPG . → ˆ ˆ  (3.8) vPG = 0 i + vPG j Also the wind is in the westerly direction, → (3.9)  vAG = (- vAG ) ˆi + 0 ˆj From Eqs. (3.7), (3.8) and (3.9), we get, → → → ˆ ˆ ˆ ˆ (3.10) v PA = vPG - vAG = { vPG j + vAG i } = ( vAG i + vPG j ) From Eq. (3.10), v 2PA = v 2AG + v 2PG , or vPG = v2PA − v2AG =

= 173 km/hr

From Eq. (3.10), the aeroplane should head in the direction , given by, v 173 Tan  = PG = , or  = 60∘ (to the east) vAG

100

Example 2: When a train has speed of 10 m/s eastward, raindrops that are falling vertically with respect to the earth make traces that are inclined 30∘ to the vertical on the window of the train. (a) What is the horizontal component of drop’s velocity with respect to the earth? (b) What is the velocity of the rain drop with respect to the earth? (c) With respect to the train? v→TE → velocity of train relative to the earth Solution: Let → vRE → velocity of rain drop relative to the earth v→ RT →velocity of rain drop relative to train → → → Then, vRE = vRT + vTE (3.11) With reference to axes in Fig. 3.8, we have, → → ˆ ˆ → ˆ ˆ ˆ ˆ + 0 j , vTE = 10 i vRE = 0 i - vRE j , vRT = - ( vRT cos 60) i - ( vRT sin 60) j (3.12) From Eq.(3.11), - vRE ˆj = {- ½ vRT ˆi - 3 2 vRT ˆj } + 10 ˆi (3.13) From Eq.(3.13), - ½ vRT +10 = 0, or vRT = 20 m/s And

- 3

2

vRT = - vRE ,

or

vRE =

3

2

 20 = 17.3 m/s

(a) From Eq.(3.12) , horizontal component of drop’s velocity is, → ( vRE )H = 0 (b) From Eq. (3.12), velocity of rain drop relative to the earth is,

19 → vRE = - vRE ˆj = - 17.3 m/s (c) Velocity of rain drop relative to train is; → ∘ vRT = 20 m/s in a direction 30 to the vertical Example 3: An airplane flies 200 km due west from a city A to city B and then 300 km in the direction of 30 north of west from city B to city C. (a) In a straight-line distance how far is city C from city A? (b) Relative to city A, in what direction is city C? Solution: (a) Referring to Fig. 3.8, straight-line distance CA is, CA = (300)2 + (200)2 + 2  200  300 cos 30∘ = 484 km (b) Direction of city C is given by the angle  in Fig. 3.8, which is given by, 150 , or  = 18 sin  = CD = CA

3.2

484

Newton’s Laws of Motion Any motion of a material body is governed by three laws of motion given by Newton.

First Law: ‘Every material body remains in its state of rest, or of uniform motion in a straight line, unless compelled by some external forces acting on it to change its state’.

According to Newton, there are only two natural states of a body: (1) state of rest, (2) State of uniform motion in a straight line. Newton puts the state of rest on equal footing with a state of uniform motion in a straight line i.e.

State of rest  Uniform motion in a straight line - This law postulates a tendency of a body to remain in either of the two natural states. This tendency is called ‘inertia’ of the body. Therefore, Newton’s first law is also called ‘Law of Inertia’.

Inertia of a body  mass of body In other words, the natural property of inertia (laziness) of a body is measured in terms of mass of a body. Mass is the quantity of matter possessed inside a material body.

Unit of mass→ 1 kilogram = 1000 gram in MKS units

Second Law: ‘The rate of change in motion of a body is directly proportional to the force acting and takes place in the direction along which the force acts’.

Quantity of motion  momentum of a body Where, momentum is defined as, → → Momentum ( p ) = mass (m)  velocity ( v), or p v (3.14)

20 - Change in motion implies change in velocity with time and is called ‘acceleration’.

Therefore, we define instantaneous acceleration as, Instantaneous acceleration→ a = Lim

v

meter (second)2

=

t→0 t

Unit of acceleration→

=

v dt m s2

(3.15)

- Newton’s second law tells that a force produces a change in→ velocity called ‘acceleration’ and the acceleration is proportional to the force applied. If F is the applied force, a  F, or F = (constant)  a = ma. The direction of acceleration is the same as that of force.

Therefore, we can write, a = F

(with proper choice of units)

(3.16)

- Unit of force is ‘Newton (N)’. One Newton is the amount of force which produces in a mass of one kilogram an acceleration of 1 m/s2.

Newton’s Third Law: ‘To every action, there is an equal and opposite reaction’ This law is true for each pair of particles inside the body. It is also true for interaction between two bodies. In Fig. 3.10, weight of → the block exerts a force W (weight) on the floor; this is what we call ‘action’. According to Newton’s third law, the floor reacts by exerting → a force R (reaction) on the block, and R = W, and R and W have opposite directions.

- Tension in a String: A string is incapable of exerting any compressive force on a body, but a string can exert a reactive force upon a body to which it is attached by reason of being under tension (Fig. 3.11). Tension is a force of reaction, and therefore a passive force. Often in problems the mass of a string or rope is much less than the masses of objects that it connects. In such cases, we can adopt the massless string approximation and assume the mass of string to be zero.

An important aspect of tension is that it pulls equally in both directions Unless the string is mass-less, the force of tension will not in general be parallel to its own length but at some angle to it. Only for a mass-less string the tension will be parallel to its length.

Strings and ropes often pass over pulleys. As the string moves, the static friction between the string and the pulley causes the pulley to turn. If we assume that (a.) the string and the pulley are both massless, and (b) there is no friction where the pulley turns on its axle, then no net force is needed to accelerate the string or turn the pulley. In other words, the tension in a massless string remains constant as it passes over a massless, frictionless pulley.

21

3.3

Friction Friction is a force of reaction which comes into play when one solid object moves or tends to move along the surface of another solid object with which it is in contact. The force of friction always tries to oppose the motion or the tendency of a body to move.

Consider a block of wood B sitting on a rough table (Fig. 3.13). If a small force F tends to move the block to the right, there will→ be a force of friction f acting on the block in the opposite direction of F . If the block does not start to move because the force is not large enough, then F = f, which means that the magnitude of the force of friction will be equal to the force acting.

- A force of static friction occurs between two surfaces where there is no sliding, but there is a tendency to slide.

- A force of kinetic friction is exerted when the two surfaces are sliding past each other while in contact.

In Fig. 3.14, as we increase the force F acting on the block by increasing weight in the pan P, there will be a point when the block will just be on the point of moving. At this point, this will be the maximum force of friction on the block B. Therefore if fmax is the maximum force of static friction before the body is just on the point of moving, experiments show that, Maximum force of static friction→ fmax  RN, where, RN is the normal reaction of the table on the block. fmax = s RN  (3.19) s→ is a constant called ‘coefficient of static friction’ Similarly, if K→ is the coefficient of kinetic friction, then The kinetic frictional force→

fK = K RN

(3.20)

- K depends on the type of material and the roughness of the two surfaces. Over a wide range, it is independent of the area of contact of the surface. It is also independent of the relative velocity of the two surfaces.

NOTE: Remember s RN gives the maximum force of static friction.

Limiting Angle of Friction: If a block of weight W on an inclined plane (Fig. 3.15) is just on the point of sliding down the plane when it is inclined at an angle , this angle is called the ‘angle of friction’ If we take the X-axis down the plane and Y-axis at right angles to the plane, the forces acting on the block are: W sin →Component of W acting along the plane (X-axis).

W cos → Component of W acting perpendicular to the plane (-ve Y-axis).

N→ Normal reaction of the plane on the block perpendicular to the plane (+ve Y-axis). sRN→ Maximum force of friction acting on the block along the plane(-ve X-axis).

22 Using the conditions of equilibrium of forces on the block, we get: W sin  - s RN = 0 (X-components) RN – W cos  = 0 (Y-components)

(3.21) (3.22)

From Eq. (3.22), W sin θ = µs W cos φ RN = W cos θ, Or, µs = tan φ (3.23) - Φ is the angle of the plane when the body is just on the point of sliding down the plane.

From Eq. (3.22),

3.4

Weight Weight is a force acting on a mass due to the gravitational pull of the earth. If m is the → mass of a body, then weight W is given by, Weight→ W = m g (3.24)

- g is the acceleration due to the gravitational pull. For the earth, g = 9.8 m/s2.

- If m = 1 kg, then the weigh is: W = (1 kg) (9.8 m/s2) = 9.8 N - Sometimes we call 9.8N as ‘1 kg-weight’.

3.5

Equilibrium of Forces When the system of forces acting on a particle is such that a particle remains at rest, or in motion at a constant velocity, the system of forces are said to be in equilibrium. An unbalanced force produces acceleration. Whenever there is unbalanced force acting on a particle, the change in motion must occur. It means that in equilibrium of a particle, there can be no unbalanced force present on the particle.

3.6

Conditions of Equilibrium of a Particle:

→ → →

→

Let a particle at O be subjected to forces F1, F2 , F3 , and F4 . To keep the particle in equilibrium, the resultant of all the forces acting at O must be zero. Otherwise, there will be unbalanced force acting at O and motion will occur.

If ˆi, and ˆj are unit vectors along X- and Y-axes (Fig. 3.12), then we can write, F1 = F1x ˆi + F1Y ˆj ,

F2 = F2X ˆi + F2Y ˆj ,

F3 = F3X ˆi + F3Y ˆj ,

F4 = F4X iˆ+ F4Y ˆj

(3.17)

For the equilibrium of point O, F1 + F2 + F3 + F4 = 0 Or, (F1X + F2X + F3X + F4X ) ˆi + (F1Y + F2Y +F3Y +F4Y ) ˆj = 0, which can only be satisfied if, (F1X + F2X + F3X + F4X ) = 0 ,

(F1Y + F2Y +F3Y +F4Y ) = 0

(3.18)

Eqs. (3.18) imply that: (1) The sum of components of all the forces along X-axis must be zero.

(2) The sum of components of all the forces along Y-axis must be zero.

Equations (3.18) are the necessary and sufficient conditions of equilibrium of a particle.

23

3.7

Solved Examples

→ → Example 1: Two forces F1 and F2 act upon a body in such a manner that the resultant force → → →  R has a magnitude equal to that of F1 and makes an angle of 90 with F1 . Let F1=R= 10N. → Find the magnitude of the second force, and its direction relative to F1 .

Solution: Taking the coordinate → system as shown in Fig. 3.16, ˆ ˆ F→1 = 10ˆ i + 0 ˆj (3.25) (3.26) Since R = F (given), R = 0 i + 10 j 1  → ˆ ˆ F2 = F2 cos (90 +θ) i + F2 sin (90 + θ) j = - F2 sin θ ˆi + F2 cos θ ˆj (3.27) → → → → Since R is the resultant, F1 + F2 = R  From Eqs. (3.25), (3.26) and (3.27), we get, (10 ˆi )+( F2 cos (90+θ) ˆi + F2 sin (90 + θ) ˆj = 10 ˆj , which gives us, 10 – F2 sin θ = 0 (3.28) F2 cos θ = 10 (3.29) From Eqs. (3.28) and (3.29),

tan θ = 1 or, θ = 45, and F2 =

200 = 14.1 N

Example 2: 10 N weights are suspended at opposite ends of a rope that passes over a light frictionless pulley. The pulley is attached to a chain that goes to the ceiling. (a) What is the tension in the rope? (b) What is the tension in the chain?

Solution: (a) Since each of the two weights suspended from the pulley (Fig. 3.17) is in equilibrium, the condition of equilibrium applied to one of these weights gives, T = w = 10 N (B) Since the pulley is also in equilibrium, we again apply the condition of equilibrium on the pulley. This gives us, T1 – W –W = 0 , or T1 = 2 W = 20 N Example 3: A picture frame is hung against a wall, suspended by two wires attached to its upper corners. If the two wires make the same angle with the vertical, what must this angle be for the tension in each wire to equal the weight of the frame?

Solution: The picture in Fig. 3.18 is in equilibrium under the effect of following forces: W→ weight of the picture acting vertically downward.

T,T→Equal tensions in the two wires making equal angles θ with the vertical Choosing the coordinate system as shown in Fig. 3.18, applying the conditions of equilibrium on the picture, we have, T cos(90-θ) – T cos((90-θ) = 0 {X-component} (3.28)   T sin(90 -θ) + T sin(90 -θ) = W {Y-component} (3.29) From Eq. (3.29), 2 T cos θ = W (3.30)

24 If the tension in each wire is to be equal to the weight of the picture, T = W  From Eq. (3.30), cos θ = ½, or θ = 60 Example 4: A safe weighing 2000 N is to be lowered at constant speed down skids 4 m long, from a truck 2 m high. (a) If the coefficient of sliding friction between safe and skids is 0.3, will the safe need to be pulled down or held back? (b) How great a force parallel to the skids is needed?

Solution: Taking X-axis down the plane (Fig. 3.19), and noting that Newton’ first law implies that State of rest ≡ State of uniform motion, There can be no unbalanced force acting on the safe. The problem reduces to static equilibrium. The forces acting on the safe are: - W sin θ→ Component of W acting down the plane - W cos θ→ Component of W along the –ve Y-axis, perpendicular to the plane - RN→ Reaction of the plane on the safe, perpendicular to the plane (+ve Y-axis) - µK RN→Frictional force up the plane in –ve X-direction.

Since the safe does not move perpendicular to the plane, RN = W cos θ (3.31)  12   = 519.6 N  Maximum force of kinetic friction → f = µK RN = (0.3)(2000)  4   -Downward component of W→W sin θ = (2000)(2/4) = 1000N Since the downward force is greater than the frictional force, the safe will have to be held back with a force = (1000-519) = 480 N Example 5: A 5.00 kg object placed on a frictionless, horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging 9.00 kg object (Fig.3.20). Find the acceleration of the two objects and the tension in the string.

Solution: Since it is an inextensible string, the acceleration, ‘a’ of the two masses will be the same. Let T be the tension in the string as shown in Fig.3.20.

The equation of motion of 9.0 kg-mass is given by: 9×a=9g–T (1) Similarly, the equation of motion of 5.0 kg-mass is: 5×a=T (2) Adding Eq. (1) and (2), we get: 14 a = 9 g,or acceleration→ a = (9/14) (9.8) = 6.30 m/s2 From Eq.(2), tension→ T= 5 × 6.3 = 31.5 N

25 Example 6: Three objects are connected on the table as shown in the Fig.3.21. The table is rough and has a coefficient of kinetic friction of 0.350. The objects have masses of 4.00 kg, 1.00 kg, and 2.00 kg, and the pulleys are frictionless. (a) Determine the acceleration of each object and their directions. (b) Determine the tension in the two cords.

Solution: Since all the three masses are connected by inextensible strings, the magnitude of acceleration of the three masses will be the same. It is clear that 4.00 kg mass will move vertically downward, 1.0 kg mass will move to the left, and the 2.0 kg mass will move vertically upward. The motion of 1 kg-mass will be opposed by kinetic friction given by: Frictional force →fk = µk m g = (0.35) (1.0) (9.8) = 3.43 N Because of this frictional force on 1.0 kg-mass, the tension on string to the right will be different from the tension on string to the left of mass as shown in Fig.(3.21). Let ‘a’ be the magnitude of acceleration of the three masses and T1 and T2 be the tensions as shown. We can now write the equation of motion of the three masses.

Equation of motion of 4 kg-mass: Equation of motion of 1 kg-mass: Equation of motion of 2 kg-mass: Addition of Eqs. (1), (2), and (3) gives: Putting the value of a in Eq.(3) gives: Putting the value of a in Eq.(1) gives:

4 a = 4 × 9.8 – T2 (1) 1 a = T2 – T1 – 3.43 (2) 2 a = T1 - 2 × 9.8 (3) 7 a = 16.17, or a = 2.31 m/s2 T1 = 2 × (2.31) + 2 × 9.8 = 24.2 N T2 = 4 × 9.8 – 4 × (2.31) = 30.0 N

Example 7: An object of mass M = 50 kg is held in place by an applied force F and a pulley system as shown in Fig.3.22. The pulleys are massless and frictionless. Find (a) the tension in each section of the rope, T1, T2, T3, T4, and T5., and (b) the magnitude of F. Solution: All the masses in the pulley system are in equilibrium.

Therefore net force on any particle of the system must be zero.

From the equilibrium of Mass M we get: T5 = 50 × 9.8 = 490 N From the equilibrium of pulley A we get: T2 + T3 = T5 = 490 N, T2 = T3, and therefore, T2 = T3 = 245 N (1) From equilibrium of pulley B we get: T4 = T1 + T2 + T3, and T1 = T3 = 245 N (2)  T4 = 245 + 245 + 245 = 735 N (3) The force exerted by hand → F = T1 = 245 N  T1 = T2 = T3 = 245 N, T4 = 735 N, and T5 = 490 N Example 8: A carton weighing 90 kg rests on a horizontal loading platform. A horizontal force of 27 kg is required to start the carton in motion, but once it is moving a force of 20 kg is adequate to keep it in motion at constant speed. Find the coefficient of static and kinetic friction.

26

Solution: Normal reaction of the platform on on carton→ RN = 90 × 9.8 = 882 N Maximum force of friction before motion starts→ fs = 27 × 9.8 = 265 N  Coefficient of static friction→ µs = fs/RN = (265)/(882) = 0.30 The maximum force of friction after uniform motion starts is→ fK = 20 × 9.8 = 196 N  Coefficient of kinetic friction → µk = (196/(882) = 0.22 Example 9: A bird weighing 0.9 kg alights on a weightless wire midway between two poles 70 m apart. Due to the weight of the bird, the wire, which was initially horizontal, sags 7.5 cm. What is the tension in the wire due to the weight of the bird?

Solution: The bird is in equilibrium under the effect of three forces: Tensions T, and T acting along CA and CB sections of the wire (Fig.3.23) Weight of the bird, mg , acting vertically downward.

Let θ be the angles as shown in the figure. Considering the forces in equilibrium we have: Horizontal components: T sin θ – T sin θ = 0 (1) Vertical components: T cos θ + T cos θ - mg = 0 (2) 7.5 7.5 = From Fig.3.23: cos θ = = 0.00214 3500 (3500)2 + (7.5)2 (0.9)(9.8) From Eq.(2), Tension in the wire →T = = = 2061 N 2(0.00214) Example 10: Find the compression in the strut and the tension in the cable of Fig.3.24 if the strut is weightless and W is 100kg. The strut is free is free to rotate about a pin at the wall so that the strut exerts a force only along its length.

Solution: There are only two forces acting on W, T upward and weight mg downward. From the conditions on the equilibrium of W we get: T – mg =0, or T = mg = 100 × 9.8 = 980 N Let S be the force of compression in the strut. The forces acting at C are T, T1, and S. Since S has no component at right angles to Ac, T sin 37∘ (980)(0.602) = T1 sin 16˚ - T sin 37˚ = 0, or, T1 = = 2138 N (0.276) sin 16∘  Compression in the strut → S = T1 cos 16 + T cos 37 = (2138)(0.961) + (980)(0.799) = 2838 N

27

Chapter 4 4.1

Rectilinear Motion

Rectilinear Motion with Constant Acceleration We know from Chapter 3 that, D Total change in displacement → = , Average Velocity→ v = Total time t D d D Instantaneous velocity→ v = Lim = → (rate of change of displacement with time) t→0 t dt When the velocity increases with time, we say the body is accelerating, we define acceleration as, Average acceleration→ a =

v change in velocity = t change in time

,

v v (rate of change of velocity) = t→0 t dt

Instantaneous acceleration→ a = Lim

(4.1)

(4.2)

If the rate of change of velocity is constant, then we say that the acceleration of the body or particle is constant. When a particle moves in a straight line, we say the motion is called ‘rectilinear’.

Note: When a motion is in a straight line, we do not generally put an arrow sign to identify a vector.

Rectilinear Motion: If a particle at time t = 0, has an initial velocity v0, and if the acceleration is constant, then we have, (Fig. 4.1) Velocity after t = 1s is v0 + a Velocity after t = 2s is v0 + 2a Velocity after t = 3s is v0 + 3a ………………………………………………

Velocity after t = t s is v0 + at  If a body is moving with a constant acceleration, the velocity at any time t is: v = v0 + a t (4.3) Where, v0 is the velocity at time t = 0 We now have, at t = 0, v = v0 At t = t, v = v0 + a t 1 v0 + (v0 + at) = (v0 + at )  Average velocity→ vave = 2 2 Let at t = 0, the particle be at x = 0, then after a time t, x = vave t = (v0 + ½ at) t  Distance traveled in time t→

x = v0 t + ½ a t2

(4.4)

28

- We now eliminate t between Eqs (4.3) and (4.4). v - v0 From Eq. (4.3), t = , and putting this value of t in Eq. (4.4) gives a 2 v(v - v0 ) a  v - v  v v v2 a v 2 − 2vv + v2  x= +  0 0 0  = 0 − 0+   a 2 a  a a 2 a2  v v v2 v2 2vv v2 (v2 − v2 ) 0 0 = 0 − 0+ − + 0 = a a 2a 2a 2a 2a 

v2 − v20 = 2ax

(4.5)

Equations (4.3), (4.4) and (4.5) will solve all the problems on motion in a straight line with constant acceleration.

4.2

Solved Examples on Rectilinear Motion Example 1: A motorist makes a round trip from city A to city B, which are 120 km apart by road. If the time required for the trip from A to B is 2 hours and the time required for the return trip is 3 hours, find (a) the average speed of the car going from A to B, (b) the average speed of the car on return trip, and (c) average speed of the car for the entire trip.

Solution: (a) Average speed in going from A to B is: Total distance 120 km = Average speed→vave = = 60 km/hr Total time 2 hr (b) Average speed for the return trip is: 120 km Average speed → vave = = 40 km/hr 3 hr (c) Average speed of the car for the entire trip is: Average speed→ vave =

Total distance Total time

=

120 km +120 km = 48 km/hr 2 hr+ 3 hr

Example 2: Over 2 second period, the distance ℓ traveled by a certain body is accurately described by the equation, ℓ = 10t + 8t 2 − 2t3 in MKS units. Find the instantaneous acceleration when t = 2s. dℓ d = (10t + 8t 2 − 2t3 ) = 10 + 16 t- 6 t2 Solution: Instantaneous speed→v = dt dt dv d = (10 +16t - 6t 2 ) = 16-12t Instantaneous acceleration →a = dt dt  dv  Acceleration at t = 2s→  dt  = 16 - 24 = - 8m/s2  t=2

29 Example 3: A body starts from rest, moves in a straight line with constant acceleration, and covers a distance of 64 m in 4 s. (a) What is the final velocity? (b) How much time was required to cover half the total distance? (c) What is the distance covered in one-half the total time? (d) What is the velocity when half the total distance has been covered? (e) What is the velocity after one-half the total time?

Solution: (a) The distance from the origin x traveled in time t is given by: x = v0 t + ½ a t2 (4.4), where v0→initial velocity at t=0, v0 = 0 at t=0 (given) Also given is: x = 64 m, and time elapsed→t = 4s 64  2 = 8 m/s2 From Eq. (4.4) we get, 64 = ½ a (4)2, or acceleration→ a = 2 4 From Eq. (4.3), velocity after 4s→ v = v0 + at = 0 + (8)(4) = 32 m/s (b) For half the distance, x = ½  64 = 32 m, v0 =0, a = 8 m/s2, t = ?

2  32 = 8 = 2.8 s Using Eq. (4.4), 32 = 0 t + ½ (8)(t2), or t = 8 (c) Now the question is: for t = 2s, a = 8 m/s2, t = ?

Again using Eq. (4.4), x = 0 + ½  8  22 = 16 m (d) Question is: x = 32 m, v0 = 0, a = 8 m/s2, v=?

2 2 2 Using Eq. (4.5)→ v – v 0 = 2 a x→ v – 0 = 2  8  32, or v =

2  8  32 = 22.6 m/s

(e) Question is: For t =2 s, a = 8 m/s2, v = ?

Using Eq. (4.3)→ v = v0 + at→ 0 + 8  2 = 16 m/s Example 4: A ball dropped from the cornice of a building takes 0.25s to pass a window 3m high. How far is the top of window below the cornice? Acceleration due to gravity, g is 9.8 m/s2.

Solution: Let t be the time taken by the ball to fall from A to B (Fig. 4.2) and h be the distance between cornice and the top of window. First we consider the motion of the ball between points A and B. At A, v0 = 0, y = h, a = g = 9.8 m/s2 , then the velocity at B→ vB is: From Eq. (4.3)→ vB = v0 + g t→ vB = 0 + g t (4.6) Also from Eq. (4.4)→ y = v0 t + ½ g t2→ h = 0 + ½ g t2 (4.7) Now consider the motion of ball between points B and C. At B, v0 = vB = gt (from Eq. 4.6), distance BC = 3 m, and t = 0.25s  From Eq. (4.4), 3 = (g t) (0.25) + ½ g(0.25)2 ,or 2 3 − (0.25) = 0.31- 0.03 =0.28, or t = 1.1s 0.25 t = 9.8 2  From Eq. (4.7)→ h = ½  9.8 (1.1)2 = 5.9 m

30 Example 5: A ball is thrown vertically upward from the ground and a student gazing out of the window sees it moving upward past him at a speed of 5 m/s. The window is 10 m above the ground. (a) How high does the ball go above the ground? (b) How long does it take to go from a height of 10 m to its highest point? (c) Find its velocity and acceleration 0.5 s after it left the ground.

Solution: Let v0 be the velocity of the ball at t = 0, when thrown from the ground. At the highest point the velocity will be zero (v = vf = 0). Height of the window above ground is → y = 10 m (Fig. 4.3) and the velocity of ball at the window is→v = 5 m/s.

Eq. (4.5) in this case can be written as: v2 – v02 = 2 (-g) y, (since the acceleration due to gravity is in the –ve y-direction Putting v = 5 m/s, y = 10 m in this equation gives us v0 : (5)2 – v02 = - 2 9.8 10, or v = 25 +196 = 14.9 m/s 0

(a) At the highest point the velocity of the ball will be zero (v = 0),y = ?, v0= 14.9m/s From Eq. (4.5)→ v2 – v02 = 2 (-g) y→ 0 - (14.9)2 = - 29.8 y, which gives us: 221 = 11.3 m Highest point the ball goes→ y = 19.6 (b) To find time to go from P (window) to the highest point Q, we have, v0(at P) = 5 m/s, v (at Q) = 0, t = ?.

5 = 0.5 s From Eq. (4.3)→ v = v0 – g t→ 0 = 5 – 9.8 t, or t = 9.8 (c) To find velocity after t = 0.5 s after it left the ground : We have v0 = 14.9 m/s at the time t = 0, v = ?  From Eq. (4.3)→ v = 14.9 – 9.8  (0.5) = 10 m/s Acceleration is constant→ a = - g = - 9.8 m/s2 Example 6: A ball is thrown vertically upward from the ground with a velocity of 30 m/s. (a) How long will it take to rise to the highest point? (b) How high does the ball rise? (c) How long after projection will the ball have a velocity 10 m/s upward? (d) Of 10 m/s downward? (e) When is the displacement of the ball zero? (f) When is the magnitude of ball’s velocity equal to half its velocity of projection? (g) When is the magnitude of ball’s displacement equal to half the greatest height to which it rises? (h) What are the magnitude and direction of the acceleration while the ball is moving upward? (j) When at the highest point?

Solution: (a) At the highest point, v = 0, also v0 = 30 m/s, from v = v0 – g t, we get, 0 = 30 – 9.8  t, or t = 30/9.8 = 3.1 s (b) To find how high the ball will rise, we use the equation, y = v0 t + ½ a t2

31 (g) To find time we use the equation, y = v0 t + ½ a t2 y – 45.9/2 = 23 m, v0 = 30 m/s, t=?  23 = 30 t – 4.9 t2 or, 4.9 t2 -30 t + 23 = 0, which is a quadratic equation in t 2  t = 30  (30) − 4.9  23 4 = 30 m 21.2 , therefore, t = 5.2 s, or, 0.9 s 9.8 2  4.9 2 Here, v0 = 30 m/s, a = - g = -9.8 m/s , t = 3.1 s, y=?  Highest point above ground→ y = 30  3.1 - ½ (9.8)(3.1)2 = 45.9 m (c) To find time, we use the equation, v = v0 + a t Here, v0 = 30 m/s, v = 10 m/s, a = - g = - 9.8 m/s2 t=?  10 = 30 – 9.8  t, or t = 20/9.8 = 2.0 s

(d) To find time we use again, v = v0 + a t v = - 10 m/s (coming downward), v0 = 30 m/s, a = -9.8 m/s2  - 10 = 30 – 9.8 t, or t = 40/9.8 = 4.1 s (e) The displacement will be zero when the ball returns to ground. (f)

We use again, v = v0 + a t to find time, Here, v = v0/2 = 30/2 = 15 m/s, v0 = 30 m/s, t = ?  15 = 30 – 9.8 t, or t = 15/9.8 = 1.5 s

(h) Acceleration of the ball when going upward is, a = 9.8 m/s2 in the – ve y-direction. (i) Acceleration when the ball is going downward is, a = 9.8 m/s2 in the - ve y-direction (j ) Acceleration when at the highest point is, a = 9.8 m/s2 in the – ve y-direction Example 7: A ball is thrown upward from the roof of a building. It has an initial velocity of 17 m/s. How high is the building if on its downward flight it just misses the thrower and falls to the ground 5 seconds after it is thrown? With what speed does it strike the ground?

Solution: Let h be the height of the building and vg be the velocity when the ball hits the ground.

Time taken for the ball to come to rest when thrown upward with an initial velocity (v - v0 ) = (17 − 0) = 1.73 s of 17 m/s is : t= g 9.8 Therefore the time taken to fall from position of rest to ground = 5 – 1.73 = 3.27 s  Velocity of the ball when it hits ground→ vg = v0 + g t = 0 + 9.8 × 3.27 = 32.0 m/s Distance covered by the ball in the first 1.73s =

(v − v ) = (17) = 14.7 m 2

2 0

2

2g 2  9.8  A distance of (h + 14.7) m is covered in 3.27 seconds  h + 14.7 = v0 t + ½ g t2 = 0 + ½ (9.8) (3.27)2 = 42.4  Height of the building → h = 52.4 – 14.7 = 37.7 m

32 Example 8: A boy riding a bicycle at a speed of 5 m/s passes a parked car. Eight seconds later the car starts and is uniformly accelerated for 10 seconds, at which time it passes the cyclist. What was the acceleration of the car?

Solution: The cyclist is traveling at a uniform speed.

Distance covered by cyclist in (8 + 10) s = 18 × 5 = 90 m Therefore, a distance of 90 m is covered by the car in 10 s with a constant acceleration ‘a’ starting from rest.  90 = 0 + ½ a (10)2, or acceleration→ a = 90/50 = 1.8 m/s2 Example 9: At t = 0 a body passes x = 0 with a velocity vx = 20 m/s. If the body has constant acceleration ax = - 5 m/s2, find (a) the velocity when t = 2 s, (b) the position of the body at t = 2 s, (c) the time at which the body returns to x = 0, and (d) its velocity at this time. dv = - 5 m/s2,  v = - 5 t + c1, c1→ constant of integration (1) dt At t = 0, v = 20 m/s, from (1), c1 = 20, and v = - 5 t +20 (2) Velocity when t = 2 s is from (2): v(t = 2) = - 5×2 + 20 = 10 m/s Solution: (a)

(b) Integration of (2) gives: x = - (5/2) t2 + 20 t + c2 , c2 → constant of integration (3) At t = 0, x = 0,  from (3) , c2 = 0, and x = - (5/2) t2 + 20 t (4) Position at t =2 s from (4): x(t = 2 s) = - (5/2) (4) + (20) (2) = 30 m (c) From (4), x will be zero after a time given by: 0 = - (5/2) t2 + 20 t , or t=8s (d)

Velocity at t = 8 s from (2) is:

v(t =8s) = - (5) (8) + 20 = - 20 m/s

Example 10: With what initial velocity will a body moving along a vertical line have to be thrown if after 5 seconds it is to be 20 m above its starting place?

Solution: Let v0 be the initial velocity of projection Putting t =5, y = 20 in y = v0 t - ½ g t2 , we get: 20 = v0 (5) - (½) (9.8) (5)2, 0r 5 v0 = 20 + 123 = 143 Initial velocity of projection→ v0 = 28.6 m/s

33

Chapter 5 5.1

Gravitation

Newton’s Law of Gravitation: It states: ‘Every particle of matter in the universe attracts every other particle with a gravitational force. The magnitude of this force between two particles is proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them, the force of attraction between the particles acts along the line joining the two particles’.

If we refer to Fig. 5.1, we can say for the magnitude of force: Force of attraction between m1 and m2→ F12  m1 m2 1  2 (r is the distance between particles) r m1m2  We can write, F = G (G is universal constant of gravitation) (5.1) 12 2 r Nm2 Value of universal constant of gravitation→ G = 6.67  10 -11 kg 2 - The universal gravitational constant G was measured in an important experiment by Henry Cavendish in 1798.

- The direction of an attractive force is taken as negative, We can write Eq. (5.1) in vector → form as, F12 = - G 2 r12 ( rˆ12 is a unit vector in the direction of r12 ) (5.2) r - From Eq. (5.2), it can be shown that the gravitational force exerted by a uniformly distributed spherical finite mass, on another particle outside this mass distribution, is the same as if the entire mass of distribution were concentrated at the center. For example, the magnitude of force exerted by the Earth on a particle of mass m near the Earth’s surface is, M m Fg = G E2 (5.3) RE where, ME is the Earth’s mass, and RE is its radius.

5.2

Free-Fall under Gravitational Force If g is the acceleration of a freely falling body under gravitational force, then from Eq. (5.3) we can write, M M m g = G E2 m g = G E2 or, (5.4) RE RE - Now consider a mass m located at a height h above Earth’s surface, or at a distance r from the center of the Earth, where, r = RE + h. The magnitude of the gravitational force Fg acting on mass m is:

34 Fg = G ME m (RE + h)2 ME g= G From Newton’ second law: m g = G ME m , or (RE + h)2 (RE + h)2 Thus, from Eq. (5.5) it follows that g decreases with increasing altitude.

5.3

(5.5)

Gravitational Field We can describe the interaction between two objects that are not in contact in a different way using the concept of a gravitational field that exists at every point in space.

When a particle of mass m is placed at a point where the gravitational field is g , the particle → → experiences a force Fg = mg . The gravitational field is defined as: → → Fg g= (5.6) m That is, the gravitational field at a point in space is the gravitational force experienced per unit mass of a test mass placed at that point.

- The field g at a distance r from the center of the Earth is : M g = −G 2E rˆ (5.7) r where rˆ is a unit vector pointing radially outward from the Earth, and the negative sign indicates that the field points toward the center of the Earth.

- Weight of a particle is the force with which the Earth attracts the particle.

The gravitational force (weight) on a particle of mass m is given by: → → → Weight→ W = Fg = mg (5.8) - A freely falling body experiences no force except that of Earth’s gravity.

5.4

Solved Examples Example 1: Calculate the gravitational force that would be exerted on a 10-kg sphere when the distance between the centers of the spheres is 60 cm. If these were the only forces acting on the spheres, what would be their acceleration? → m m → =G 1 2 F12 r12 r2 Here, m1 = 10 kg, m2 = 10 kg, r = 60 cm = 0.6 m -11 (6.67 10 )(10)(10)  F12 = = 1.8510-8 N (0.6)2 Solution: From Eq. (5.2)→

Acceleration→ a =

F12 1.85 10−8 = = 1.8510-9 m/s2 m 10

35 Example2: Calculate the average density of Moon in kg/m3, approximating g of the moon by a sphere of radius 1700 km and mass 7.31022 kg. Compare with the average density of the Earth.

Solution: Mass of the Moon→m = 7.31022 kg, Radius of the Moon→RM = 1700 km = 1.7106 m 4 3 4 Volume of the Moon→ VM = R = (1.7 106)3 = 2.06 1019 m3 3 M 3 V 7.3 1022 3 3  Density of the Moon→ M = M = = 3.5410 kg/m 19 2.06 10 m Radius of the Earth→ RE = 6371 km = 6.371106 m Mass of the Earth→ ME = 5.9751024 kg Volume of the Earth→ VE = (4π/3)(6.371106)3 = 1.081021 m3 M 5.975 1024  Density of the Earth→ E = = 5.53103 kg/m3 21 1.08 10 VE Density of the Earth 5.53 103  = = 1.56 Density of the Moon 3.54 103 Example 3: A rocket is 637 km above the surface of the Earth. What is the value of g at this height? Find the weight of a 5 kg-mass at this point.

Solution:

g(r) at a distance r from the center of the Earth is given by: M g(r) = - G 2e rˆ (1) r where, Me → mass of the Earth = 5.98 × 1024 kg, G = 6.67 × 10-11 Nm2/kg2 If y is the height of a body above the surface of the Earth, Eq.(1) can be written as: Me ŷ (2) g(y) = - G 2

(

R e + y) where, Re = Radius of the Earth = 6.37 × 106 m Magnitude of g at height y = 637 km → g (y) = (6.67 × 10

-11

)

5.98 1024

(6.37 106 + 637 103 )

2

= 8.1 m/s2 Weight of 5 kg-mass at a height of 637 km→ W = m g(y) = 5 × 8.1 = 40.5 N Example 4: A body of mass 10 kg is moving with a constant velocity of 5 m/s on a horizontal surface. The coefficient of sliding friction between body and the surface is 0.2. (a) What is the horizontal force required to maintain the motion? (b) If the force is removed, how soon will the body come to rest?

Solution: (a) Normal reaction on the body→RN = 10g = 109.8 = 980 N Force of kinetic friction→ fK = (0.2)(980) = 19.6 N

(Fig. 5.3)

36 Therefore, 19.6 N- force is required to maintain the motion. (b) If the force is removed, the force of friction will oppose the motion and deceleration will occur.

Force 19.6  Deceleration due to friction→ af = = = 1.96 m/s2 Mass 10 Using the Eq. (4.3)→ v = v0 + at , and putting into it, v = vf =0, v0 = 5 m/s, a = - 1.96 m/s2 (negative sign is due to deceleration), we get, 0 = 5 – 1.96  t, or t = 5/1.96 = 2.6 s Example 5: An elevator weighing 4000 kg rises with acceleration of 4 m/s2. What is the tension in the supporting cable?

Solution: Forces acting on the elevator are shown in the Fig. 5.4 → We use Newton’ second law of motion equation→ m a= F Mass of elevator→ m = 4000 kg, a = 4 m/s2, Weight→F = 4000g, Tension→T? Putting these values in the equation of motion we get, (4000)(4) = T– (4000)(9.8) →T = (16000 + 39200) = 55200 N or 5633 kg-weight Example 6: A block weighing 10 N is supported by a string that can be moved up and down. What conclusion can you draw regarding magnitude, direction, and velocity of the upper end of string when the tension in the string is (a) 5N, (b) 10 N, and (c) 15 N? Solution: Forces acting on the block are: Tension→T in the positive y-direction, and Weight→W = 10 N (downward)  10  (a) For T = 5 N, Newton’s equation of motion gives:  9.8  a = 5 - 10   (-5)(9.8)  a= = - 4.9 m/s2. This means that the direction of acceleration is downward in 10 the negative y-direction. In other words, the block will decelerate (will slow down).

-The direction of velocity will be upward, i.e. in the positive y-direction.  10  (b) For T = 10 N, the equation of motion becomes,  9.8  a = 10 - 10 , or a = 0,   and there will be no acceleration of the block. The direction of velocity can be up or down.  10  (c) For T = 15 N, the equation of motion gives, 9.8 a = 15 - 10 ,or a = 4.9 m/s2.     The direction of acceleration is now along positive y-axis. The direction of velocity is also along the positive y-axis.

Example 7: If the coefficient of friction between tires and road is 0.5, what is the shortest distance in which the automobile can be stopped when traveling at 30 m/s.

37

Solution: If m is the mass of the automobile, then its weight is→W = m g (Downward), and the normal reaction of ground on automobile is→RN = m g (upward). The force of kinetic friction on the auto is→ fK = µK RN = (0.5) (mg).

This force of friction will resist the motion of the auto. (Fig. 5.6) (−fK ) (−0.5mg) =  acceleration→ a = = - 0.5 g = - 4.9 m/s2 m m The shortest distance in which the auto is stopped is given by the equation, v2 – v02 = 2 a x Here, v0 = 30 m/s, v = vf = 0, a = - 4.9 m/s2 , x = ?

900 = 91.8m The above equation gives: 0 – (30)2 = 2(-4.9) x, or x = 9.8 Example 8: A 5 kg block slides down a plane inclined at 30 to the horizontal. Find the acceleration of the block (a) if the plane is frictionless, (b) if the coefficient of friction is 0.2.

Solution: (a) If the plane is frictionless, there is no force of friction on the block. Take the x-axis down the plane, and y-axis at right angles to the plane (Fig. 5.7(a)).

We can resolve the weight (m g) of the block into two components: (1) (x- comp.)→mg sin 30 (down the plane), (2) (y-comp)→ m g cos 30 (⊥ to the plane) - Since the block does not move in the y-direction, Reaction of plane on the block→ RN = mg cos 30 (5.9) -The only force acting in the x-direction down the plane is→ mg sin 30. To find the acceleration ‘a’ of the block down the plane, we write the equation of motion along the x-axis: m a = Fx→ 5 a = 59.8sin 30, or acceleration down the m/s2 (b) When the force of friction is present (Fig.5.7(b)), Magnitude of force of friction in the negative x-direction is: Kinetic friction→ fK = µK RN = (0.2)(5 9.8 cos 30) = 8.49 N - The equation of motion can now be written as: 5  a = m g sin 30 - fK  5 a = 5 9.8  (0.5) – 8.49)  acceleration along x-axis→ a = 3.2 m/s2 Example 9: A 20 kg box is pushed up a rough sloping ramp, inclined at 30 to the horizontal having a coefficient of kinetic friction of 0.3, by a horizontal force of 300 N. (a) What is the normal force? (b) What is the frictional force? (c) What is the acceleration of the box? (d) If the force is reduced until the acceleration becomes zero, what is the magnitude of force?

38 Solution: → If F is the horizontal force acting on the block (Fig. 5.8), we break it up into two components: F cos θ along the positive x-direction, which we have taken up the plane, and F sin θ perpendicular to the plane (negative y-direction) - Since the box does not move in the y-direction is: (5.10) RN = F sin θ + m g cos θ - In this problem F = 300 N, m = 20 kg, θ = 30 (a) From Eq. (5.9), Normal reaction→ RN is: 300  (0.5) + 20  9.8  3 = 320 N 2 (b) Frictional force on the block→ fK = µK RN = (0.3)  (320) = 96 N (down the plane) (c) If ‘a’ is the acceleration up the plane (positive x-direction), then the equation of motion of the block along the plane is written as: m a = - fK – m g sin θ + F cos θ (5.11) Eq. (5.11) then gives→ 20 a = - (96) – (20  9.8  sin 30) + (300  cos 30)  20 a = - 96 – 98 + 260, or acceleration→ a = 3.3 m/s2 (d) Let F be the horizontal force needed to make the acceleration of the block zero, The force of friction → fK = µK RN = 0.3( F sin 30 + 20  9.8 cos 30) - From Eq. (5.11), the equation of motion of the block now is: m  0 = - fK - m g sin θ + F cos θ, or 0 = - 0.3( F  0.5 + 20  9.8  0.866) - 20  9.8  0.5 + F  0.866  0 = - 0.15  F - 50.9 – 98 + 0.866  F , or 0.72 F = 148.9 148.9 Required horizontal force to make acceleration zero is → F = = 207 N 0.72

39

Chapter 6 6.1

Motion in a plane

Motion of a projectile → A particle is projected at O with a velocity v making an angle  with the horizontal 0 → (Fig. 6.1). We can write v0 as, → Initial velocity of projection→ v = (v0 cos ) ˆi + (v0 sin ) ˆj 0

Acceleration of the particle→ a = (0) ˆi + (-g) ˆj Since the forces act on a particle independent of each other, the motion in a plane can be described by considering the motion as a result of two motions under the effect of forces acting in x- and y-directions, respectively.

Considering the motion of particle along x-axis, Distance traveled in x-direction→x =(v0 cos ) t + ½ 0  t2, or x = v0 t cos  y = (v0 sin ) t - ½ g t2 Distance traveled in y-direction→ Equations (6.1) and (6.2) describe the motion of a projectile under force of gravity.

(6.1) (6.2)

If we eliminate t between Eq,s (6.1) and (6.2), we get the equation of the resultant path taken by the projectile under the effect of forces acting in x and y directions. The equation of the path of projectile can be shown to be: 1 g x2 y = x tan  sec 2 (6.3) 2 v20 It can be shown that the path described by Eq. (6.3) is an inverted parabola. Equations (6.1) and (6.2) are the parametric equations of the parabola in Eq. (6.3).

In other words, because of independent nature of forces, the actual path (OVP) of a projectile from O to point P(x,y) during a certain time t, is the same as if: (1) the particle first moves along the path (OQ) with a uniform speed v0 cos  for time t (since there is no force acting in the direction of x-axis), and (2) then, during the same time t, particle moves vertically against gravity with initial velocity v0 sin  along QT, goes past point P, attains a maximum height (QT) at T, which is at the level of point V (vertex of parabola), comes to rest at point T, and then falls back to point P under gravitational acceleration.

Range of a Projectile: The distance (OR) in Fig. 6.1 is defined as horizontal ‘range’ of a projectile. During the time t the particle goes with a uniform velocity (v0 cos ) from point of projection O to the point R where the projectile hits the horizontal line OX, during the same time t, the particle goes up vertically against gravity with an initial velocity v0 sin  , comes to rest at a certain height and falls back accelerated by gravity and hits the horizontal line. We use this idea to calculate the range of a projectile.

40 Time t1 taken by the particle to go up and come to rest is given by, v sin  t1 = 0 (6.4) 0 = v0 sin  - gt1 or, g The same time t1 will be taken by the particle to fall from the position of rest under gravity and hit the horizontal plane.

2v0 sin  The time taken by the particle to go up and come down→ t = 2 t1 = (6.5) g  From Eq. (6.5), the range of the projectile → r = (v0 cos )(2t1) r=

(v0 cos )(2v0 sin  ) v20 sin(2 ) = g g

(6.6)

To find maximum range attainable with a given velocity of projection, we can find the maximum value of r in Eq. (6.6). r will be maximum when, sin (2) = 1, or  = 45. v2 rmax = 0 Maximum range→ (6.7) g A maximum horizontal range is attainable with a given velocity of projection when the angle of projection is 45.

6.2

Solved Examples on Projectiles Example 1: A gun pointed 30 above the horizontal fires a shell with muzzle velocity of 400 m/s. Neglecting air resistance, (a) find the maximum height reached, (b) time of flight, and (c) range.

Solution: Muzzle velocity of gun→ v0 = v0x ˆi + v0y ˆj = (400 cos 30) ˆi +(400 sin 30) ˆj x-component of projection velocity→ v0x = 400 cos 30 = 400 0.866 = 346.4 m/s y-component of projection velocity→ v0y = 400  sin 30 = 400  ½ = 200 m/s (a) Maximum height attained can be calculated by using : vy2 – v0y2 = -2 gy In the above equation, vy = 0 (for maximum height), v0y= 200 m/s,  0 – (200)2 = - 2  (9.8) y, (200)2 or, maximum height→ ymax = = 2041 m 2  9.8 (b) Let t1 be the time taken to attain the maximum height, then the total time taken to go up and come down→ t = 2 t1, where t1 is given by the equation, vy = v0y – g t1 In this equation, vy = 0 (for maximum height), v0y = 200 m/s  o = 200 – 9.8 t1, or

t1 =

200 9.8

= 20.4 s,

 Time of flight → t = 2 t1 = 2  20.4 = 40.8 s

41 (c)

The range will be given by the distance traveled along x-axis in 40.8 seconds with a uniform velocity of 346 m/s.  Range → r = (346.4)  (40.8) = 14133 m

Example 2: A block slides off a horizontal table top 1 m high with a speed of 3 m/s, find (a) horizontal distance from the edge of the table at which the block strikes the floor, and (b) the horizontal and vertical components of its velocity when it reaches the floor. → Solution: Velocity of projection→ v0 = v0x ˆi + v0y ˆj In this problem, v0x = 3 m/s, v0y = 0 (a) We calculate the time taken by the block to fall through 1 m to ground. We use the equation, y = v0y t + ½ g t2 , which gives us, 2 = 0.45 s 1 = 0  t + ½ (9.8)  t2, or t= 9.8 During 0.45 s, the horizontal distance the particle moves is: Horizontal distance from the edge of table→ r = v0x  t = 3  0.45 = 1.35 m (b) The horizontal component of velocity v0x remains the same throughout the motion of the block. Horizontal component of velocity when it hits the ground→ vx = 3 m/s The vertical component of velocity will be the velocity of block when it hits the floor in a vertical fall under gravity. We use the relation, vy = v0y + g t In the problem, v0y = 0, t = 0.45, Vertical component of velocity when the block hits ground→vy = 0+9.8  0.45 = 4.4 m/s Example 3: A baseball leaves the bat at a height of 1.0 m above the ground, traveling at an angle of 45 with the horizontal with a speed such that the horizontal range would be 140 m.

At a distance of 120 m from the home plate is a fence 10 m high. Will the ball be home run?

Solution: Since the base ball is projected at 45 angle, the horizontal range will be maximum for a given speed of projection. We can there use Eq. (6.7) maximum range given as, v20 , which gives us v , the magnitude of velocity of rmax = 0 g projection.  speed of projection→v0 = rmax  g = 140  9.8 = 37.0 m/s Time taken by the ball to travel a horizontal distance of 120 m to the fence is given by, x x 120  2 120 = = t= = = 4.59 s 37 37  cos 45∘ v0x v0 cos  After t =4.59 s, the vertical height of ball is given by,

42 y = v0y  t- ½ g t2

y = (37  sin 45) (4.59) – (0.5)(9.8)(4.59)2 37  4.59 - 103.2 = 120.1 – 103.2 = 16.9 m Height of ball at 120m→ y = 2 Since the ball is higher than the fence by 16.9 – 9 = 7.9 m, it will be a home run. or

Example 4: A block sliding towards the edge of a horizontal table top has a speed of 4 m/s at a point 3 m from the edge. It slides off the edge of the table, which is 1 m high, and strikes the floor 1 m from the edge of the table. What is the coefficient of friction between the block and the table?

Solution: Let v0 be the speed of block when it is at the edge of the table, then the time t1 taken to travel a horizontal distance of 1m is, 1 t1 = . This is also the time taken by the block to fall through a distance of 1m v0 under gravity, starting with initial downward velocity→ v0y = 0.

1 1 + ½ (9.8) 2 ,  1 = v0y t1 + ½ g t12 → 1 = 0  v0 v0 or v0 = 4.9 = 2.21 m/s To find the deceleration of the block on the table top from an initial speed of 4 m/s to 2.21 m/s within a distance of 3m, we use, vx 2 – vx02 = 2 a x, Here vx = 2.21 m/s, vx0 = 4 m/s x = 3 m.  (2.21)2 – (4.0)2 = 2  a  3 Or 4.88 – 16.0 = 6 a, or a = - 1.85 m/s2 Force of kinetic friction between block and the table surface→ fk =  m g  The equation of motion of the block is, m a = - µk mg→ - 1.85 = - µk  9.8,

or coefficient of friction→  k =

− 1.85 = 0.19 − 9.8

Example 5: A stone is thrown from the edge of a cliff with a velocity of 17 m/s at an angle of 30 above the horizontal. Four seconds later it strikes the ocean below. Find the height of the cliff above the ocean and the horizontal range of the stone.

Solution: Taking the axes as shown in Fig. 6.6, we can write velocity of projection as: → Velocity of projection→ v = v0x ˆi + v0y ˆj = v0 cos  ˆi + v0 sin  ˆj 0

= (17 cos 30) ˆi + (17 sin 30) ˆj = (14.7) ˆi + (8.5) ˆj The vertical distance down to ocean in 4 s will be the height of the cliff.

Height of cliff→ h = v0y t - ½ g t2 = (8.5)(4) - ½ (9.8)(4)2 = - 44.4 m

43 The negative sign above means that the ocean is 44.4 below the top of the cliff where the origin of the coordinate system is. → Since the horizontal component of v0 remains constant throughout the motion, the horizontal range is→ RH = 4  (v0x) = 4 14.7 = 58.8 m Example 6: Find the minimum speed with which the baseball must be hit at an angle of 53˚ with the horizontal to leave the bat at a height of 1m and clear a 12-m screen at a distance of 100 m from the home plate. Assume air fraction is negligible.

Solution: Let v0 be the minimum velocity of projection. Then the coordinates of any point P (Fig. Ex-6) on the top of the screen are (100,11). The point P must lie on the parabolic trajectory of the ball just to clear the screen.  100 = (v0 cos 53˚) t (1) ˚ 2 11 = (v0 sin 53 ) t - ½ g t (2) Substituting the value of t in (1) into (2), we get: 2 11 = 100 tan 53 ˚ - (½) (9.8)  100    v0 cos 53∘  49000 , or 121 v2 = 135208, or v0 = 135208 = 33.4 m/s  11 = 133 0 2 2 121 (0.602) v 0 Example 7: A baseball is given a velocity of 33 m/s at an angle 53º above the horizontal. It leaves the bat at a height of 1 m above the ground. What are its height and distance from the home plate (a) after 2 seconds, and (b) after 4 seconds? Will the ball clear a 7-m fence 95 m from the plate for a home run? By how much is the clearance?

Solution: x,y coordinates of a projectile at any time t projected with a velocity vo at an angle α is given by: x = (vo cos α) t (1) y = (vo sin α) t - ½ g t2 (2) (a) Height and distance at t =2 s from the home plate from (2) and (1) are: Height above x-axis→ y = (33 sin 53º) (2) - ½ (9.8) (2)2 = 52.7 – 19.6 = 33.1 m  Height above ground = 33.1 + 1 = 34.1 m above ground Distance→ x = (33 cos 53º) (2) = 39.7 m (b) Height above x-axis at t = 4s is: y = (33 sin 53º) (4) - ½ (9.8) (4)2 = 105.4 – 78.4 = 27.0 m Height above ground = 27.0 + 1.0 = 28.0 m Distance from home plate→ x = (33 cos 53º) (4) = 79.4 m 95 Time taken to cover a distance of 95 m= = 4.78 s (33)(cos53∘ ) Height above x-axis of the ball at t = 4.78 s is: y = (33 sin 53º) (4.78) - ½ (9.8) (4.78)2 = 126 – 112 = 14m, or 15 m above ground  It will clear the fence by 15-7 = 8 m

44

Example 8: A dive bomber has a velocity of 280 m/s at an angle θ below the horizontal.

When the altitude of the aircraft is 2.15 km, it releases a bomb, which subsequently hits a target on the ground. The magnitude of displacement from the point of release of the bomb to the target is 3.25 km. Find the angle θ.

Solution: The coordinates of bomb in (Fig.Ex-8) at any time during the flight are given by: x = (vo cos θ) t (1) 2 y = (vo sin θ) t + ½ g t (2) Horizontal distance covered by bomb→ x = (3.25)2 − (2.15)2 = 2.437 km From (2) time taken by bomb from the point of release to target is: x 2.437 t= = = 8.704 sec θ (3) (0.28) cos vo cos During this time t in (3), the bomb travels a vertical distance of 2.15 km.  From (2), 2150 = (280 sin θ)(8.704 sec θ) + ½ (9.8)(8.704 sec θ)2  2150 = 2437 tan θ + 371.2 (1 + tan2 θ), or 5.792 = 6.565 tan θ +(1 + tan2 θ)  tan2 θ + 6.565 tan θ – 4. 792 = 0 is a quadratic equation in tan θ, whose solution is: - 6.565  (6.565)2 + 4  4.792 − 6.565  7.891 = = 0.663, or -14.46 2 2 Choosing the positive root, θ = tan-1 0.663 = 33.5 ˚ tan θ =

Example 9: A boy throws a water filled balloon at an angle of 53.1º with a speed of 10 m/s. a car is advancing toward the boy at a constant speed of 5 m/s. If the balloon is to hit the car, how far away the car should be when the balloon is thrown?

Solution: Let us imagine the car is brought to rest by superimposing on the car and water balloon, a speed of 5 m/s away from the boy. The x,y-components of velocity of projection are: x-component→ vox = 10 cos 53.1º + 5 = 11.00 m/s y-component → voy = 10 sin 53.1º = 8.00 m/s Let D be the distance of the car if it is going to be hit by balloon.

D. (1) Time taken by balloon to hit the car→ t = 11 This is also the time balloon will take to go vertically up and come back to the point of projection when projectedvoyalong y-axs  8 withDa speed 8 m/s .  from (1), t=2 =2  = , g 9.8 11   16 11 Distance→ D = = 18.0 m 9.8

45 Examole 10: A projectile shot at an angle of 60º above the horizontal strikes a building 30 m away at a point 15 m above the point of projection. (a) Find the speed of projection. (b) Find the magnitude and direction of the velocity of projectile when it strikes the building.

Solution: The coordinates of a projectile projecte with a velocity vo at an angle α to the horizontal at any time are given by:: x = (vo cos α) t (1) y = (vo sin α) t - ½ g t2 (2) (a) Let vo be the speed of projection (Fig.Ex.10). The coordinates of point P in the figure are (30,15) which must satisfy (1) and (2).

From (1), time taken by the projectile to coverdistance 30 x = OQ→ t = (3) vo cos v o cos 60∘ 1 (9.8)(30)2 17640 ˚ Putting t in Eq.(2), 15 = 30 tan 60 ,or 15 = 51.96 v 2o 2 v2o (cos 60o )2  Speed of projection → vo =

17640 = 21.8 m/s 36.96

(b) The x-component of velocity at the point P→ vx = vo cos3060˚ =21.8 × (½) = 10.9 m/s y-component of velocity at P→ v = v sin 60˚ - g y o  ∘   v o cos 60  = (21.8) (0.866) – (9.8) (2.75)= - 8.07 m/s Speed at P→ v = (10.9)2 + (−8.07)2 = 13.6 m/s  − 8.07  Dirction of velocity at P→ θ = tan-1 = - 36.4˚  10.9 

6.3 Motion in a Circle A particle in Fig. 6.7a moves from point A to point B on a circular path of radius R with a constant speed v. If t is the time interval in which the particle moves from A to B, then AB in Fig. 6.7b is the displacement of the particle in time t.

From Fig. 6.7b, vB = vA + v , and, magnitude→ v = v  → → v v The instantaneous acceleration of particle at A→ a = Lim = Lim t→0 t t→0 t → → vdθ  Magnitude of a → a = a = Lim v = , (6.8) dt t→0 t

46 d is the angular speed of the particle. dt R  d - Speed of the particle→v = Lim =R  t→0 t v dt 2  From Eq. (6.8), a= R As  → 0 , B approaches A and the direction of  v becomes perpendicular to the → → direction of vA or vB . Therefore the direction of acceleration a is towards the center of the circular path. This is called the ‘centripetal acceleration’. v2 ˆ Therefore we can write, centripetal acceleration→ aR = R (6.9) R R̂ is a unit vector in the direction of along the radius away from the center of the circular path. mv 2 is a force directed towards the center and is called ‘centripetal force’ R mv 2 -A reaction to this force is directed away from the center. This reactionary R force is called ‘centrifugal force’

where,

6.3

Solved Examples on Motion in a Circle Example 1: A 1000 kg car moves around a circular race track at a constant speed of 20 m/s.

If the radius of the track is 200 m, (a) find the centripetal acceleration of the car, and (b) magnitude of the centripetal force exerted on the car. (c) What exerts the force on the car?

Solution: (a) Speed of the car→v =20 m/s Radius of the curve→R = 200 m v 2 (20)2 = Centripetal acceleration of the car→a = = 2 m/s2 R 200 mv2 (1000)(20) = (b) centripetal force→ FR = = 100 N (toward center) 200 R (c) This force is as a result of reaction of the fictitious centripetal force Example 2: A highway curve of radius 500 m is to be banked so that a car traveling 90 km/hr will have no tendency to skid sideways. At what angle should it be banked?

Solution: Speed of the car→ v =

(90)(1000) = 25 m/s (60)(60)

47 Radius of the highway curve→ R = 500 m If m is the mass of the car, then the equation of motion of the car is: m(25)2 mv2 = m a = F(centripetal) = = (1.25)(m) N R 500 If RN is the normal reaction of the highway on the car (FIG.6.9), RN can be resolved into two components: 1. RN sin θ (Horizontal component towards the center of highway curve).

2. RN cos θ (Vertical component downward) If the car is not to skid, then the following conditions must be satisfied: RN cos θ = m g (6.10) RN sin θ = 1.25 (m) (6.11) R N sin  1.25m 1.25 From equations (6.10) and (6.11) , we get: = , or tan θ = = 0.13 9.8 R N cos  mg  The angle at which highway should be banked→θ = tan-1(0.13) = 7 to the horizontal.

Example 3: A cord is tied to a pail of water and the pail is swinging in a vertical circle of radius 1 m. What must be the minimum velocity of the pail at the highest point of the circle if no water is to spill from the pail?

Solution: Let vm is the minimum speed of pail at the highest point (Fig. 6.10). Water in the pail will not spill if there is no net force acting on water in the pail. There are two forces acting on water, (1) Centrifugal force acting along the radius away from the center of the mv2 m circle, which is given by, F(centrifugal) = (upward) (R=1m) R (2) The second force is the weight of water mg (downward) If the water is not to spill with minimum speed, the two forces must be equal  m vm2 = m g, or vm = g = 3.13 m/s Example 4: A ball of mass 1 kg is attached to the end of a cord whose length is 1.50 m.

The ball is whirled in a horizontal circle. If the cord can withstand a maximum tension of 50 0 N, what is the maximum speed the ball can have before the cord breaks?

Solution: If T is the tension acting on the ball, the equation of motion of the ball  v2  can be written as, m a→ m R  = T   The maximum tension the cord can withstand = 50 N, Mass of ball = 1 kg (Tmax )(R) (50)(1.5) = = 8.66 m/s  The maximum speed ball can have→ vmax = m 1

48

Example 5: How many revolutions per hour would the Earth have to make in order that the weight of a body at the equator may become zero?

Solution: For the body to have zero weight , the centrifugal force due to the rotation of the Earth must be equal to Earth’s gravitational field g.  If ω is the angular speed of the Earth, 9.8 1.24 10−3 (Re )2 = g , or ω = g = -3 = 1.24 × 10 rad/s = rev/s 6.37 106 2 Re Re 1.24 10−3 × (3600) = 0.71 rev/hr Number of revolutions per hour = 2 Example 6: With what speed must a 2-kg pendulum bob swing in the circular path of the accompanying figure Fig.Ex-6.6 if the supporting cord is 1.60 m long and θ is 30o? Find the tension in the cord.

Solution: Radius of the circular path→ r = 1.6 sin 30º = 0.8 m Let v be the required speed to maintain θ = 30º mv2  = T sin 30º (1) r Also considering the equilibrium of forces in the vertical direction we have: m g = T cos 30º (2) v2 = tan30º, or v = rg tan30∘ Dividing (1) by (2) we get: rg 

The required speed→ v = (0.8)(9.8) tan 30∘ = 2.13 m/s mg 2  9.8 = From (2) tension → T = = 22.6 N ∘ cos30 cos 30∘ Example 7: It is proposed to put a space station in a circular orbit at a distance of one Earth’s radius above the Earth’s surface. (a) What will be the acceleration due to gravity at this station? (b) Find the speed of this station if it is to go around the center of the Earth in a circular orbit. (c) How long will it take to make a complete revolution?

Solution: (a) The value of g(y) at a height y above the surface of the Earth is given by: g(y) = − GMe (1) (re + y)2 Put in (1): G = 6.67×10-11 N m2/kg2, Me→ mass of the Earth = 5.98×1024 kg, and height of satellite abuve the Earth’s surface→ y = re = 6.37×106 m, we get, (6.67 10-11)(5.98 1024 ) g= − = -2.45 m/s2 (6.37 106 + 6.37 106 )2

49

(b)

speed v of the station is given by:

re + y)

 from (2),

(c)

(

v2

v=

=

GM e ,or v = (re + y) 2

GMe (re + y)

(2)

(6.67 10−11)(5.98 1024 ) (6.37 106 + 6.37 106 ) = 5600 m/s

2 (re + y) v 6 6) 2 (6.37 10 + 7.37 10 = 1.43×104 s  From (3), we get: T = 5600 Time taken to complete one revolution→ T =

(3)

Example 8: A pilot of mass m in a jet aircraft executes a loop-theloop, as shown in Fig.Ex-6.8. In this maneuver, the aircraft moves in a vertical circle of radius 2.70 km at a constant speed of 225 m/s.

Determine the force exerted by the seat on the pilot (a) at the bottom of the loop and (b) at the top of the loop. Express your answers in terms of the weight of the pilot mg.

Solution: (a) Let RT be the reaction of the seat on the pilot.

There are three forces acting on the pilot when the aircraft is at the top of the loop. (see Fig.Ex6-8) Reaction of seat on the pilot→ RT (vertically downward) Weight of pilot → mg (acting downward) mv2 Centrifugal force due to circular motion→ (acting vertically upward) R From equilibrium condition we have:  v2  mv2 Force exerted by the seat on pilot→ RT = - mg = mg − 1 R  Rg   v2  RT In terms of weight of pilot→ mg =  (1) − 1 Rg   2 Putting R = 2.7 km, v= 225 m/s in (1): R T =  (225) − 1 = 0.913 (downward) mg

(

)

3  2.7 10 (9.8)



(b) Let RB be the reaction of seat on the pilot when the aircraft is at the bottom of loop. The forces keeping the pilot in equilibrium are shown in Fig.Ex6-8,  v2  RB In terms of weight of the pilot → mg =  (2) + 1  Rg  2 + 1= 2.91 (upward) Putting the values of R and v in (2): R B =  (225) mg

(

)

3  2.7 10 (9.8)



50

51

Chapter 7 7.1

Work and Energy

Work:

When a force, acting on a body causes a displacement in the body, work is said to → be done by the force. Work done by force F acting at an angle  to the displacement

→ D (Fig. 7.1), caused by the force is defined as ‘the product of the displacement and the

component of the force in the direction of the displacement’. The work done by the force is defined by the equation: Work done→ W = FD = F D cos 

(7.1)

If the force does not produce a displacement, no work is done in the sense the work is used in physics. → → When F is perpendicular to D , then, from Eq. (7.1), work done is zero. Therefore no work is done when the force causing the displacement is perpendicular to the displacement. When a man walks at a constant horizontal velocity carrying a suitcase, he does no work on the → → suitcase, since the force F he exerts on the suitcase and the displacement D are mutually perpendicular. → → → is parallel to D , i.e. the displacement is in the same direction as the force F , then When F W = F D→ → When F is acting at an angle  to the direction of D , then W = F D cos  Unit of Work: Unit of work→ Joule = (1 Newton) (1 meter) When a force of one Newton produces a displacement of one meter in a body in the direction of the force, then one joule of work is said to be done by the force.

Example: How many joules of work are done by a force in lifting a mass of 2 kg upward a distance of 3 meters?

Solution: Work in joules = (force in newtons) (distance in meters) Force→ F = (2 kg)(9.8 m/s2) = 9.6 N Work done→ W = (19.6 N)(3 m) = 58.8 J

7.2

Power In physics, power is the rate (time) of doing work. Thus, power is the equal to the work divided by time: D P = = = (7.2) Average Power→ Time t t An engine of high power can do work more rapidly.

52 Instantaneous Power→

P = Lim

W

=

dW

(7.3) t dt Unit of Power: In the mks system power is measured in terms of joules per second called ‘watts’. One watt of power is expanded when one joule of work is done each second.

- In British engineering system power is often expressed in terms of horsepower; 1 horsepower = 746 watts t→0

Example 1: An electric motor exerts a force of 400 N on a cable and pulls it a distance of 30 meters in 1 minute. Find the power supplied by the motor.

Work (400 newtons)  (30 m) = = 200 w t (60 s) When a constant force acts on a body in the direction of body moving with a velocity v , the power can be expressed in the form : Solution:

Power→ P =

work PD = =Fv (7.4) time t Equation (7.4) is useful in resolving such practical questions as how powerful a motor must be provided to lift an elevator at a specified speed.

P=

Example 2: An aircraft has four 2000-hp engines and flies 650 km/hr when all engines are delivering rated power. Find the drag.

From Eq. (7.3), P=Fv (650) 103 8000  746  60  60 (8000)  (746) = F  or F = = 60  60 650 103

Solution:

7.3

33050 N drag

Energy and its Conservation Energy is defined as the ability or capacity to do work. It occurs in many forms, like heat, electricity, sound, gravitational and chemical energy. There are two kinds of mechanical energy: (1) Energy associated with motion is called ‘kinetic energy’. A stone thrown at a certain speed possesses kinetic energy. (2) The ability of a body to do work by virtue of its position is called ‘potential energy’. A raised body above the surface of the Earth can do work when dropped because of its elevated position.

The study of various forms of energy, and of the transformation of one kind of energy into another has led to the statement of a very important principle known as ‘law of conservation of energy’: ‘Energy cannot be created or destroyed; it may be transformed from one form to another, but the total amount of energy remains constant’. This principle is one of the great generalizations of physical science.

53 Early in the twentieth century it was found that mass itself could be converted into energy. We know that sun’s mass is decreasing because mass energy is converted into radiant energy in the sun. Einstein showed that, whenever mass is converted into another form of energy, the equivalence between mass and energy could be expressed by the equation: (7.5) E = m c2 E is the amount of some other form of energy appearing or disappearing, m is the mass appearing or disappearing, and c is the speed of light.

Unit of energy: Since energy is nothing but stored work, the unit of energy is the same as that of work (joule).

In atomic and nuclear physics another smaller unit of energy called ‘electron volt’ is used.

One electron volt is the amount of energy acquired by an electron in falling through a potential of one volt.

Electron volt→ ev = (1.610-19 C)(1 V)= 1.610-19 J Example: What is the total energy in a gram of matter at rest?

Solution: From Eq. (7.4), Total energy→ E = m c2 = (1 10-3 kg)(3102)2 = 91013 J This energy is sufficient to lift a mass of one million kilograms through a distance of 9000 km against the force of gravity.

7.4

Potential and Kinetic Energy The energy that a body has by virtue of its position or configuration is called ‘potential energy’. When a mass is lifted above the surface of the earth, it has energy because of its position. When a spring is compressed, potential energy is stored up. The measure of potential energy of a body because of its position is the work done against gravity in lifting he body. The upward force required is equal to the weight of the body and the work done in lifting a body through a height h is given by, Potential energy→ EP = W h = m g h (7.6) Example 3: Find the potential energy given to a 50 kg hammer when it is raised to 4 m.

Solution: Potential energy→ EP = m g h = (50 kg)(9.8 m/s2)(4 m) = 1960 J The energy possessed by a body by virtue of its motion is called ‘kinetic energy’. The bullet leaving the muzzle of a gun has kinetic energy and can do work in penetrating a board. To find the kinetic energy of a body, we consider the work which must be done on the body to give it its speed. When the body is stopped, it gives up this amount of energy. It can be easily shown that, (7.7) Kinetic energy→ EK = ½ m v2 The unit of kinetic energy is joule.

54 Example 4: What force is required to stop a bullet that has a mass of 15 g and a velocity of 400 m/s in a distance of 20 cm?

Solution: If F is the required force, the work done by the force in a displacement of 20 cm is: W = F D = F  0.2 = 0.2 F, this work done by the force must be provided by the kinetic energy of the bullet, which is given by, Kinetic energy→ EK = ½ m v2 = ½ (0. 015 kg) (400 m/s)2 = 1200 J 1200  0.2 F = 1200, or F = = 6000 N 0.2

7.5

Transformation of Potential and Kinetic Energy The potential energy of a body depends upon a height h, and this in turn depends on the zero of height. A book on a table has no potential energy relative to the table top, but it does have potential energy relative to the floor. Potential energy is always determined relative to some level or position which is assigned the value of zero. The choice of this level is at the disposal of the person working the problem.

In many situations we have a simple transformation of potential into kinetic energy with no other form of energy involved. Consider water going over a dam (Fig. 7.2). just above the dam, the water has potential energy, and if its velocity is negligible, no significant kinetic energy . As the water falls its potential energy decreases and the kinetic energy increases. When the water strikes the bottom, part of this kinetic energy is transformed into heat.

The swinging of a pendulum illustrates the way in which the potential energy may be transformed into kinetic energy and then back to potential energy (Fig. 7.3). When the pendulum is pulled on one side, the mass has no velocity and no kinetic energy, but it does have potential energy. When the mass is released, it swings downward acquiring kinetic energy and losing potential energy. At the bottom of the swing, the pendulum bob has maximum kinetic energy. As it swings upward, the kinetic energy is retransformed into potential energy.

Example 5: A pendulum bob is pulled to one side until its center of gravity has been raised 10 cm above the equilibrium, position. Find the speed of the bob as it swings through the equilibrium position.

Solution:

Potential energy at the top = Potential energy at the bottom  m g h = ½ m v2 or v2 =2 g h Substituting numerical values, v2 = 2  9.8  0.01 = 1.96 m2/s2  v = 1.96 = 1.4 m/s Note: For sake of convenience, symbol V for potential energy, and T for kinetic energy is sometimes used in the notes

55

7.7

Gravitational Potential Energy Let m be the mass of a body, Me the mass of the Earth, and r the displacement vector of m from the center of the Earth (Fig. 7.4). From Newton’s law of gravitation, the gravitational force on m is given by: Fg = - G

mM e rˆ r2

(7.9)

Nm2 kg2 The gravitational potential energy of mass m can be calculated ase follows: mM V = dV = r − dW = r −  - G mMe  = − G dr   g   r2  r    mMe Vg = - G (7.10)  Gravitational potential energy→ r If a satellite of mass m is in orbit around the Earth, the total mechanical of the satellite in the orbit is: Total energy→ E = Kinetic energy→T + Potential energy→ Vg T = ½ m v2, where v is the speed of the satellite mMe  E = ½ m v2 - G r Note: r is the distance of the satellite from the center of the Earth.

G is the universal constant of gravitation, and G = 6.6726 10 −11

7.8

Elastic Potential Energy If s is the extension in a wire due to tension T caused by a weight w (Fig. 7.5), then Hooke’s law gives us: Ts, or T = k s, where k is the elastic constant of the wire.

The elastic potential energy of a wire can be calculated as follows: s

s

s

0

0

0

Elastic potential energy→ EP =  − dw =  − (− T ds) =  (k s)ds = ½ k s2 (7.11) EP(elastic) = ½ k s2 In the same way, the potential energy of a stretched spring is given by, Potential energy of a spring→ (7.12) EP(spring) = ½ k x2 k is the spring constant, x is the extension of the spring

7.9

Solved Examples on Work and Energy Example 1: An external force of 4.8 N is applied to a 0.5 kg body initially at rest. (a) What is the kinetic energy of the body after it has been displaced 8 m in the direction of the applied force? (b) How much work is done on the body by the force?

Solution: First we calculate the acceleration of the mass using the equation of motion.

56 m a = F→ (0.5) a = 4.8, or a = 9.6 m/s2 (a) If v is the speed of the body after a distance of 8 meters, then v2 – v02 = 2 a x→ v2 – 0 = 2 9.6  8, or v2 = 153.6  Kinetic energy→ EK = ½ m v2 = ½ (0.5) (153.6) = 38.4 J (b) Work done by the force→ W = F D cos θ = (4.8) (8) (1) = 38.4 J Example 2: An automobile weighing 12500 kg moves at a speed of 20 m/s along a level road. When the car reaches an upward slope, the driver allows the car to coast uphill until it comes to rest. If the final position of the car on the hill is 15 m above the level of the approaching roadway, how much mechanical energy was dissipated during the car’s ascent of the hill?

Solution: Kinetic energy of the car before hitting the slope is : Kinetic energy→ EK = ½ m v2 = ½ (12500)(20)2 = 2500,000 J Energy spent in raising the car by 15 m along the plane against gravity and transforming into potential energy is: Potential energy→ EP = m g h = (12500)(9.8)(15) = 1837500 J  Mechanical energy dissipated→ EK – EP = 2500,000 – 1837500 = 662,500 J Example 3: A small cart on the circular loop-to-loop track in Fig. 7.4 can be approximated as a mass m sliding on a frictionless track. (a) calculate the smallest height h in terms of the radius r from which the mass can be released and still remain in contact with the track throughout the path. (b) If the car starts from a height of 4r, what is its acceleration at the top of the loop? (c) With what force does the track press down on the cart at the top?

Solution: (a) Let hm is the smallest height at which the cart of mass m remains in contact with the track throughout the path.

Potential energy of the cart at point H→ VH = m g hm, when it reaches the point Q (Fig.7.7), the whole potential energy has transformed into kinetic energy, and as it reaches the point P, part of the kinetic energy has transformed back into potential energy.

Therefore the kinetic energy of the cart at point P is: Kinetic energy of cart at P→ TP = ½ m vP2 = m g h – m g (2 r) (7.8), Where, vP is the speed of the cart at the point P. At P the forces acting on the cart are: mv2 P Centrifugal force→ (acting away from the center) r Weight of the cart→ m g, (acting downward toward the center) Reaction of track on the cart→ R (acting toward the center, which will be equal to the mv2 P resultant of centrifugal force and weight). That means, - mg = R. When the height is r least, R=0, and the least speed of the cart must be: Least speed so that cart remains in contact→ v L2= r g. By putting this value of vL in Eq. (7.8) we get: ½ m (r g) = m g hL- 2 m g r  The smallest height for the cart to remain in contact→ hL = 2.5 r

57

(b) If the cart starts from the height h = 4 r, the potential energy of cart = m g (4r). Ffrom Eq. (7.7), speed of the cart at the top of loop is given by: ½ m vP2 = m g (4 r) – m g (2 r) = 2 m g r, or vP2 = 4 g r v2 4gr  Acceleration at the top→ ar = P = = 4g r r v2P - mg = m (4 g) – mg = 3 m g (c) The reaction on the cart at the top→ R = m r This is the force with which the track presses against the cart Example 4: An electron strikes the screen of cathode-ray tube with a velocity of 107 m/s.

Calculate the kinetic energy in joules. The mass of electron is 9.1110-31 kg.

Solution: Kinetic energy→ T = ½ m v2 = ½ (9.1110-31)(107)2 = 4.56 10-17 J Example 5: A force of magnitude 5 N acts on a 2-kg body moving initially in the direction of the force with a speed of 4 m/s. Over what distance must the force act in order to change the body’s speed to 6 m/s.

Solution: Initial kinetic energy→ Ti = ½ m v2 = ½ (2)(4)2 = 16 J Kinetic energy when the speed changes→ Tf = ½ (2)(6)2 = 36 J The work done by the force to change speed→ Tf – Ti = 36 – 16 = 20J Work done → W = F  D W 20 The force acts over a distance→ D = = =4m F 5 Example 6: A man whose mass is 70 kg walks upto the third floor of a building. This is a vertical height of 12 m above the street level. (a) How many joules of work has he done? (b) By how much he has increased his potential energy? (c) If he climbs the stairs in 20 s, what was his rate of working in watts? → → Solution: (a) Work done →W = F  D = F D = (m g)(h) = (70)(9.8)(12) = 8232 J (b) Increase in potential energy→∆V = 8232 J W 8232 = 411.6 W = (c) Rate of working→ P = t 20 Example 7: A pump is required to lift 800 kg (about 200 gallons) of water per minute from a well 10 m deep and eject it with a speed of 20 m/s. (a) How much work is done per minute in lifting the water? (b) How much in giving it kinetic energy? (c) What horse power engine is needed?

Solution: (a) Work done per minute in lifting water→W = m g h = (800)(9.8)(10) = 78,400 J/min (b) Kinetic energy per minute→T = ½ m v2 = ½ (800)(20)2 = 160,000 J/min

58

(c) Total energy supplied by pump per minute: Total energy→E = T + V = (160,000) + (78400) = 238,400 J E 238,400 = = 3973 W Average power of engine→ P = t 60 3973 = 5.3 hp Horse power of the engine = 746 Example 8: A 2 kg-mass stretches a spring by 5 cm. (a) What is the spring constant of the spring? (b) What is the potential energy of the spring when stretched by 10 cm?

2  9.8 Tension T mg = = = Solution: (a) constant of spring→k = = 392 N/m 0.05 Extension x x (b) Potential energy of spring→ EP = ½ k x2 = ½ (392)(0.1)2 = 2.0 J Example 9: A particle of mass 1-kg is suspended from a rubber band and the extension in the band is 10 cm when in equilibrium. It is pulled downward by another 5 cm from the equilibrium position and then released. What is the speed of the particle when it is at the equilibrium position?

(1)  (9.8) = 98 n/m 0.1 Taking the potential energy to be zero at the equilibrium position, the change in potential after it is extended by 5 cm from equilibrium position is: Solution: Elastic constant of the band→ k =

Change in potential energy→ ∆EP = ½ k s2 = ½ (98)(0.05)2 = 0.123 J When it passes the equilibrium point after release, the potential energy ∆EP is all transformed into kinetic energy.

2EP 2  0.123 ½ m v2 = ∆EP = 0.123, or v = = = 0.5 m/s m 1 Example 10: A block of mass0.25 kg is placed on top of a light vertical spring of force constant 5000 N/m and pushed downward so that the spring is compressed by 0.10 m.

After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise?

Solution: The potential energy stored in the spring when compressed by 0.10 m is: Potential energy of spring→ EPS = ½ k s2 = ½ (5000)(0.10)2 = 25 J (1) When the block is released, the potential energy of spring is trnsferred to the block as initial kinetic energy of the block, which again trnsforms into gravitational potential energy of the block when it attains maximum height. If h is the maximum height attained, the potential energy will be mgh.

25 25 = = 10.2 m Therefore from energy conservation, m g h = 25, or h = (0.25)(9.8) mg

59 Example 11: The coefficient of friction between the 3.00-kg block and the surface in ˚ has fallen 1.50 m?

Solution: Energy loss due to friction in the movement of 3-kg mass by a distance of 1.5 m: Energy loss → E = µ m g d = (0.4)(3.0)(9.8)(1.5) = 17.6 J Total energy gained by 3-kg and 5-kg masses when 5-kg mass falls under gravity by 1.5 m: Total energy gained by the two masses→ E = m g h = (5)(9.8)(1.5) = 73.5 J Net energy gained = E - E = 73.6 – 17.6 = 55.9 J This 55.9 J of energy goes into the two masses as kinetic energy.

55.9  ½ (3.0 + 5.0) v2 = 55.9, or v = = 3.74 m/s 4 Example 12: A 5.0-kg block is set into motion up an inclined plane with an initial speed of 8.0 m/s. The block comes to rest after traveling 3.0 m along the plane, which is inclined at an angle of 30˚ to the horizontal. For this motion determine (a) the change in the block’s kinetic energy, (b) the change in potential energy of the block-Earth system and (c) the friction force exerted on the block (assumed to be constant). (d) What is the coefficient of kinetic friction?

Solution: (a) Initial kinetic energy of the block→ EK = ½ m v 2o = ½ (5.0)(8.0)2 = 160 J After it comes to rest the kinetic is zero.

The change in block’s kinetic energy→ EK = 0 - 160= - 160 J (b) The potential energy after it comes to rest → EP = m g (3 sin 30˚) (5)(9.8)(1.5) = 73.5 J (c) Loss of energy due to friction→ Ei – Ef = 160 – 73.5 = 86.5 J Let f be the force of friction.

The work done against friction in a displacement of 3 m→ W = f × (3) = 3f = 86.5 86.5  The frictional force exerted on the block→ f = = 28.8 N (1) 3 (d) Coefficient of kinetic friction is given by:

f = µK (m g cos 30 ) = 28.8 28.8 Coefficient of kinetic friction→ µK = = 0.679 (5)(9.8)(0.866)

(2)

Example 13: A 2.0-kg block situated on a rough incline is connected to a spring of negligible mass having a spring constant of 100N/m (Fig.Ex.7.13). The pulley is frictionless.. The block is released from rest when the spring is unstretched . The block

60 moves 20 cm down the incline before coming to rest. Find the coefficient of kinetic friction between the block and incline.

Solution: The change in energy of the block in moving down the plane by 20 cm is: Change in energy→ E = EK + EP = 0 – (m g sin 37˚)(0.2) = - (2)(9.8)(0.602)(0.2)= - 2.36 J 2.36 J of energy is spent on stretching the spring by 20 cm and the work done on friction between the block and the plane.

Increase in potential energy of spring → EPS = ½ k s2 = ½ (100)(0.2)2 = 2.0 J The work done against friction→ Wf = µK (m g cos 37˚)×(0.2) =2.36 – 2.0 = 0.36 J 0.36 0.36 The coefficient of kinetic friction→ µK = = 0.115 = (mgcos37∘ )(0.2) (2)(9.8)(0.799)(0.2)

61

Chapter 8 8.1

Momentum and Impulse

Momentum We know from Newton’s laws that an unbalanced force produces acceleration which results in motion of a body. It is clear that the motion depends on the mass and velocity of a body. The quantity of motion called ‘momentum’ of a body is a vector defined as the product of its mass m and its velocity v . We can express it as: momentum→

p =m v (8.1) → → → → → → dv (m v) m(v− v ) 0 = From Newton’s second law, F = m a = m = , which gives: t t dt → → → → → (8.2) = ( − ) Ft m v v0 = p - p0 The right side of Eq. (8.2) is just the momentum at time t minus the initial momentum, or the change in momentum of the body.

The left side of the equation, the product of the force and the time during which the force acts, is called the ‘impulse of the force’. From equation (8.2), we see that the impulse of a force is equal to the change in the momentum.

In many situations the force acting on a body varies so rapidly that it is not practical to know what the force is at a particular instant, but the change in the momentum of the body is equal to the total impulse. If we know how long the impulsive force acts, we can calculate the average value of the force even though we cannot know the instantaneous value, i.e. the exact shape of the force curve.. We have impulsive forces in play when a pitched ball is hit by a bat or when two billiard balls collide..

Example: A baseball of mass 0.145 kg is thrown by the pitcher with a speed of 30 m/s.

The bat is in contact with the ball for 0.01 s and gives it a speed of 40 m/s in a direction straight toward the pitcher. Find the impulse and average value of force.

Solution:

8.2

Impulse = Change in momentum = (0.145){(30) - (- 40)} = 10.1 N.s Average force→ F = Impulse = 10.1Ns = 1,010 N Time 0.01s

Conservation of Momentum

→ on the bullet. By Newton’s When a trigger of a gun is pulled, a force Fb is exerted →

third law the bullet exerts an equal and opposite force Fg on the gun. We therefore write

62 → → → → Fb = - Fg . Both the forces Fg and Fb act for exactly the same length of time, and impulse of the gun and bullet are equal in magnitude but opposite in direction. Therefore, vB = - mg vg , (mB, mg are the masses of bullet and gun) (8.3) The momentum gained by the bullet is equal in magnitude but opposite in direction to that received by the gun. The resultant of the two momenta is zero, just as it was before the trigger was pulled. The momentum of the system composed of gun and the bullet has not changed. This leads us to a very important principle called ‘the law of conservation of momentum’. It may be stated as follows.

‘In the absence of any external force, the total momentum of any system of bodies is unchanged by any actions which occur among the different constituents of the system.

When two billiard balls collide, momentum is transferred from one ball to the other, but the resultant momentum after the collision is equal to the momentum before the collision.

Example 1: (a) What is the momentum of a 10,000 kg truck whose velocity is 20 m/s? (b) What velocity must a 5000 kg truck attain in order to have the same momentum and (c) the same kinetic energy? → → Solution: 4 5 (a) Momentum of the truck→ ptruck = mtruck vtruck = (10 )(20) = 210 kg.m/s (b) Let v→2 be the velocity of 5000 kg truck, to have the same momentum (210 5 kg.m/s), → → 5000  v2 = 2 105 or v2 = 40 m/s (c) To have the same kinetic energy as the first truck, ½ m2 v22 = ½ mtruck v 2truck , or ½ (5,000) v22 = ½ (104)(20)2 v2 =

104  400 = 800 = 28.3 m/s 5000

Example 2: A baseball has a mass of o.2 kg. (a) If the velocity of pitched ball is 30 m/s, and after being batted it is 50 m/s in the opposite direction, find the change in momentum of the ball and the impulse of the blow. (b) If the ball remains in contact for 0.002s, find the average force of the blow.

Solution: Suppose the ball is thrown in the negative x-direction (a) Momentum before hit→ pi = mB vB = (0.2) (-30)= - 6 kg.m/s (-ve x-dir.)

Momentum after hit→ pf = (0.2) (50) = + 10 kg.m/s (+ve x-dir.)

Change in momentum→∆p = pf – pi = (10) – (- 6) = 16 kg.m/s (+ ve x-dir)

63

Change in momentum = Impulse of the blow→ P = 16 N.s (+ ve x-dir) (b)

Impulse p 16N.s Average force→ F = Time = t = 0.002s = 8000 N

Example 3: A bullet having a mass of 0.05 kg moving with a velocity of 400 m/s, penetrates a distance of 0.1 m into a wooden block firmly attached to the Earth. Assume the accelerating force to be constant, calculate (a) the acceleration of the bullet, (b) the accelerating force, (c) the time of acceleration, and (d) the impulse of the collision.

Solution: The kinetic energy of the bullet→EK = ½ mB vB2 = ½ (0.05)(400)2 = 4000 J If F is the magnitude of the accelerating force, then the work done by the force is: Work done → W = F D = F(0.01) Assuming that no energy is lost, the kinetic energy of the bullet must be equal to the work done by the bullet in penetrating wood, 4000 = 4 104 N 0.01 F= 4000, or F = 0.01 F 4000 (a) Acceleration of bullet→a = = = 8 105 m/s2 m 0.05 (b) Accelerating force→ F = 4  104 N (c) Momentum of bullet before penetration→pi = mB vB = (0.05)(400) = 20 kg.m/s Change in momentum of bullet→ p = pf − pi = 0 − 20 = 20kg.m/s

(c)

Magnitude of impulse of force → P =  p = 20 kg.m/s Impulse P 20 = = The time of acceleration→ t = = 5104 s Force F 4 104 Impulse of force→P =  p = 20 kg.m/s

Example 4: A car of 2000 kg mass has a velocity of 15 m/s eastward. (a) What is the momentum of the car? (b)What would be the velocity of 150-g bullet if its momentum were the same as that of car? (c) Which would have greater kinetic energy- the car or the bullet?

Solution: (a) Momentum of the car→ p = m v = (2000)(15 m/s) = 30,000 kg.m/s in the easterly direction. (b) If vB is the speed of bullet with the same momentum as the car, 0.15 vB = 30,000, or vB = 200,000 m/s (eastward) (c) Kinetic energy of bullet→TB = ½ mB vB2 = ½ (0.15)(2 105)2 = 3  109 J Kinetic energy of the car→Tc = ½ (2000)  (15)2 = 225,000 J = 2.25  105 J Obviously, the kinetic energy of the bullet is greater than the kinetic energy of the car.

8.3

Center of Mass

64 A point inside or outside a body where the total mass of the body can be assumed to be concentrated is called ‘center of mass’. The x-coordinate of the center of mass xc of a system of particles of masses m1, m2, m3………mN is defined at any instant by the relation, N

xc =

m1x1 + m 2 x 2 +  +m N = m1 + m 2 +  +m N

 mixi i=1 N

(8.3)

 mi

i=1

where, x1 is the x-coordinate of m1, etc. Similarly, m1 y1 + m2 y2 +  + mN yN

, and zc = m1z1 + m 2 z 2 +  + m N z N m1 + m2 +  + mN m1 + m2 +  + mN The sum of the individual masses in the denominator is simply the total mass M of the system of particles. y-coordinate of c.m.→yc =

Center of Gravity: Every point of an extended body possesses weight, so that the pull of the earth on the body is composed of a large number of forces directed towards the center of the Earth. For a body of ordinary size, these forces are parallel to one another. The body can however be supported in equilibrium by a single upward force, provided its line of action passes through a point called the ‘center of gravity’ of the body. When we are dealing with an extended body, we may consider all the weight of the body as concentrated at the center of gravity, and we may thus replace the individual forces acting on many particles of the body with a single force equal to the weight.

‘The center of gravity of an object is a point at which we can consider the entire weight of the object to be concentrated’.

-If (xc, yc, zc) are the coordinates of the center of gravity of a system of weights.

N

 xi wi

xc = i=1N = i=1 w  wi i=1

N

N

 xi w i

,

 yiwi

yc= i=1N  wi i=1

N

N

 yiwi

= i=1

w

,

 ziwi

N

 ziwi

zc= i=1N = i=1 w  wi

(8.4)

i=1

Example: Find the center of gravity of two spheres A and B of weights 0.1 and 0.4 newton, respectively, connected by a light (negligible mass) rod 20 cm long.

Solution: We take the coordinate axes as shown in Fig. 8.3. Referred to this system, the coordinates of the two spheres are, A(0,0) and B(20 cm, 0). From Eq. (8.4), we get:

65 w A x A + w Bx B wA+ wB (0.1)(0) + (0.4)(20) = 8.0 = 16 cm = (0.1+ 0.4) 0.5 The x-coordinate of center of gravity is at a distance of 16 cm from o. yc is zero. Therefore the coordinates of center of gravity are (16,0) with reference to the coordinate system chosen. x-cordinate of center of gravity→ xc =

8.4

Collision Phenomena When two bodies collide, the laws of conservation of momentum and conservation of energy are both always applicable. However, in some collisions part of the kinetic energy of bodies is transformed into non mechanical form of energy as heat or sound. In this case the application of the law of conservation of energy cannot be applied because the energy loss is difficult to measure. Thus there are two types of collisions: (1) elastic collision, and (2) inelastic collision.

(1) In ‘elastic collision’ it sis assumed that no mechanical energy is dissipated or dissipated energy is negligible. In such cases the total kinetic energy of bodies colliding is conserved before and after the collision.

(Elastic):

Kinetic energy before collision = Kinetic energy after collision

In general there is always some energy loss (as heat) when bodies collide. If the loss is small compared to the kinetic energy, the collision is assumed to be elastic.

(2) In other extreme, if the bodies stick together after collision, such collisions are called ‘perfectly inelastic’ In practice most of the collisions are neither perfectly elastic, nor perfectly inelastic.

Therefore in such cases, the principle of conservation of energy cannot be used unless the process can be considered as elastic collision.

In every kind of collision, the momentum is always conserved; kinetic energy is conserved only in perfectly elastic collisions.

Collision of two spheres: We consider collisions between two spherical →bodies →of masses m and m (Fig. 8.4) which have initial velocities and v2i v1i 1 2 → → respectively, before collision, and final velocities v1f and v2f after collision.. In the equations that follow, we adopt the convention that velocities to the right are positive, and velocities to the left negative.

In the collision of two spheres on a horizontal surface, there

66 are no external forces acting on the spheres in the direction of motion. principle of conservation of momentum holds.

Therefore, the

v + m 2 v2i = m1v1f + m 2 v2f



(8.5)

1 1i

In elastic collision, the kinetic energy is also conserved:  Or,

m1 2

½ m1v1i2 + ½ m 2 v22i = ½ m 1 v21f + ½ m 2v22f 2 1i

2 1f

(8.6)

m2 2 m m v 2f - v 22i → 1 (v1i + v1f )(v1i − v1f ) = 2 (v2f + v 2i )(v2f − v 2i ) (8.7) 2 2 2

From Eq. (8.5),

1

From equations (8.7) and (8.8),

1

v1i

v1f

2

v2f

v2i

(8.8)

(v + → ) = 1 (→ + → ) → (→ − → ) = (→ − → ) 2

1i

v1f

2

v2f

v2i

v1i

v2i

v2f

v1f

(8.9)

It means that in elastic collision: Relative velocity before collision = Relative velocity after collision In an inelastic collision, conservation of energy cannot be applied.

Instead, we apply Newton’s rule of collision, which can be stated as: → → → → (v1f − v2f ) = −e (v1i −v2i ) (8.10) e is called the coefficient of restitution. e  1. In case of perfectly elastic collision, e = 1. In case of perfectly inelastic collision, e = 0.

If Ti and Tf are the kinetic energies before and after the collision, it can be shown that the energy loss in the collision is given by, E = Tf – Ti = - (1-e2) Ti

(8.11)

Example 1: A 3.0 kg steel ball strikes a wall with a speed of 10 m/s at an angle of 60 with the surface. It bounces off with the same speed and angle. If the ball is in contact with the wall for 0.20 s, what is the average force exerted on the ball by the wall?

Solution: → → Let pi and and final momenta of the ball, respectively. Then → pf be the initial  ˆ  ˆ pi = (pi sin 60 ) i + (pi cos 60 ) j , p→i = (3.0)  (10) = 30 kg.m/s where,  ˆ  ˆ Similarly, pf = (- pf sin 60 ) i + (pf cos 60 ) j , where , pf = (3.0)  (10) = 30 kg.m/s

67 → →

Change in momentum→

p = pf - pi

= (-30 sin 60 ˆi + 30 cos 60 ˆj ) – (30 sin 60 ˆi + 30 cos 60 ˆj )

= (- 25.98 ˆi + 15 ˆj ) – (25.98 ˆi + 15 ˆj ) = - 51.96 ˆi ˆi = Impulse = F→  t   p = - 51.96 → ˆ →  ˆ p - 51.96 i = = 259.8 i N  F= t 0.20 The average force exerted by the wall is 259.8 N in the negative x-direction.

Example 2: A 12g-bullet is fired into a 100-g wooden block initially at rest on a horizontal surface. After impact, the block slides 7.5 m before coming to rest. If the coefficient of friction between block and the surface is 0.65, what was the speed immediately before impact?

Solution: The collision is inelastic, the conservation of energy principle cannot be applied. However, the conservation of momentum principle applies.  (8.12) mB vB = (mB + mw) vf mB→ mass of the bullet, vB→speed of bullet, mw→ mass of wooden block, and vf→ is the speed of wooden block and the bullet lodged inside the block. After impact the kinetic energy of the block and the bullet is: Tf = ½ (m B + m w )v2f (8.13) Kinetic energy after impact→ This kinetic energy (Tf) is lost due to friction between the block and the surface. Therefore the kinetic energy in equation (8.13) must be equal to the work done against the force of friction or the energy lost due to friction.

Energy lost due to friction→ Wf = µ (mB + mw) g 7.5 From Eqs (8.130 and (8.14) we have,

(8.14)

½ (mB + mw )v2f = µ (mB + mw) g 7.5

 vf = 2g  7.5 = 2  0.65  9.8  7.5 = 9.77 m/s From Eq.(8.12) we get, 0.012 vB = (0.1 + 0.012)  9.77, 0.112  9.77 = 91.2 m/s Or vB = 0.012 Example 3: A billiard ball moving at 5.00 m/s strikes a stationary ball of the same mass.

After collision, the first ball moves at 4.33 m/s at an angle of 30 with respect to the original line of motion. Assuming an elastic collision (and ignoring the frictional and rotational motion), find the struck ball’s final velocity.

Solution: Since the balls have the same mass, m1 = m2 = m

68 Before collision → Velocity of m1 → v1i = v1i ˆi + 0 ˆj = 5.0 iˆ , velocity of m2 →v2i = 0 After collision → Velocity of m1→ v1f =(v1f cos 30) ˆi +(v1f sin 30) ˆj = (0.8664.33 ) ˆi +(0.54.33) ˆj = 3.75 ˆi +2.17 ˆj → Velocity of m2→ v2f = (v2f cos ) ˆi - (v2f sin ) ˆj → → Since the momentum is conserved in any collision, m v i = m +m v1f v2f 1i → → 0r , v1i i = v1f + v2f  5.0 ˆi = (3.75 ˆi +2.17 ˆj ) + (v2f cos ) ˆi - (v2f sin ) ˆj = (3.75 + v2f cos ) ˆi + (2.17 + v2f sin ) From above we get, 5.0 = 3.75 + v2f cos , or 1.25 = v2f cos  And 0 = 2.17 + v2f sin , or - 2.17 = v2f sin  → Final velocity→ v2f = (v2f cos ) ˆi - (v2f sin ) ˆj = (1.25 iˆ - 2.17 ˆj ) Example 4: On a uniform plank 5 m long weighing 25 kg. is placed a concentrated weight of 20 kg at one meter from one end, and another concentrated weight of 15 kg, at 3.5 m from the same end. Find the distance of the center of mass of plank plus weights from the same end.

Solution: If we take the x-axis along the plank with one end as origin (Fig. 8.6), the coordinates of the three masses, including the weight of the plank at the center of mass of the rod are: Mass of the rod→ m1 = 25 kg, at coordinates (x1 = 2.5 m, y1 = 0) Concentrated mass→ m2 = 20 kg, at coordinates (x2 = 1.0 m, y2 = 0) Concentrated mass→ m3 = 15 kg, at coordinates (x3 = 3.5 m, y3 = 0) The x-coordinate of center of mass of m1, m2, m3 is given by: m x + m2 x 2 + m 3 x 3 (25)(2.5) + (20)(1.0) + (15)(3.5) 135 xc = 1 1 = = =2.25 m 25 + 20 +15 60 m1 + m2 + m3 The coordinates of center of mass are: (xc = 2.25, yc = 0) Example 5: A ball of mass 0.15 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.96 m. what impulse was given to the ball by the floor?

Solution: Let vi and vf be the be the velocities of the ball before, and after hitting the floor, respectively. vi is given by: (negative y-direction) vi = 2 g y = 2(9.8)(1.25) = 4.95 m/s vf = 2(9.8)(0.96) = 4.34 m/s Similarly, (positive y-direction) Change in momentum of the ball→ p = m (vf – vi) = (0.15)(4.34 + 4.95) = 1.39 kg.m/s ( j)  Impulse given by floor = 1.39 N.s in the y-direction

69 Example 6: A 45-kg girl is standing on a plank that has a mass of 150 kg. The plank, originally at rest, is free to slide on a frozen lake, which is flat, frictionless supporting surface. The girl begins to walk along a plank at a constant speed of 1.5 m/s relative to the plank. (a) What is her speed relative to the ice surface? (b) What is the speed of the plank relative to the ice surface?

Solution: Let vg and vf be the velocities in the x-direction of the girl and plank with respect to ice, respectively.

Velocity of the girl relative to plank→ vgp = vg – vp = 1.5 m/s (x-direction) (1) Since there is no external force acting on the girl and plank in the x-direction, the total momentum of the girl and plank in the x-direction will be conserved before and after the girl starts walking with a constant speed. The total momentum in x-direction before the girl started walking is zero. Therefore, it must be zero even when the girl stated walking. This will only be possible if the plank, moves opposite to girl’s direction with the same momentum as the momentum of the girl.  mg vg + mp vp =0 (2) (a) Substituting vp in (1) into (2), we get mg vg + mp (vg - 1.5) = 0, or (45)(vg ) + (150)(vg – 1.5) = 0  195 vg = (150) (1.5) (150)(1.5) Velocity of girl relative to ice→ vg = = 1.154 m/s (positive x-dir) 195 (b) From (2), velocity of plank relative to ice→ vp = vg – 1.5 = 1.154 – 1.5 = - 0.346 m/s Example 7: A 10.0-g bullet is fired into a stationary block of wood (m = 5.00 kg) The relative motion of the bullet stops inside the block. The speed of the bullet-plus-wood combination immediately after the collision is 0.600 m/s. What was the original speed of the bullet?

Solution: This is a cse of inelastic collision. The energy will not be conserved in this collision. However, since no external force is acting on the bullet and the block, the momentum will be conserved.

Let vo be the original speed of the bullet, then from momentum conservation principle, mB vo = ( mB + MBL) vf, where MBL→ mass of the block, vf→ speed of bullet + block (5.01)(0.6)  (0.01) (vo) = (0.01 + 5.0) (0.6), or vo = = 301 m/s 0.01 Example 8: Consider a system of two particles in the xy-plane: m1 = 2.0 kg is at the → location r1 = (1.0 ˆi + 2.0 ˆj ) m and has a velocity of (3.0 ˆi + 0.5 ˆj ) m/s; m2 = 3.0 kg is at → ˆ ˆ ˆ ˆ r2 = (- 4.0 i - 3.0 j ) m and has velocity (3.0 i - 2.0 j ) m/s. (a) Find the position of the center of mass of the system. (b) Determine the velocity of the center of mass. (c) What is the total linear momentum of the system.? → Solution: (a) The center of mass coordinate rc is given by:

70 → rc =

 m →r n

(2.0)(1.0 ˆi + 2.0 ˆj) + (3.0)(- 4.0 ˆi - 3.0 j)

i i

1

=

n

m

=

(2.0 + 3.0)

i

(−10ˆi - 5ˆj) 5.0

= ( - 2 ˆi - ˆj ) m

→ The velocity of center of mass vc is given by; n → m v  i i (2.0)(3.0 ˆi + 0.5 ˆj) + (3.0)(3.0 ˆi - 2.0 ˆj) (15 ˆi - 5ˆj) → vc = 1 n = = = (3.0 ˆi - ˆj ) m/s 5.0 (2.0 + 3.0)  mi

(b)

1

1

→ (c) The total linear momentum of the system is given by: ⇀ Total momentum→ p =(m +m ) = (2.0 +3.0)( 3.0 ˆi - ˆj ) = (15.0 ˆi - 5 ˆj ) kg.m/s 1 2 vc Example 9: A spherical mass of 2 kg moving along the x-axis with a velocity of 5 m/s collides elastically with a spherical mass of 4 kg initially at rest. After the collision the 2kgmass moves off at an angle of 37˚ with the x-axis. Find the speed of both masses after the collision and the angle which the path of 4 kg-sphere makes with the x-axis.

Solution: Using the principle of conservation of omentum: X-comp: m1v1i + m2v2i = m1v1f cos θ + m2 v2f cos φ Given: m1 = 2kg, m2 = 4 kg, v1i = 5 m/s, θ = 37º  (2)(5) + (4)(0) = (2)(v1f cos 37º) + (4)(v2f cos φ) or, 1.6 v1f + 4 v2f cos φ = 10 (1) Y-comp: 0 = v1f sin θ - v2f sin φ  1.2 v1f = 4 v2f sin φ (2) Since the collision is elastic, the energy before collision is equal to energy after bthe collision;  ½ m1 v1i2 + ½ m2 v 22i = ½ m1 v1f2 + ½ m2 v 22f  ½ (2) (5)2 + 0 = ½ (2)( v1f2 ) + ½ (4)( v 22f ) or,

2 v1f2 + 2 v2f = 25

(3)

Squaring (1) and (2) and adding, we get: 16 v 2f2 = 100 + 4.0 v1f2 -32 v1f (4) 2 2 (25 − v 2 1f Putting the value of v2f = from (3) in (4), we get: 3 v1f - 8 v1f –25= 0 2 Equation (4) is a quadratic in v1f, whose roots are:



Taking the positive root gives: v1f = 4. 51 m/s

(5)2 − (4.51)2 = 1.53 m/s

Putting the value of v1f in (30 gives:

v2f =

Putting v1f and v2f in (2) gives:

1.2  4.51 = 0.88 or φ = 62º 4.0 1.53

sin φ =

2

(5)

71 Example 10: A tennis ball bounces down a flight of stairs, striking each step in turn and rebounding to the height of step above. If the height of each step is 20 cm, find the coefficient of restitution.

Solution: from Fig-Ex-9.10, it is clear that the ball falls under gravity by 40 cm and rebounds to a height of 20 cm against gravity.

The speed of tennis ball when it hits the step: vi = 2gh = (2)(9.8)(0.4) = 2.8 m/s Speed of ball after rebound to a height of 20 cm: vf = = 1.98 m/s v  The coefficient of restitution→ e = f = 1.98 = 0.707 2.8 vi Example 11: A 5-g bullet is fired into a 2.995-kg block of wood which is the bob of a ballistic pendulum. If the bob leaves its equilibrium position with a speed of 1 m/s, find the speed of the bullet and height above the equilibrium position reached by the center of gravity of the block.

Solution: Let v1i be the speed of the bullet.

From the principle of conservation of momentum we get: m1 v1i + m2 v2i = (m1 + m2) vf  (0.005) (v1i) = (2.995 +0.005) (1.0), 3.00 or v1i = = 600 m/s 0.005 Kinetic energy of block and bullet → EK = ½ (m1 + m2) v 2f = ½ (3.00)(1.0)2 = 1.5 J This kinetic energy transforms into potential energy by gaining a height h against gravity.  EK = (m1 + m2) (g) (h) , 0r 1.5 = (3.00) (9.8) (h) 1.5 = 0.051 m  Height above equilibrium position→ h = (3)(9.8) Example 12: A 4-g bullet is fired horizontally with a speed of 1000 m/s into a 4-kg block of wood at rest on a table. If the coefficient of friction between the block and the table is 0.25, how far will the block slide? What fraction of the bullet‘s energy is dissipated in the collision? (20.4 cm, 99.9%%) Solution: Let vf be the speed of block and bullet after the bullet hits the block.

From momentum conservation: (0.004) (1000) = (0.004 + 4.00) vf 4.00  vf = = 0.999 m/s 4.004 mg Acceleration due to friction→ a = = - µ g = - (0.25) (9.8) = - 2.45 m/s2 m

72

The distance d the block will slide before coming to rest is:

vf2 (0.999)2 d= = = 0.204 m 2  2.45 2a

Energy before collision→ Ei = ½ (0.004) (1000)2 = 2000 J Energy after collision → Ef = ½ (4.004) (.999)2 = 1.998 J Loss of energy→ ∆E = Ei – Ef = 1.998 – 2000 = 1999 J 1999 = 0.999, 0r 99.9 % Fraction of bullet energy lost = 2000 Example 13: A water molecule consists of an oxygen atom with two hydrogen atoms bound to it (Fig-Ex-8.13).. the angle between the two bonds is 106˚. If the bonds are 0.100 nm long, where is the center of mass of the molecule?

Given: Mass of hyhrogen atom = 1.008 u, Mass of oxygen atom = 15.999 u Solution: Taking the center of oxygen atom origin and x and y-axes as shown in Fig-Ex-8.13, the coordinates of cm of O, H, and another H atoms are: O(0,0), H( ℓ cos 53˚, ℓ sin 53˚), H( ℓ cos 53˚, - ℓ sin 53˚) ℓ is the length of bond between H-atom and O-atom. The coordinates of center of mass are given by: n

m x

i i

x-cordinate of cm→ xc =

1

n

m

i

1

(15.999)(0) + (1.008)(0.1 cos 53∘ ) + (1.008)(0.1cos53∘ ) (15.999 + 1.008 + 1.008) 0 + 2(1.008)(0.1)(0.6018) = = 0.00673 nm 18.015

=

n

my  1 i

y-coordinate of cm→ yc =

i

n

m

i

1

=

(15.999)(0) + (1.008)(0.1 cos 53∘ ) + (1.008)(0.1cos53∘ )

or,

(15.999 + 1.008 + 1.008) yc = 0

The cm lies at a distance of 0.00673 nm from the oxygen atom on the x-axis.

73

Chapter 9

Equilibrium

Statics is a branch of mechanics which deals with the balance of forces on an object that remains at rest or in uniform motion. A body of finite size can undergo two types of motions: (1) A body can move as a whole with its center of mass. Such motion is called ‘translation’. (2) In the other type of motion called ‘rotation’, a body can rotate about an axis which may be fixed or moving. When all the effects of forces acting on a body are balanced in such a way that the state of rest or of uniform motion (translation and/or rotation) does not change, the body is said to be in ‘equilibrium’. Thus, we have two kinds of equilibriums; static and dynamic equilibriums.

9.1

Translational Equilibrium

→ → → → Let F1, F2, F3, and F4 be forces acting on a body (Fig. 9.1), the condition of translational equilibrium is: → → → → F1 + F2 + F3 + F4 = 0 (9.1) It means that the vector sum of all the external forces acting on a dynamical system at rest or in uniform motion must be zero. If resolve the forces in a 3-dimensional coordinate system, Eq. (9.1) can be written as: (F1xˆi + F1yˆj + F1zk̂) + (F2xˆi + F2yˆj + F2zk̂) + (F3xˆi + F3yˆj + F3zk̂) + (F4xˆi + F4yˆj + F4zk̂) =0 0r (F1x + F2x + F3x + F4x )ˆi + (F1y + F2y + F3y + F4y )ˆj + (F1z + F2z + F3z + F4z ) k̂ = 0 Therefore, the conditions of translational equilibrium can be written as, F1x + F2x + F3x + F4x = 0,

F1y + F2y + F3y + F4y = 0,

F1z + F2z + F3z + F4z =0

(9.2)

In general, the conditions of translational equilibrium are:  Fix = 0,  Fiy = 0,  Fiz = 0

9.2

(9.3)

Rotational Equilibrium Force produces translational acceleration in a body. To have rotational acceleration one must apply ‘torque’ on a body.

Torque (Moment of force) about an axis is defined as the product of magnitude of force and the perpendicular distance from the point of rotation to the direction of applied force. If a plane body (Fig. 9.2) is rotating about an axis passing through the point O, moment of force about the point O is: → Moment of force (torque) about O→ M = F  (OM) (in the anti→ clockwise direction). In vector form, torque can be written as a cross product of F , and r : M = r F (9.4) Torque→

74 Torque is a vector quantity. For a plane body there are only two directions of torqe; clockwise and anti-clockwise. By convention, the anti-clockwise moment is talen as positive.

For Rotational equilibrium in a plane, the sum of counter-clockwise torques about any axis must be equal to the sum of clockwise torques about the same axis. i.e.

(9.5) (about any axis)  Mi = 0 The necessary and sufficient conditions of equilibrium of a body are:  Fi = 0,

9.3

 Mi = 0 (about any axis)

(9.6)

Rotational Motion Pure rotation is the rotation of a rigid body about a fixed axis. A rigid body consists of a large collection of particles (atoms and molecules) where the inter-particle distance remains fixed. An ideal rigid body is completely inflexible.

Angular displacement: It is the angle through which any line fixed in a body turns with respect to a fixed line in space. Counterclockwise direction of rotation is taken as positive and clockwise direction as negative. Angular displacement is a vector quantity.

Angular velocity: The time rate of change of angular displacement is called ‘angular velocity’. The unit of angular velocity is radian/second. We can write it as: Instantaneous angular velocity→  = Lim t→0

t

=

d dt

(9.7)

If angular velocity is constant, the angular displacement θ at any time is given by:

 = 0 +  t

(0 is the angular displacement)

(9.8)

Angular acceleration: It is defined as the time rate of change of angular velocity. i.e.

Angular acceleration→

 = Lim t→0

t

=

dt

(9.9)

For constant angular acceleration, as in rectilinear motion, we can write the following relations between angular displacement, angular velocity, angular acceleration, and time.  = 0 +  t,

 = 0 t + ½  t2,

2 - 02 = 2  

(9.10)

75 Example: A motor driven grinding wheel starts from rest and receives a constant angular acceleration of 3 rad/s2 for 12 seconds. Determine (a) the angular velocity at the end of this interval and (b) the angle through which it has turned.

Solution: Given: 0 =0,  = 3 rad/s2, t = 12 s (a) From  = 0 +  t , we get, (b)

9.4

From

 = (0) + (3)(12) = 36 rad/s

 = 0 t + ½  t2 , we get,  = 0 + ½ (3)(12)2 = 216 rad 216 = 34.4 revolutions 216 rad = 2

Motion of a Point on a Rotating Body Let  be the angular velocity of the point P (Fig. 9.3) in the counterclockwise direction, then the linear velocity of P along the tangent at P is: v = r  , r being the radius of circular path. We can write v in terms of its components as: Velocity of point P→ v = (- r  sin θ) ˆi + (r  cos θ) ˆj (9.11) Or, x-component of v → vx = - r  sin θ, y- component of v → vy = r  cos θ Let  be the angular acceleration of P. The tangential component of acceleration ( a ) of P → at = r  The radial component of a → ar = - r 2 We can write a as: a = at ˆt + ar rˆ = (r ) ˆt + (- r 2) rˆ (9.12) ˆ Where, t is a unit vector along the tangent at P, and rˆ is a unit vector in the radial sense away from the center.

Example: An automobile on a circular track of radius 1500 m is traveling at 30 meters per second and increasing its speed at a rate of 3 m/s2. What is its acceleration?

Solution: Here, radius of track→ r = 1500m, Tangential speed→ v = 30 m/s, and Tangential acceleration → at = 3 m/s2 (30)2 v 2 2 (towards the center)  Radial acceleration→ ar = 1500 r = 0.6 m/s  Magnitude of resultant acceleration→ a =

a 2t + a2r

= (3)2 + (0.6)2 = 3.06 m/s2 a Direction of a is given by: tan  = t = 3 = 5, or  = tan-15 = 78.7 0.6 ar -  is the angle which the resultant a makes with the radial direction.

76

9.5

Moment of Inertia Consider a plane body rotating about an axis passing through O on the plane (Fig. 9.6) and the axis being perpendicular to the plane. If the plane body consists of particles of masses m1, m2,…………..mi,…..mn, and the perpendicular distances of those masses from the axis are r1, r2,…..ri….rn, respectively. The moment of inertia of the body consisting of n particles about the axis passing through o is defined as: Moment of inertia→ I = m r2 + m r2 + ........m r2 = n m r2 = M k2  ii 1 1 2 2 n n i=1

k is called the ‘radius of gyration’, total mass→nM = 2m1 + m2 +………+ mn n m r2 mr  i i  i i k is given by: k2 = i=1n m = i=1 M ,  i i=1

 Moment of inertia about an axis→ I =

n



i=1

m r2 = M k2

(9.13)

ii

The MKS unit of moment of inertia is ‘kg. m2’ There are two important theorems to calculate the moment of inertia about other axes which are parallel or perpendicular to the present axis.

Parallel axis theorem: ‘The moment of inertia of a rigid body about any given axis equals the moment of inertia about a parallel axis through the center of mass plus product of mass and the square of the distance between the two axes’ Example: The moment of inertia of a circular disc of radius ‘a’ about an axis passing through the center of mass of disc and perpendicular to the plane of the disc (Fig. 9.7) of mass M is: Moment of inertia about an axis through center of mass→ Ic = M kc2 = ½ M a2 By parallel axis theorem, the moment of inertia Io of the disc about an axis passing through ‘o’ and parallel to the axis passing through the center of mass of disc is: 3 3 Moment of inertia about axis through o→ Io =½ M a2 + M a2 = Ma 2 , or k 2 = a2 o 2 2 Perpendicular axis theorem: It states: ‘The moment of inertia of any plane lamina (Fig. 9.6) about an axis normal to the plane of the lamina is equal to the sum of moments of inertia about any two mutually perpendicular axes passing through the given axis and lying in the plane of lamina’ In Fig. 9.6

Iz = Ix + Iy

(9.14)

77 We can use the perpendicular axis theorem to find the moment of inertia of a circular disc of radius ‘a’ about one of its diameters..

Moment of inertia about an axis passing through the center of disc and perpendicular to the disc is: Ic = ½ M a2.. If ID is the moment of inertia about a diameter coincident with x-axis, then using Eq. (9.14), we get: ID + ID = ½ M a2, or

ID = ¼ M a2

1. Moment if Inertia of a Thin Rod Moment of inertia of a thin rod of mass ‘M’ and of length ‘ ℓ ’ about an axis passing through the center of mass of the rod and perpendicular to the rod (Fig. 9.9) is given by; +ℓ

+ℓ

x3  2 ℓ3 (ℓ)ℓ2 Mℓ2 IC =  dI =  ( dx) x = = =  = ℓ 3 12 12 12 − 2 −ℓ 2 2

2

λ→ is the mass per unit length of rod.  I of thin rod about an axis through c.m.→

IC =

12

,

kc=

12

(9.15)

By using parallel axis theorem, we can calculate the moment of inertia about an axis perpendicular to the rod through one of its ends (Q in Fig. 9.9)  ℓ2 ℓ2  Mℓ 2 2 ) = M Io = IC + M ( 2 12 + 4  = 3 ,   ℓ

ko2 =

3

2. Moment of Inertia of a hoop or Cylindrical shell Moment of inertia of a hoop or cylindrical shell of radius ‘R’ about an axis perpendicular to the plane and passing through the center of mass ‘C’ (Fig. 9.11) is given by: (9.17) Ic = M R2, kc2 = R2 From parallel axis theorem, the moment of inertia about a parallel axis passing through the point D is: ID = M R2 + M R2 = 2 M R2 ,

kD2 = 2 R2

(9.18)

3. Moment of inertia of a circular disc Moment of inertia of a circular disc of radius ‘a’ about an axis perpendicular to the plane passing through the center ‘C’ is given by a2 Ma 2 (9.19) Ic = , kc2 = 2 2

(9.16)

78 From parallel axis theorem, the moment of inertia about a parallel axis passing through the point D (Fig. 9.11) on the disc is given by: ID = ½ M a2 + M a2 = (3/2) M a2,

kD2 = (3/2) a2 (9.20)

4. Moment of inertia of a sphere Moment of inertia of a sphere of mass ‘M’ and radius ‘R’ about one of its diameters is: Ic =

9.6

2 5

MR 2 ,

kc2 =

2 5

R2

(9.21)

Kinetic Energy of Rotation Let a plane body rotate with angular velocity  about an axis passing through O perpendicular to the plane (Fig 9.12). All the particles of the body, except those lying on the axis move in a circular motion of varying radii, about the axis. If mn is the mass of nth particle of the body, and rn the radius of the corresponding circular path, the kinetic energy of rotation of the rigid body is given by: 1 2 2 m r2  nn Kinetic energy of rotation→ TR =  m n(rn) = n 2 2 n 2  m r2 Put I = (9.13) n n 2 n

 Kinetic energy of rotation→ TR = ½ I 2 9.7

(9.22)

Dynamics of Pure Rotation The equation of motion of a body rotating about a fixed axis (no translation) is given by: (moment of inertia about an axis)(angular acceleration) = (Torque about the same axis) We can express it as:

I  =  or

I

dt

=

(9.23)

Angular Momentum: The product of moment of inertia about an axis of rotation and the angular velocity → is called ‘angular momentum’ about that axis. We represent it by symbol ‘ J ’ We can write:

Angular momentum→

J =I

(9.24)

79 The equation of rotational motion of a rotating body can also be written in the form:

dt



(9.25)

Table 9.1 below shows that there is a lot of similarity between translational and motions.

rotational

Table 9.1 TRANSLATION

ROTATION

Displacement → D Linear Velocity→ v Mass→ m Linear Acceleration→ a → Force→ F Momentum→ p

Angular Displacement→ Angular Velocity →  Moment of Inertia →I Angular Acceleration→  → Torque→  → Angular momentum→ J → Angular Momentum→ J 2= I  Rot. K.E.→ TR = ½ I  → I =  → dJ =  dt

Linear Momentum→ p = m v Kinetic Energy→Tt = ½ m v2 → m a = F → → dp = F dt

Conservation of Angular Momentum If the sum of external torques acting on a rotating body is zero, the eqaion of rotational motion of the body can be written as, → → dJ = J = constant.  i =0, or dt It means that the total angular momentum a body remains constant, if the sum of external torques acting on the system is zero. This is called the ‘law of conservation of angular momentum’.

9.8

Solved Examples: Example 1: A flywheel with 1 m radius has an angular velocity of 30 rad/s and angular acceleration of 3 rad/s2. Find the magnitude of radial and tangential components of acceleration of a point on the rim of wheel.

Solution: Given:  = 30 rad/s, R = 1m,  = 3 rad/s2 Radial acceleration→ ar = - R 2 = - (1)(30)2 = - 900 m/s2 Tangential acceleration→ at = R  = (1)(3) = 3 m/s2

80

Example 2: A 50 kg- flywheel with moment of inertia of 115 kg.m2, has an angular velocity of 50 rad/s. (a) What is the rotational kinetic energy? (b) What would be the velocity of a 50g bullet having translational kinetic energy equal to the kinetic energy of the wheel?

Solution: (a) Given: I = 115 kg.m2 ,  = 50 rad/s Rotational Kinetic energy → Tr = ½ I 2 = ½ (115)(50)2 = 1.44 105 J (b) Mass of bullet→ mB = 50 g = 0.05 kg, velocity of bullet→vB = ?

Translational kinetic energy of bullet→ Tt = ½ mB v B2 = ½ (0.05) v B2 We have to find vB such that, Translational kinetic energy of bullet = Rotational kinetic energy of flywheel 

½ (0.05) vB2 = 1.44  105, or

vB =

1.44 105 = 2.4  103 m/s (0.5)(0.05)

Example 3: Masses of 2 kg and 3 kg are connected by a light cord that passes over a pulley of 12-cm radius (Fig. 9.13). The downward acceleration of the 3 kg-mass is observed to be 1.2 m/s2. Assuming that the cord does not slip and that the bearing of the pulley are frictionless, calculate the rotational inertia of the pulley.

Solution: Let T1 and T2 be the tensions on the two sides of the cord as shown in Fig. 9.12. The equation of motion for the 3 kg-mass is: m a = - T1 + m g, or T1 = m g – m a → T1 = (3)(9.8) – (3)(1.2) = 25.8 N (9.26) Similarly writing the equation of motion for 2 kg-mass we have, m a = T2 – m g, or T2 = m g + m a → T2 = (2)(9.8) + (2)(1.2)= 22 N

(9.27)

Resultant torque on the pulley→ =T1 R – T2 R = (25.8 – 22.0)(0.12) = 0.46 N.m Tangential acceleration The angular acceleration of the pulley→  = = 1.2 = 10 rad/s 0.12 Radius From the rotational equation of motion→ Ic  = , we get:  0.46 = = 0.046 kg.m2 Moment of inertia of the pulley→ Ic =  10 150  2  100h = 1502, or h= =3m 100 Example 4: A cubical box 4 m on side rests on the floor with an edge against a small cleat, as shown in Fig. 9.14. If the box weighs 150 kg and has its center of mass at its center, at what height must a 100 kgforce must be applied to just tip the box?

81 Solution: Let h be the height at which a force should be applied to just tip the box. When the box is just on the point of tipping, the moment (torque) due to 100 kg force in the anticlockwise sense, must be equal to the torque due to weight of the box in the clockwise direction. i.e.

Torque due to 100 kg force = Torque due to the weight of box at center of gravity.

Example 5: A uniform meter stick with a mass of 200 g has 400g- mass suspended at the 20 cm mark and a 1200g-mass at 60 cm mark, and is supported in a horizontal position by a laboratory spring scale attached to the ends. What is the reading of the spring scale suspended at the 20 cm mark and a 1200g-mass at 60 cm mark, and is supported in a horizontal position by a laboratory spring scale attached to the ends. What is the reading of the spring scale at zero centimeter mark?

Solution: Let w1 and w2 be the readings at 0 and 100 centimeter mark (Fig. 9.15).

We can get from conditions of equilibrium, w1 + w2 – 0.4 – 0.2- 1.2 = 0, or w1 + w2 = 1.8

(9.28)

Taking moments about point O gives: - (0.4)(0.2) – (0.2)(0.5) –(1.2)(0.6) + (1.0)(w2) = 0, which gives w2 = 0.08 +0.1 + 0.72 = 0.9 kg (9.29) Putting the value of w2 in Eq. (9.28) gives: w1 = .8 – 0.9 = 0.9 kg Example 6: A horse capable of exerting a pull of 800 kg force attempts to drag a weight of 10,000 kg along a rough road. The coefficient of friction is 0.4. A crane (Fig. 9.16) is available to help the horse. If the vertical cable from the crane is attached to the weight, find the tension in the cable such that the horse will just be able to start the weight in motion.

Solution: The maximum force the horse is capable of exerting is 800 kg. To start moving the weight the horse has to overcome the force of friction between block and the rough road.

Therefore, the maximum force of friction that the horse can overcome is 800 kg.

Maximum force of friction → fmax =  RN =  (10,000 – T), where (9.30) RN → Normal reaction of road on the block, T→ Tension in the cable of the crane  From Eq. (9.30),

0.4 (10,000 – T) = 800, or T = 10,000 -

800 0.4

= 8000 kg

82

Example 7: Two parallel forces of magnitude 5 kg and 15 kg act, respectively, at one end and the mid point of the rod 1 m in length. Their lines of action are perpendicular to the rod.

A force of magnitude 10 kg is acting at the other end in a direction opposite to the other forces. Find the magnitude, direction, and line of action of the resultant.

Solution: → If R is the resultant force acting downward at P, then a → force of (- R ) at P in Fig. 9.17 will bring the system in equilibrium.

Magnitude of the resultant→ R = 15 + 5 – 10 = 10 kg To find the position of application of this resultant force, we take moments of all the forces about the point P, and the sum of those moments must be zero if the system is going to be in equilibrium with the resultant acting upward. Let the distance of point P from O be x.  - (10  x) + (0  R) + 15  (0.5 – x) + (1-x)  5 = 0 - 10 x + 7.5 – 15 x + 5 – 5 x = 0, or 30 x = 12.5, or 0.42 m = 42 cm from O  A vertical resultant force of 10 kg acts downward at the point which is 42 cm from one end of the rod.

Example 8: A flywheel has a mass of 500 kg, all concentrated on the rim, and the radius is 1.2 m. It is rolling with an angular velocity of 2o rad/s. How much work was necessary to give the flywheel this angular velocity? If the wheel is stopped in one minute by a friction brake, what torque is applied.

Solution: Mass of wheel→ M = 500 kg, Radius of wheel→ R = 1.2 m  Moment of inertia of the wheel about an axis passing through the center of wheel and perpendicular to trhe plane of rim→ I = M R2 = (500) (1.2)2 = 720 kg.m2 Angular velocity of wheel (given)→ ω = 20 rad/s Work which is necessary→ W = ½ I ω2 = ½ (720) (20)2 = 1.44 × 105 J 20 Angular acceleration of the wheel in braking→ α = = - ⅓ rad/s 60  Torque applied by friction brake→  = I α = (720) (- ⅓) = - 240 n.m (in the opposite direction to that of  ) Example 9: A yo-yo has a moment of inertia about its center of mass of 2 × 10-5 kg-m2 and a mass of 0.05 kg. If the string is wrapped around a center post of 5 mm radius and the yo-yo is released from rest, find the angular acceleration and the tension in the string.

Solution: Let T be the tension in the string, and α the angular acceleration The linear acceleration of yo-yo→ a = α R = (α) (0.005) = 5 ×10-3 α (m/s2)

83 The translational equation of motion of yo-yo is: m a = m g – T, or (0.05) (5 ×10-3 α) = (0.05) (9.8) – T, or (1) The rotational equation of motion of yo-yo is: I α = , or ( 2 × 10-5) (α) = T (0.005) , or α = 250 T

(2)

(2.5 × 10-4 ) (250 T) = 0.49 – T 0.49  6.25 × 10-2 T + T = 0.49, or T = = 0.461 n 1.0625 Putting the value of T in (2) we get: angular acceleration→ α = (250) (0.461) = 115 rad/s2 Putting (2) in (1) we get:

Example 10: A meter stick has a small hole bored at 10 cm mark so it may be hung from a horizontal nail. If the meter stick is rotated about the nail until it makes an angle of 37˚ with the vertical and is then released, find its angular velocity as it passes through its equilibrium position, assuming that the friction is negligible.

Solution: The center of gravity of the meter stick is at 50 cm mark.

Therefore, the distance from the point of rotation is 40cm.

The potential energy of the stick when at an angle of 37˚ to the vertical is: EP = m g h = m g (0.4) (1 – cos 37˚ ) = m g (0.4)(1- 0.799) = (0.080) m g (1) Moment of inertia of meter stick about 10-cm mark is: 1  I = m  + (0.4)2  = 0.243 m  12 The potential energy completely transforms into kinetic energy at the equilibrium position. 

½ I ω2 = EP,

or

½ (0.243) m ω2 = 0.0804 m (9.8) (0.080)(9.8) ω= = 2.54 rad/s (0.5)(0.243)

Example 11: A uniform ladder weighs 30-kg and is 5 m long. It leans against a smooth (frictionless) wall at a point 4 m above a cement driveway with its base 3 m from the wall.

A 90-kg man stands 0.6 of the way up the ladder. (a) Find the horizontal and vertical components of the force exerted on the ladder by the driveway. (b) If the coefficient of static friction between the ladder and the driveway is 0.3, what fraction of the way to the top can the man climb before the ladder slips?

Solution: Forces acting on the ladder are (Fig-Ex-9.11): - Weight of ladder acting = 30 kg -wt (vertically downward) - Weight of the man = 90 kg-wt (vertically downward) - Reaction of ground on one end of ladder at P = RV (vertically upward) - Reaction of wall on the other end of ladder at Q = RH (positive-xx) - Force of friction between ground and ladder = f (negative-xx))

84 (a) Considering the equilibrium of vertical forces we get: RV – 30 – 90 = 0, or RV = 120 kg-wt (1) Considering the equilibrium of horizontal forces we get: f - RH = 0, or RH = f (2) Considering the equilibrium of moments of all the forces about point P, we get: (30) (2.5 cos θ) + (90) {(5 × 0.6) cos θ} = RH × 4 (3) Or (30) (2.5 × 0.6) + (90) {(5 × 0.6) (0.6)} = 4 RH  45.0 + 162 = 4 RH, or RH = 51.8 kg-wt From (2), the force of friction→ f = RH = 51.8 kg-wt (b) The maximum force of friction between ladder and ground is: fmax = µ RV = (0.3) (120) = 36 kg-wt Let x be the distance along the ladder the man can go up without slipping.

Considering the equilibrium of moments of forces about P taking RH = fmax = 36, we have: (30) (2.5 × 0.6) + (90) (x × 0.6) = (36)(4), or 54 x = 144 – 45, or x = 1.83 m x 1.83 The distance as fraction of ladder-length = = = 0.366 5 5 Example 12: A 100-kg horizontal bar AB in the Fig-Ex-9.12 is non-uniform, being 10 m long with center of gravity 4 m from the wall. A 120-kg weight is suspended on the bar 7 m from the wall.

The angle θ is 60˚. What is the tension in the cord BC? Find the horizontal and vertical components of the force exerted on the bar by the pin at A. Solution: Considering the equilibrium of vertical forces, we have: V + T sin 60º - 100 – 120 = 0, or V+ 0.866 T = 220 (1) Considering the equilibrium of horizontal forces we have: H – T cos 60º = 0, or 2H=T (2) Considering the equilibrium of moments of forces about the pin A we get: T (10 sin 60º) – (100) (4) – (120) (7) = 0, or 8.66 T = 1240, or T = 143 kg-wt  Tension → T = 145 kg-wt, or 143 × 9.8 = 1401 n From (2), the horizontal component of force at pin A→ H = T/2 = 143/2 = 72 kg-wt = 902 n From (1), the vertical component of force at A→ V = 220 – (.866) (143) = 96 kg-wt = 941 n Example 13: A 60.0-kg woman stands at the rim of a horizontal turntable having a moment of inertia of 500- kg.m2 and a radius of 2.00 m. The turntable is initially at rest and is free to rotate about a frictionless, vertical axle through its center. The woman then starts walking around the rim clockwise (as viewed from above the system) at a constant speed of 1.50 m/s relative to the Earth. (a) In what direction and with what angular speed does the turntable rotate? (b) How much work does the woman do to see herself and the turntable into motion?

85 (a) Magnitude of linear momentom p of the woman in the tangential direction is: p = m v = (60) (1.5) = 90 kg.m/s  Moment of momentum (angular momentum) of the woman in clockwise direction is: Angular momentum→J = r × p = (2) (90) = 180 kg.m2/s Since there is no external torque acting on the turn table, the total angular momentum of the turn table and woman must be conserved  (angular momentum of woman) + (angular momentum of turntable) = 0 180 180 Or, 180 + I ω = 0, or ω = == - 0.36 rad/s I 500 Therefore the turntable will rotate in the counterclockwise direction with an angular velocity of 0.36 rad/s. (b) Work done → W = T = (kinetic energy of woman) + (rotational energy of turntable) = ½ (m) (v)2 + ½ (I) (ω)2 = ½ (60) (1.5)2 + ½ (500) (0.36)2 = 67.5 + 32.4 = 99.9 J Problem 14: A projectile of mass 100 g moves to the right with a speed 10 m/s (Fig-Ex-9.14). The projectile strikes and sticks to the end of a stationary rod of mass 1.0 kg, and length 40 cm pivoted about a frictionless axle through its center. (a) Find the angular speed of the system right after the collision. (b) Determine the frictional loss in mechanical energy due to the collision.

Solution: Mass of the projectile→ m = 0.1 kg, Mass of the rod → M = 1.0 kg Length of the rod → d = 0.4 m, Speed of the projectile→ vi = 10 m/s Moment of inertia of the rod about the axle→ IR =1/12 M d2 = (1/12) (1) (0.4)2 = 1.33 × 10-2 kg.m2 (a) Since there is no external torque acting the angular momentum will be conserved before and after the collision.  Md2   d 2  (mvi )d  + m    =  2  12   2     (1)(0.4)2   0.4 2  (0.1)(10)(0.4)  + (0.1) =  Or,    = (1.33 × 10-2 + 0.4 × 10-2) ω 2  12   2   0.2 Or, ω= = 11.6 rad/s 1.73 10−2 (b) The initial energy→ Ei = ½ (m)(vi)2 = ½ (0.1) (10)2 = 5.0 J

86 2

 d  2 Energy after→ Ef = ½ I ω + ½ m   = ½ (1.33×10-2) (11.6)2 + ½ (0.1)(0.2×11.6)2  2  = 0.895 + 0.269 = 1.164 J Loss of energy = Ei – Ef = 5.0 – 1.164 = 3. 84 J

87

Chapter 10

Dynamical Systems

10.1 Momentum of a Dynamical System The total momentum of a dynamical system consisting of N particles (Fig. 10.1) is : → N → → , where → → are the velocities of N Total momentum→ → , P =  pi =  mivi v1 v2 ,…… vN i=1 i=1 → particles masses, m1, m2,……mN of the system. The velocity vc of the center of mass of the system is: N mv → →  i i → i=1 m v + m v + ..... m v 2 2 N N = (10.1) Velocity of center of mass v = 1 1 c N + m + ...... m m1 2 N  mi i=1

N

But,  m i = M→Total mass of particles of the system. i=1

 From Eq. (10.1) we have:

vc

P

(10.2)

Eq. (10.2) tells us that the total momentum of the system of N particles is equal to : Total momentum = (mass of the system)  (velocity of the center of mass of system) The equation of motion of the system of N particles is given by: dP dt

N → N → =  Fi (ext) +  Fi (int) i=1

i=1

From Newton’s third law of motion which states that in any action of a force, there is an N → equal and opposite reaction, we can conclude that the sum of internal forces (  Fi (int) ) i=1

→ within a system of particles is zero. i.e.  Fi (int) =0. i=1 → N → If we put: sum of external forces→  Fi (ext) = F , then the equation of motion of a system N

i=1

of particles can be written as, → → → dP d (M v ) = F→ , or = M ac = F c dt dt → ac is the acceleration of the center of mass (c.m.) of the system

(10.3)

- It means that the translational motion of a system of particles can be treated like the motion of a single particle at the c.m. of the system whose total mass is concentrated at the c.m.

88

10.2 Conservation of Momentum → If there are no net external forces acting on a system of particles i.e. if F = 0, from Eq. (10.3), it follows, N → dP d N → =  pi = 0, or  pi = constant i=1 dt dt i=1 This means that the total momentum of a system of particles remains constant when there is no unbalanced external force acting on the system. This is the ‘law of conservation of momentum’.

10.3 Angular Momentum of System of particles → The total angular momentum J of the system of particles is: N N → N → → → → → J =  Ji =  ri p i =  ri  (mi vi ) 1

1

1

N N → → d → d N → → → → → =  ri  (mi vi ) =  v  (mi v ) + r  (m v ) = 0 +  i r  (m i i ai ) i i i i 1 dt dt 1 dt 1 N → → → → → dJ But, mi ai = Fi ,  =  ri  Fi = L → (Total moment of all the forces about O) dt 1 dJ = L (8.8) dt → N → → → dJ = 0, or, J = constant If L =  ri  Fi = 0, 1 dt It means that in the absence of any external torque, the total angular momentum remains constant. This is called ‘Principle of Conservation of Angular Momentum’

dJ

10.4 Rolling Motion of a Rigid Body Consider a uniform cylinder of radius R rolling without slipping on the horizontal surface (Fig-10.2). As the cylinder rotates through an angle θ, its center of mass moves a linear distance R θ. Therefore the linear speed of the center of mass for pure rolling motion is given by: d d vc = (R ) = R = R (rolling condition) (10.4) dt dt ω is the angular speed of the cylinder.

Eq.(10.4) holds whenever a cylinder or a sphere rolls without slipping and is the condition for pure rolling motion.

The magnitude of linear acceleration of the center of mass for pure rolling motion is: d (vc ) = R d = R ac = (10.5) dt dt α is the angular acceleration of the cylinder.

89 Note that the linear velocity of any pont is in a direction perpendicular to the line from that point to the contact point P. At any instant , the part of the rim that is at point P is at rest relative to the surface because the slipping does not occur.

Kinetic Energy of a Rolling Object We can express the total kinetic energy of the rolling cylinder as: Kinetic energy of rolling cylinder→ T = ½ IP ω2 where IP is the moment of inertia about a rotation axis through P. Using parallel axis theorem and (10.4) in (10.6) give: T = ½ (Ic + M R2) ω2 = ½ Ic ω2 + ½ M vc2

(10.6) (10.7)

Thus, the total kinetic energy of a rolling object is the sum of the rotational kinetic energy about the center of mass and the translational kinetic energy of the center of mass.

Using the fact that vc = R ω for pure rolling motion, we can express Eq.(10.7) in the form: 2 1 v  1 1 I  T = I  c  + Mv2 =  c + M v2 (10.8) c c 2 cR 2 2  R2 

10.5 Solved Examples Example 1: Two → forces act simultaneously on a 4.0 kg mass. ˆ ˆ One force is F1 = (−3.0 N) i + (6.5N) j , and the other has a magnitude 5.0 N directed at 240 counter-clockwise from the positive x-axis. Find the magnitude and direction of acceleration.

Solution: F1 = (-3.0) ˆi + (6.5) ˆj (given)  ˆ  ˆ → From Fig. 10.2a, F2 = (- 5 sin 30 ) i - (5 cos 30 ) j  F + F→ = (-3.0 – 5 sin 30) ˆi + (6.5 - 5 cos 30) ˆj 1

2

= (- 3.0 – 2.5) ˆi + (6.5 – 4.3) ˆj = (-5.5) ˆi + (2.2) ˆj →F − 5.5ˆi + 2.2 ˆj  Acceleration→ a = = (−1.4) ˆi + (0.6) ˆj = m 4

(

)

(−1.4)2 + (0.6)2 = 1.5 m/s 2  0.6  4  = - 23, or 157 with Direction of acceleration→ = tan-1  − 1.4   the positive x-direction  Magnitude of acceleration→ a =

90

Example 2: The end of a thread wound around a reel is held fixed and the reel is allowed to fall so that the thread is unwound. Find the acceleration of the reel, assuming its axis remains horizontal.

Solution: Let m be the mass and R the radius of the reel and mkc2 its moment of inertia about its axis. Let vc be the velocity of its center and ω its angular velocity at any time.

The part A of the reel in contact with the straight thread is the instantaneous center of rotation and has no velocity, but the velocity relative to the center is Rω, therefore vc – R ω = 0 (1) The equation of motion of translation of the center of mass is: m ac = m g – T (2) The equation of motion of rotation of the reel is: (m kc2) 2˙ = R T (3) 2  mkc   v̇c  mk2 v̇ mk a   = From (1) and (3), T =  c c c c (4) = 2 2  R R R R   mk 2c a c Eliminating T between (4) and (2) we have: m ac = m g – R2 2 2  + kc  = g, or  ac 1 ac = g R . For a cylindrical reel kc 2 = R 2/2, and ac= 2g  R2  (R 2 + k2c ) 3   Example 3: A 4.00-m length of light nylon cord is wound around a uniform cylindrical spool of radius 0.500 m and mass 1.00 kg. The spool is mounted on a frctionless axle and is initially at rest. The cord is pulled from the spool with a constant acceleration of magnitude 2.50 m/s2. (a) How much work has been done on the spool when it reaches an angular speed of 8.00 rad/s? (b) Assuming there is enough cord on the spool, how long does it take the spool to reach this angular speed? (c) Is there enough cord on the spool? (4.0 J,1.6 s, 3.2 m<4.00) Solution: (a) Moment of inertia of the spool→ I = ½ m R2 = ½ (1) (0.5)2 = 0.125 kg.m2 Work done→ W = TR = ½ I ω2 = ½ (0.125) (8)2 = 4 J a 2.5 = = 5.0 rad/s2 R 0.5 Time taken by the spool to reach an angular speed of 8 rad/s is given by: 8.0 ω = ω0 + α t, or 8 = 0 + (5.0) (t), or t= = 1.6 s 5.0

(b) Angular acceleration→ α =

(c) length of cord unwound after 1.6 s is given by: ℓ = ℓ 0 + ½ a t2 = ½ (2.5) (1.6)2 = 3.2 m < 4.0 m Yes, there is enough cord

91

Example 4: A uniform straight rod of 1.00 m- length is held at an inclination 60˚ to the vertical with its lower end in contact with a smooth horizontal plane and let go. Find the angular velocity when the rod becomes horizontal Solution: Since there is no horizontal force acting on the road its center of gravity C does not acquire any horizontal velocity and therefore remains in the same vertical line OY. If m is the mass, ℓ is the length, and ω be the angular velocity when the rod becomes horizontal,.the kinetic energy T of the rod is given by: 2 2  m  1 ℓ  + k2 2  T = T (translation of C) + T(rotation) = ½ m     + ½ I ω2 =    (1)  2  2 2   

ℓ2 1 = (k→ radius of gyration), 12 12 m 1 1  2 m 2  +  =  From (1), T = (2) 6 2 4 12  The work done as C descends from a height ℓ cos 60˚ to zero ℓ mg Work done→ W = m g (cos 60˚ ) = (3) 2 4 m 2 mg = From (2) and (3) we have: (4) 6 4  ω2 = 9.8 × 0.5 × 3, or ω = 3  4.9 = 3.83 rad/s  Angular velocity when the rod becomes horizontal→ ω = 3.83 rad/s Where, k2 =

Example 5: A meter stick rotates about an axis passing through 90 cm mark. What is the angular velocity if the stick rotates in a vertical plane?

Solution: The potential energy of stick when it is horizontal is: Potential energy → EP = m g h = m g (0.4) The potential energy transforms completely into kinetic energy when the rod becomes vertical. Let ω be the angular velocity of the rod when it is vertical, and IQ be the moment of inertia about an axis through Q (Fig-Ex-10.5). 1 2 2 IQ (by parallel axis theorem) is given by: IQ = IC + m (0.4) = m  + (0.4)  12  1  2 2 2  Kinetic energy of the rod in vertical position→ EK = ½ IQ ω .= ½ m  + (0.4)  ω 12 

92

From the principle of conservation of energy we have: EP = EK 1  2  0.4  9.8  ½ m  + (0.4)2  = m (9.8) (0.4) , or ω = = 5.68 rad/s (0.083 + 0.16)  12 Example 6: Mass of 2 kg and 3 kg are connected by a light cord that passes over a pulley of 12-cm radius. The downward acceleration of the 3-kg mass is observed to be 1.2 m/s2. Assuming that the cord does not slip and the bearings of the pulley are frictionless, compute the rotational inertia of the pulley.

Solution: Let T1, T2 be the tensions on the two sides of the cord (Fig-Ex10.6) The equation of motion of 3-kg mass can be written as: m a = - T1 + m g, or

T1 = m (g – a) = 3 (9.8 – 1.2) = 25.8 N

(1)

Similarly, the equation of motion of the 2-kg mass which is accelerating upward with the same acceleration a, can be written as: m a = T2 – m g, or T2 = m (a + g) = 2 (1.2 + 9.8) = 22 N (2) Rotational equation of motion of the pulley can be written as: I α =  = (T1 – T2) (0.12) = (25.8 -22) (0.12) = 0.456 N.m (3) 0.456 From (3) the rotational inertia of the pulley→ I = (4)  a 1.2 = But, angular acceleration of the pulley α = =10 rad/s2 R 0.12 0.456 From (4), inrtia of the pulley→ I = = 0.046 kg. m2 10 Example 7: A particle of mass 0.400 kg is attached to the 100-cm mark of a meter stick of mass 0.100 kg. The meter stick rotates on a horizontal frictionless table with an angular speed of 4.00 rad/s. Calculate the angular momentum of the system when the stick is pivoted about an axis (a) perpendicular to the table through the 50.0-cm mark and (b) perpendicular to the table through the 0-cm mark.

Solution: (a) Moment of inertia of the system in this case is: I = Moment of inertia of the stick about cm + moment of inertia of 0.4 kg mass 1 1  ℓ 2 = mℓ2 + m1   = (0.100)(1.00)2 + (0.400) (0.500)2 = 0.108.3 kg.m2 12 12 2 Angular momentum of the system→ J = I ω = (0.108.3) (4.00) = 0.433 kg.m2/s (b) Moment of inertia in this case is: I = Moment of inertia of stick about one end + Moment of inertia of 0.4-kg mass

93 1 1  ℓ 2  =  mℓ2 + m   + m1 ℓ2 = (0.100)(1.00)2 + (0.400) (1.00)2 = 0.433 kg.m2 3 12  2   Angular momentum of the system→ J = I ω = (0.433) (4.00) = 1.73 kg.m2/s Example 8: A 15-kg object and a 10.0-kg object are suspended, joined by a cord that passes over a pulley with a radius of 10.0 cm and a mass of 3.00 kg (Fig-Ex-10.8). The cord has a negligible mass and does not slip on the pulley. The pulley rotates on its axis without friction. The objects start from rest 3.00 m apart. Treat the pulley as uniform disc, and determine the speeds of the two objects as they pass each other.

Solution: Since the two masses are connected by a string, the two masses will move with the same acceleration and velocity, with 15-kg mass moving down and 10-kg mass moving up. Therefore, both masses will move a distance of 1.5 m when they pass each other with the same velocity , let us say, v.

Let us take the passing point of the two masses as the point of zero level of potential energy of the system, The potential energy of the system when the two masses are at position of rest is given by: Potential energy of system→ EP = (15) (9.8) (1.5) + (10) (9.8) (- 1.5) = 73.5 J

(1)

When the two masses pass each other, the potential energy transforms into kinetic energy of the two masses plus the rotational energy og the pulley.  Kin. Energy of 15-kg mass + Kin. Enrgy of 10-kg mass + Rot. Energy of pulley = 73.5 v v or, ½ (15) v2 + ½ (10) v2 + ½ I ω2 = 73.5, but ω= = = 10 v R 0.10 1 2 2 (0.1)2 2 2 2  ×(10 v) = 73.5, 7.5 v + 5.0 v + × (3.0) × (I = m k = 3 R /2) 2 2 73.5  13.25 v2 = 73.5, or v = = 2.36 m/s 13.25 The speed of two objects when they pass each other→ v = 2.36 m/s Exqample 9: Suppose a meteor of mass 3.00 × 1013 kg, moving at 30.0 km/s relative to the center of the Earth, strikes the Earth. What is the order of magnitude of the maximum possible decrease in the angular speed of the Earth due to the collision?

Solution: The mateor will slow the rotation of the Earth by the largest amount if its line of motion passes farthest from the Earth’s axis. The Earth rotatatefrom west to east, therefore the mateor should be headed to west and strike a point on the equator tangentially.

94 Let us take the z-axis along the axis of rotation of the Earth pointing northward.

Since there is no external torque acting on the system consisting of the Earth and the meteor, the angular momentum must be conserved before and after the collision. Let ωi be the angular velocity of the Earth before collision, and ωf after the collision.

(1) From conservation of angular momentum of the system,  L i =  Lf Angular momentum before collision→  L i = I ωi + (m v)× R 2 = M(earth) R2 ωi + {m(meteor)} v × R 5 2 M R 2 ωf [m(meteor)<<M(earth)] Angular momentum after collision→  L f = 5 2 M R2 ωf = 2 M R2 ωi + (m v)× R  From (1) we have: 5 5 (3.00 1013)(30.0 103 ) mv -14 = = 5.91×10 rad/s or, (ωf – ωi) = 24 6 (0.4)(5.98 10 )(6.37 10 ) 0.4 M R 

max

 10-13 rad/s

95

Chapter 11

Elasticity and Periodic Motion

11.1 Elastic Properties We have so far assumed that objects remain rigid when external forces act on them.

In reality all objects are deformable. It is possible to change the shape and/or size of an object by applying external forces.

Hooke’s Law: The law states that ‘the deformation of an elastic body is directly proportional to the magnitude of the applied force provided the elastic limit is not reached’.

It means that if a force F = T (tension) in a wire in Fig. 11.1 of length, OA = ℓ causes a deformation, AB = s, then the Hooke’s law tells us that: Tension→ T  s→Extension, or (11.1) T= ks - k is called the elastic constant of wire or material Hooke’s law holds in the part OA of the curve in Fig. 11.2. In the part AB of the curve, there is little deformation with the increase of force (tension).

After the B on the curve there is a very rapid increase of deformation without increase in force, and the breakage in wire occurs.

We now define stress, which is a quantity that is proportional to the force causing the deformation.

Stress =

Deforming force F = , unit of stress→N/m2 = Pascal Area of cross section A

The result of a stress is strain, which is a measure of degree of deformation. For small stress, it is found that the strain is proportional to the stress. The proportionality constant depends on the material being deformed and on the nature of deformation. We call the proportionality constant the ‘elastic modulus’.

Elastic modulus =

Sress Strain

(11.2)

In the following we consider three types of deformations, and define modulus of elasticity for each: 1. Young’s modulus: Elasticity in length It measures the resistance of a solid to a change in length of long (thin) wire-like body. Let L be the initial length of a long wire of area of cross-section A, clamped at one

96 end (Fig. 11.1). When an external force is applied perpendicular to the cross section of the wire, let L be the extension of the wire.

F L Deforming force parameter→Stress = , Deformation parameter→ strain = A L Young’s modulus of elasticity→

Y=

Stress FL = strain A L

(11.3)

2. Bulk Modulus It measures the elastic property of a body which undergoes a change in volume when a force is applied e.g. when a force is applied on a volume of gas by piston of a pump (Fig. 11.3). Let the force F applied on the piston cause a change in volume V. F Then, Stress = (A is area of cross section of the piston) A V Volume strain = (V is the initial volume) V Stress FV Bulk modulus of elasticity→ B= = (11.4) Volume strain A V

3. Shearing modulus It measures the elastic property of a material or body which undergoes a change in shape on the application of a tangential force as shown in the Fig. 9.4.

Shearing strain is defined as the angle  in radians.

Tangential force F = Shearing stress = Area A F Shearing stress = Shearing modulus→  = Shearing strain A

(11.5)

11.2 Spring Force Constant of a Spring: It is defined as the force per unit elongation of a spring. If a force F increases the length of a spring by ‘ s’ (Fig. 11.5) .

From Hooke’s law, F  s, and therefore, F = k s, k is called the ‘spring constant’ F  Force constant of a spring→ k = , unit of spring constant→ N/m s

97 Potential Energy of a stretched spring When a spring is stretched it stores energy in the form of potential energy. The potential energy of a spring with extension‘s’ is given by: s

s

EP =  dV =  - dW =  - (-T ds) =  ks ds = ½ k s2 0

Potential energy of a spring→

0

2

EP = ½ k s

(11.6)

11.3 Solved Examples on Elasticity Example 1: When a body weighing 4 kg is attached to the end of a spring hanging vertically, the spring is stretched 5 cm. What is the force constant of the spring?

Solution: The force on the spring→ F = 4 kg = 4  9.8 = 39.2 N Elongation of the spring→ s = 5 cm = 0.05 m F 39.2 = = 784 N/m  Force constant of he spring→ k = s 0.05 Example 2: A 4200 kg- weight is supported by a steel rod 8 m long and 2.5 cm in diameter.

What is the resulting elongation? (Young’s modulus of steel= 2.01011 Pa) Solution: Given: Force→ F = 4000 kg = 4000  9.8 N = 39200 N Diameter of the rod → D = 2.5 cm = 0.025 m, initial 2length→L = 8 m  0.025  2  = 4.91  10-4 m2  Area of cross section→ A =  (D 2) =    2  From Eq. (11.13), we have: FL (39200)  (8) = Extension→L = = 3.2  10-3 m = 3.2 m m YA (2.0 1011) (4.9110−4 ) Example 3: What increase in pressure is required to decrease the volume of a cubic meter of water by 0.005% ? Given (Modulus of water→ B = 2.1  109 N/m2 Force → Stress Solution: Pressure→ P = Area V = 0.005% = 5  10-5 Strain = V P Stress VP = Bulk modulus of elasticity→ B = Strain (V V) B  Pressure required→ P = B  (strain) = (2.1  109) (5  10-5) = 1.05  105 Pa→ (N/m2) Example 4: A solid brass sphere is initially surrounded by air, and the air pressure exerted on it is 1.0  105 N/m2 (normal atmospheric pressure). The sphere is lowered into the ocean to a depth where the pressure is 2.0  107 N/m2. The volume of sphere in air is o.5 m3. by

98 how much does this volume change once the sphere is submerged?

(Bulk modulus for 10 2 brass = 6.1  10 N/m ) P Solution: From the definition of bulk modulus, B = , we have V V V P (0.50)(2.0 107 −1.0 105 ) Change in volume → = = = 1.6  10-4 m3 10 (6.110 ) B

11.4 Periodic Motion Simple Harmonic Motion: It is a motion of a particle under a force such that: (1) The force is proportional to the displacement at any time.

(2) The force acts in a direction opposite to the displacement.

If x is the displacement of a particle of mass m from O (Fig. 11.6), then the force, F = -k x, acting on that particle will result in a simple harmonic motion. The equation of a particle under a force, F = - k x is given by: ma= -kx

(11.7)

The acceleration of the particle in Eq. (11.7) is, a = - (k/m) x, i.e. acceleration is proportional to the negative of displacement at that instant.

By solving Eq. (11.7), it can be shown that such a motion will be periodic and will repeat itself about the point O. This repetitive motion is called ‘oscillation’. The time taken to complete one oscillation is called ‘time period of oscillation’.

The maximum displacement from the center of oscillation ‘ O’ is called ‘Amplitude’ - The maximum acceleration of the particle is: Maximum acceleration→

amax = - kA , A→Amplitude m

(11.8)

11.5 The total energy of a particle executing simple harmonic motion Total energy of SHM→ E = EK + EP = ½ m v2 + ½ k x2

(11.9)

When the particle is at the end of amplitude, x = A, v = 0  From Eq. (11.9), total energy→

E = ½ k A2

 Eq. (11.10) can be written as, E = ½ m v2 + ½ k x2 = ½ k A2

(11.10) (11.11)

99 From Eq. (11.11) we have, Velocity at any point →

v=

k m

(11.12)

The displacement ‘x’ of point P from center of oscillation ‘O’ along the x-axis at any time is given by:  k  Displacement→ x = A cos  t+   , ( is the initial phase angle)  m  From Eq. (11.13), the time period of a simple harmonic motion is: Time period of oscillation→ T =

2 m 2 = , = 2 k  k m

(

k m

=)

(11.13)

(11.14)

 is called ‘angular frequency’ The frequency of oscillation→

f=

 

=

1

k

=  2 m 2

(11.15)

Unit of frequency ‘f’→ Hertz (Hz), Unit of angular frequency ‘’ → rad/s The relation between the two frequencies is:

=2πf

(11.16)

11.6 Simple Pendulum A simple pendulum consists of a mass suspended at one end of a thread or wire. The mass will execute simple harmonic motion when pulled sideways a short distance and released. The time period of a simple pendulum (Fig. 11.7) for small amplitude of vibration is given by: Time period→

T = 2

g

( ℓ is the length of the pendulum)

(11.17)

Note: EQ. (11.17) holds only for small amplitude.

11.7 Simple Harmonic Motion in Spring A mass ‘m’ suspended from a spring of spring constant k (11.8) also executes simple harmonic motion, and the time period of oscillation for small amplitude is given by:

100

Time period → T = 2

m k

(k→ spring constant)

(11.18)

If s is the extension in a spring when a mass m is suspended from the spring mg . of constant k, m g = k s, or k = s

11.8 Examples on SHM Example 1: A body of mass 5 kg is suspended by a spring which stretches 0.1 m when a body is attached. It is then displaced downward an additional 0.05 m and released. Find the amplitude, period, and the resulting SHM. mg (5)(9.8) = = 490 N/m Solution: The spring constant→ k = s (0.1)  Time period→ T = 2

m 5 = 2 = 0.64 s k 490

Amplitude → A = 0.05 m 1 1 = = 1.6 Hz Frequency→ f = T 0.64 Example 2: A body is vibrating with simple harmonic motion of amplitude 15 cm and frequency 4 Hz. Calculate (a) the maximum value of acceleration and velocity, (b) acceleration and velocity when the coordinate is 9 cm, and (c) the time required to move from equilibrium position to a point 12 cm distant from it.

Solution: Amplitude→ A = 15 cm = 0. 15 m (given) Frequency→ f = 4 Hz (given) The equation of a simple harmonic motion is: m a = - k x (A) (a) Maximum acceleration will occur when x is maximum, i.e. when x = A = 0.15 m kA (0.15)x =  From Eq. (A), maximum acceleration→ amax = m m k But, = 2 = (2  f)2 = (2   4)2 m  amax = (2   4)2 (0.15) = 94.8 m/s2 -

Velocity at any time→ v =  A2 − x 2 , it will be maximum when x = 0.  vmax =  A = 2f  A = 2  (4)(0.15) = 3.8 m/s

(b) When x = 9 cm = 0.09 m, kx a== -  2x = -(2 f)2 x = - (2   4)2  (0.09) = -56.9 m/s2 m v =  A2 − x 2 = 8 (0.15)2 − (0.09)2 = 3.0 m/s

101

(c)

The displacement at any time is, x = A cos ( t + ) For x = 12 cm = 0.12m, A = 0.15 m,  = 8 , and  = 0, we get  0.12  0.12 = 0.15 cos (8  t) or, 8  t = cos-1  0.15     Time taken to move from equilibrium position to 12cm distant from it is: 4 1 cos-1 = 0.64 rad = 0.03 s t= 8 5 8

Example 3: A mass of 0.500 kg stretches a spring 20.0 cm. Find the force constant of the spring. If the mass is pulled down an additional distance of 5.00 cm and then released, find the initial restoring force, the period, and the total energy of the resulting simple harmonic motion.

Solution: Here, F = m g = 0.500  9.8 = 4.90 N S = 20 cm = 0.2 m mg 4.9 =  Spring constant→ k = = 24.5 N/m s 0.2 0.5 m Time period→ T = 2 = 2 = 0.09 s 24.5 k Amplitude→ A = 5 cm, since this is the maximum displacement Total energy → E = ½ k A2 = ½ (24.5)(0.05)2 = 0.031 J Example 4: An object oscillates with a simple harmonic motion along the x axis its position varies with time according to the equation, x = (4.00 m) cos (  t + ¼ ) (a) Find the acceleration of the object at any time. (b) Determine the position, velocity, and acceleration of the object at t = 1.00 s. (c) Determine the maximum speed and (d) maximum acceleration of the object. (e) Find the displacement of the object between t = 0 and t = 1.0 s.

Solution: (a) comparing the given equation with:

x = A cos ( t + ), we have, k  =  = 2  f =, or f = Amplitude→ A = 4.00 m, = 0.5 Hz m 2  Period of oscillation→ T = 1/f = 1/(0.5) = 2.0 s



(

)

d dx d (4.00) t +  = A( t +  ) = 4 dt dt dt v = -(4.00 ) sin (  t + ¼ )

(b) Velocity at any time→ v =

102



(

)

dv d − 4.0 sin t +  2 = 4 = - 4.0  cos ( t + ¼ ) dt dt (c) Position at t = 1 s→ x(1s) = (4.00) cos( + ¼ ) = - (4.00)(-0.707) = - 2.83 m Velocity at t =1 0 s→ v(1.00) = - (4.00) sin ( + ¼ ) = - (4.0)(-0.707) = 8.89 m/s Acceleration at t = 1.0 s→ a(1.0 s)= - (42) cos ( + ¼ )= -(42)(-0.707) = 27.9 m/s2 Acceleration→ a =

(d)

Maximum speed→ vmax = (4.00)() = 12.6 m/s Maximum acceleration→ amax = 4.0 2 = 39.5 m/s2

(e)

Displacement at t = 0→ x(0) = (4.00) cos (0 + ¼ ) = (4.00)(0.707) = 2.83 m Displacement at t = 1→ x(1) = (4.00) cos ( + ¼ ) = (4.00)(- 0.707) = -2.83 m Displacement from t = 0 to t= 1s is: x = -2.83 – 2.83 = - 5.66 m

Example 5: A 1.00-kg object is attached to a horizontal spring. The spring is initially stretched by 0.100 m, and the object is released from rest there. It proceeds to move without friction. The next time the speed of the object is zero is 0.500 s later. What is the maximum speed of the object?

Solution: The spring is initially stretched by 0.1 m before the object is released, therefore, the amplitude of oscillation → A = 0.1 m It comes to rest at the other end of amplitude after 0.5 s, therefore, the time period of oscillation → T = 2 × 0.5 = 1.0 s 2 2 = 2  =  Angular frquency→ ω = T 1 Maximum speed of the object will be attained at the center of oscillation.  maximum speed→ vmax = A × ω = (0.1) (2 ) = 0.628 m/s

Some useful expressions

103

→→

→ →

Scalar Product→ A.B  AB cos ( A, B)

→ → → → Vector Product→ A B  AB sin ( A, B)

B B A

Cosin law: R 2 = A2 + B 2 − 2 AB cos

tan =

B sin

iˆ → → = A →A B x Bx

Cross Product

A + B cos

kˆ A

ˆj A y

z

By

Bz

→ → Scalar Product→ A.B= Ax Bx + A y By + Az Bz

Density→ = M V (Projectile): x = v0x t + ½ a t2 y = v0y t - ½ g t2 vx = v0x + a t, vy = v0y t- gt

s = tan  (→angle of friction) ,

fmax = s Rn ,

fmax =  k Rn,

v = v0 + a t,

s = v0 t + ½ a t2,

v2 – v02 = 2 a s,

→ v2 ) acent = − r, r

2

2

2

2

p=mv vx − v0 x = 2ax vy − v0 y = −2gy 2 → Gm m ) GM mM -11 Nm 1 2 E E F=r , (Grav.

Pot) U = − G ), g = g s 2 (Gravitation) (G = 6.67x10 Kg 2 2 R r RE E → → → 2 2 K.E.→ EK = ½ m v , U(elastic) = ½ k s (s→extension), I(impulse) = Fxt = p , v1f – v2f = - e (v1i – v2i) (collision rule),

(Rotation):

 = 0 +  t,

→ → L = I ,

→ → → → ⇀ 2 (Torque)→  = r x F , I (m of inertia) = m i ri , I =  , TR = ½ I 2,

2

I = Ic + m p2 (ll-axis),

Iz = Ix + Iy

(⊥- axis),

→ v = r  ˆt ,

2 - 02 = 2   → a = ( r) ˆt - ( 2 r) rˆ

2 kc2 = L (rod), kc2 = R 2 (Sphere) 5 12 mx m y mz (c.m): x c =  i i , y c =  i i , z c =  i i mi mi mi

kc2 = a (circ. disc), 2

kc2 = a2 (circ. wire),

 = 0 t + ½  t2,

2

F F F A A Stress A , (Bulk Mod.): B = , (Shear Mod):  = = (Young’s Mod): Y = L V Strain  L V 2 = 2 f , k = 2 , T = 2 m , v =  A 2 − x 2 , Pot. E: EP=½ kx2 (SHM): x = A sin (  t +  ),  = m k T 1 H.P = 746 W,

1 eV = 1.6x10-19 J,

M(earth) = 5.98x1024 kg, 1 in.= 2.54 cm,

1 MeV = 1.6x10-13 J,

R(earth) = 6.37x106 m,

1 m=3.281 ft,

1 km= 0.621mi,

1 L = 1000 cm3 = 1.0576 qt = 0.0353 ft3, 1 N = 0.2284 lb,

1 lb = 4.448 N,

e = 1.6 x 10-19 C, me = 9.11x10-31 kg Nm2 ke = 9.0x109 , 1 Pa  1 N/m2 C2 1 mi= 1.609 km, 1 m =10-6m = 103 nm

1 gal = 3.786 L = 231 in.3, 1 cal = 4.186 J,

1 m3 = 106 cm3 = 6.102104 in3

1 eV =1.610-19 J,

1kWh =3.6106 J