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1 Solutions to Chapter 1 Student Exercises

1.1 Think back to how you derived the equations of motion for 1D motion with constant velocity and 1D motion with constant acceleration. Now derive the equations of motion for rotational motion with constant angular velocity and rotational motion with constant angular acceleration, shown in Table 1.1. Also, construct graphs for the position, velocity, and acceleration (both angular and linear) as functions of time for all combinations of positive and negative velocity, and positive and negative acceleration. Compare the graphs for the linear and angular quantities as functions of time. Why should you not be surprised that the graphs look the same? Let’s start with the equations for rotational motion with constant angular ⃗ . Because the angular velocity is constant, then the average angular velocity ω ⃗ avg is the same as the instantaneous angular velocity ω ⃗ . That is: velocity ω ∆⃗θ . (1.1) ∆t Realizing that when a quantity is constant its average and instantaneous values are equal is the foundation of such derivations where we do not use calculus. Now, we can easily see that ⃗ ∆t, ∆⃗θ = ω (1.2) ⃗ =ω ⃗ avg = ω

which is the equation of motion with constant angular velocity. Of course, we could derive the same equation using integration. By definition, we know that d⃗θ ⃗ = ⃗ dt. ω ⇒ d⃗θ = ω (1.3) dt 1

Solutions to Chapter 1 Student Exercises

We can integrate both sides of this equation having ⃗θ as the variable of integration on the left-hand side while t is the variable of integration on the right-hand side. Also, we are assuming that at the initial time t0 the initial angular position is ⃗θ0 while at some arbitrary time t the angular position is ⃗θ. We have: Z ⃗θ Z t ⃗ dt d⃗θ = ω ⃗θ0 t0 (1.4) Z t ⃗ ⇒ ⃗θ − ⃗θ0 = ω dt. t0

⃗ is constant, we have taken it out of the integral on the rightwhere, since ω hand side. Therefore, we can now also evaluate the integral on the right-hand side: ⃗θ − ⃗θ0 = ω ⃗ (t − t0 ) (1.5) ⃗ ∆t, ⇒ ∆⃗θ = ω which is, of course, the same as Equation 1.2. We can also derive the equations for rotational motion with constant accel⃗ using calculus. Again, by definition, we know that eration α ⃗= α

d⃗ ω ⃗ dt. ⇒ d⃗ ω=α dt

(1.6)

⃗ as the variable of inteWe again integrate both sides of this equation having ω gration on the left-hand side while t is the variable of integration on the righthand side. Also, we are assuming that at the initial time t0 the initial angular ⃗ 0 while at some arbitrary time t the angular velocity is ω ⃗ . We have: velocity is ω Z ω⃗ Z t ⃗ dt α d⃗ ω= ⃗0 ω t0 (1.7) Z t ⃗ −ω ⃗0 = α ⃗ ⇒ω dt. t0

⃗ is constant, we have taken it out of the integral on the right-hand where, since α side. We can now evaluate the integral on the right-hand side: ⃗ −ω ⃗0 = α ⃗ (t − t0 ) ω ⃗ =ω ⃗0 + α ⃗ ∆t, ⇒ω

(1.8)

which is the equation for the angular velocity as a function of time when the angular acceleration is constant. Given the definition of Equation 1.3, we can now use our result of Equation

Solutions to Chapter 1 Student Exercises

1.8 to get an equation for the angular position as a function of time when the acceleration is constant. We again make the assumptions that at times t0 and t the angular positions are ⃗θ0 and ⃗θ, respectively. We have: d⃗θ dt ⃗ dt ⇒ d⃗θ = ω Z ⃗θ Z t ⃗ dt ⇒ d⃗θ = ω ⃗ = ω

⇒

⃗θ0

Z ⃗θ

Z t

⃗θ0

d⃗θ =

(1.9)

⃗ ∆t)dt. (⃗ ω0 + α

⃗ 0 and α ⃗ are constants and we Both integrals can be evaluated easily since ω obtain: 1 ⃗ ∆t2 . ⃗ 0 ∆t + α (1.10) ∆⃗θ = ω 2 Figure 1.1 shows the graphs for the displacement, velocity, and acceleration as functions of time when we have rotational motion with constant angular velocity and when we have linear motion with constant linear velocity. If we assume that the initial position is zero (i.e., ⃗θ0 = 0 and ⃗x0 = 0, as we have done here) then the displacement is the same as the position at time t. Figure 1.1a shows the graphs when the velocity is positive while Figure 1.1b shows the graphs when the velocity is negative. As expected, the graphs are identical since the positions, velocities, and acceleration, whether angular or linear, are described by equations of the same form. Similarly, Figure 1.2 shows the graphs for the position, velocity, and acceleration as functions of time when we have rotational motion with constant angular acceleration and when we have linear motion with constant linear acceleration. Again, if we assume that the initial position is zero (i.e., ⃗θ0 = 0 and ⃗x0 = 0, as we have done here) then the displacement is the same as the ⃗ 0, α ⃗ > 0, which is the same position at time t. We present four scenarios: a) ω ⃗ 0 > 0, α ⃗ < 0, which is the same as when ⃗v0 > 0, ⃗a < 0, as when ⃗v0 , ⃗a > 0, b) ω ⃗ 0 < 0, α ⃗ > 0, which is the same as when ⃗v0 < 0, ⃗a > 0, and d) ω ⃗ 0, α ⃗ < 0, c) ω which is the same as when ⃗v0 , ⃗a < 0. As expected, in each case the rotational motion graphs are identical to the corresponding linear motion graphs since the positions, velocities, and acceleration, whether angular or linear, are described by equations of the same form. Of course, if θ0 , 0 and x0 , 0 then the graphs of position as a function of time will be shifted up or down, depending on the sign of ⃗θ0 and ⃗x0 . In

Solutions to Chapter 1 Student Exercises

a) θ t x

x t

ω t

t a

v t

b) θ

t a

v t

Figure 1.1 The graphs of position, velocity, and acceleration as functions of time when the velocity is constant and a) positive or b) negative. The acceleration is always zero, the velocity is constant, and the displacement changes linearly with time. Of course, the slope of the displacement vs. time graph is the velocity and the slope of the velocity vs. time graph is the acceleration.

addition, keep in mind that the displacement and/or velocity can change sign. For example, in the ω vs. t graph of Figure 1.2b, the angular velocity ω will become negative (i.e., the object changes its direction of motion) if we continue drawing the graph for longer time. This would result in the slope of the θ vs. t graph becoming negative after that time.

1.2 An object is moving counterclockwise around a circle of radius r = 1.0 m with constant speed. Within time ∆t = 10 ms it traverses an angle of 120◦ . Find a) the object’s angular and linear velocities for an arbitrary position of the object along the circle and b) the period of the object’s motion. c) Draw a diagram to show the directions of the linear and angular velocities for an arbitrary position of the object along its trajectory. a) First we need to convert the angular displacement into units of radians. We know that since an angle of 360o corresponds to 2π radians, then an angle of 120o corresponds to 2π/3 radians. To find the velocities, we can again use the definitions. The speed is constant so the average and instantaneous angular velocities are equal. Therefore, ω=

2π rad 200π ∆θ = 3 = rad/s ∆t 0.01 s 3

(1.11)

Solutions to Chapter 1 Student Exercises

a) θ

t x

c) θ

α t

t a

d) θ

v t

x t

ω t

t a

v t

b) θ

v t

t a

Figure 1.2 The graphs of position, velocity, and acceleration as functions of time when the acceleration is constant. Depending on whether the acceleration and ⃗ 0, α ⃗ > 0 velocity are positive or negative, we have the following scenarios: a) ω ⃗ 0 > 0, α ⃗ < 0 (also ⃗v0 > 0, ⃗a < 0), c) ω ⃗ 0 < 0, α ⃗ > 0 (also (also ⃗v0 , ⃗a > 0), b) ω ⃗v0 < 0, ⃗a > 0), and d) ω ⃗ 0, α ⃗ < 0 (also ⃗v0 , ⃗a < 0). The acceleration is constant, the velocity changes linearly with time, while the position vs. time curve is a parabola. Of course, the slope of the displacement vs. time graph is the velocity while the slope of the velocity vs. time graph is the acceleration.

and v = ωr =

200π 200π rad/s · 1 m = m/s. 3 3

(1.12)

b) For the period, now that we know ω, we have: T=

2π 2π rad = 200π = 0.03 s. ω 3 rad/s

(1.13)

Of course, we could also reason to get the period. We are given that the mass travels 1/3 of a circle (120◦ ) in 0.01 s so, naturally, it will travel the whole

Solutions to Chapter 1 Student Exercises

v2 t2 O

ω 120o 1m

v1 t1

Figure 1.3 A mass is executing rotational motion with constant angular velocity and is found to have traversed an angle of 120o in 0.01 s. The angular velocity is along the axis of rotation while the linear velocity is tangent to the trajectory.

circle in three times that time (i.e., in 0.03 s). c) Based on the indicated rotation, we see from the right-hand rule that if we grasp the axis of rotation with the four fingers of our right hand curling counterclockwise along the plane of the circle, our extended thumb points up. ⃗ , as shown in Figure 1.3, where This is the direction of the angular velocity ω v1 = v2 = v. The instantaneous direction of ⃗v can again be found via the right⃗ ×⃗r or the fact that it is tangent to the trajectory hand rule of the equation ⃗v = ω at the point where the mass is located, pointing in the direction of motion.

1.3 While moving at a constant frequency of 0.5 Hz along the circumference of a circle of radius 0.5 m, an object passes through point A. At the same time, another object starts from rest from the same point A to move along the diameter AB. The second object moves with constant acceleration. a) Find the second object’s acceleration given the fact that the two objects meet at point B at the moment when the first object reached B for the third time (but the second object reached B for the first time). b) What is each object’s velocity at the moment they meet at point B? When we have two objects, we study the motion of each object separately and then look for clues that will allow us to relate the motion of one object to that of the other. a) Figure 1.4 shows the two objects. Object 1 is executing rotational motion

Solutions to Chapter 1 Student Exercises

+x +y +z

v1 A

Figure 1.4 Object 1 moves around a circle with constant angular velocity while object 2 moves along the circle’s diameter with constant linear acceleration. Both objects start at point A at the same time and meet at point B when object 1 reaches B for the third time.

with constant angular velocity. This is because the frequency is constant and, since ω = 2π f , the angular velocity is also constant. Therefore, we have: ω1 = 2π f = 2π × 0.5 Hz = π rad/s.

(1.14)

Since the motion is rotational with constant angular velocity, the equation that describes the motion of object 1 is ⃗ 1 ∆t1 ∆⃗θ1 = ω ⇒ ∆θ1 = ω1 ∆t1 ,

(1.15)

where we have assumed that object 1 is moving counterclockwise and, thus, ⃗ are positive. both ∆⃗θ and ω Object 2 is moving linearly with constant acceleration. Therefore, the equations that describe its motion are, given that v0 = 0: ⃗v2 = ⃗a2 ∆t2 ⇒ v2 = a2 ∆t2

(1.16)

and 1 ∆⃗x2 = ⃗a2 ∆t22 2 (1.17) 1 ⇒ ∆x2 = a2 ∆t22 , 2 where we have defined the direction of motion of object 2 along the diameter to be the +x direction. Because point B is diametrically across from point A, object 1 will have

Solutions to Chapter 1 Student Exercises

reached point B for the third time when it has completed 2.5 revolutions around the circle (i.e., it reaches point B for the first time after it has travelled half a circle and then two full revolutions will result in it reaching point B again for the third time). Therefore, ∆θ1 = 2.5 × (2π) = 5π rad. Object 2, on the other hand, has only moved along the diameter so it has travelled ∆x2 = 2R, where R is the circle radius. However, both objects have travelled for the same amount of time ∆t1 = ∆t2 = ∆t. Using these clues, we now have for Equation 1.15: 5π rad = π rad/s ∆t ⇒ ∆t = 5 s,

(1.18)

which we can now substitute in Equation 1.17 using also that ∆x = 2R = 2 · 0.5 m = 1 m: 2·1m 1 = 0.08 m/s2 . (1.19) 1 m = a2 × (5 s)2 ⇒ a2 = 2 25 s2 b) Now that we know the time interval of the motion and the acceleration of object 2, we can use Equation 1.16 to find the velocity of object 2 when it reaches point B: v2 = 0.08 m/s2 × 5 s = 0.4 m/s.

(1.20)

As for object 1, since its angular velocity is constant, its linear velocity has a constant magnitude of π v1 = ω1 R = π rad/s × 0.5 m = m/s. (1.21) 2

1.4 Two cars are moving around the same circle of radius R = 15 m. Their linear, constant speeds are v1 = 3.0 m/s and v2 = 2.0 m/s. Assuming at time t = 0 the cars are at the same point along the circumference, find the first time that these two cars will meet again when a) they move in the same direction around the circle and b) they move in opposite directions around the circle. Again, we have two objects, so we need to study each separately. Both cars are moving with constant speed, so they execute circular motion with constant angular velocity. We have for each angular velocity that: v1 3 v2 2 = rad/s, ω2 = = rad/s. R 15 R 15 Therefore, the equations that describe the cars’ motion are: ω1 =

⃗ 1 ∆t1 , ∆⃗θ2 = ω ⃗ 2 ∆t2 , ∆⃗θ1 = ω

(1.22)

(1.23)

for cars 1 and 2 respectively. Both cars travel for the same amount of time, and,

Solutions to Chapter 1 Student Exercises thus, ∆t1 = ∆t2 = ∆t.

v2 v1

Δθ1

Δθ2

R v1

Δθ1

Figure 1.5 Two cars move around a circle with constant angular velocities in a) the same direction and b) opposite directions. In each case, the relations between the magnitude of their angular displacements are different.

a) As shown in Figure 1.5a, when the two cars move in the same direction, in order for car 1 (the faster car) to catch up with car 2, it must complete an additional full revolution (compared to car 2). In other words, the relation between the magnitudes of the angular displacements is ∆θ1 = ∆θ2 + 2π. If we substitute Equations 1.23 for the magnitudes ∆θ1 and ∆θ2 we have: ∆θ1 = ∆θ2 + 2π ⇒ ω1 ∆t = ω2 ∆t + 2π ⇒ (ω1 − ω2 )∆t = 2π 2π ω1 − ω2 2π rad ⇒ ∆t = 3 = 30π s. 2 15 rad/s − 15 rad/s ⇒ ∆t =

(1.24)

b) If the cars are moving in opposite directions, as shown in Figure 1.5b, then the two angular displacements’magnitudes will add up to a full revolution once they meet. In other words the relation between the magnitudes of the angular displacements is ∆θ1 + ∆θ2 = 2π. Again, we substitute Equations 1.23 for the

Solutions to Chapter 1 Student Exercises

magnitudes ∆θ1 and ∆θ2 : ∆θ1 + ∆θ2 = 2π ⇒ ω1 ∆t + ω2 ∆t = 2π ⇒ (ω1 + ω2 )∆t = 2π 2π ⇒ ∆t = ω1 + ω2 2π rad ⇒ ∆t = 3 = 6π s. 2 rad/s + 15 rad/s 15

(1.25)

1.5 In Equation (1.31) we showed that at = αr by starting from Equation (1.16) and taking its time derivative. Instead, now use Equation (1.14) as the starting point. Take its time derivative to find the total acceleration. a) Show that the total acceleration consists of two terms: • The tangential acceleration ⃗at given by ⃗at = α ⃗ × ⃗r, • and the centripetal acceleration ⃗ac given by the triple cross product ⃗ac = ω ⃗ × (⃗ ω × ⃗r). b) Explain why the result of this triple cross product is consistent with the magnitude (Equation (1.25)) and the direction of the centripetal acceleration we discussed in Section 3.1. You might need to use the identity that ⃗ × (B ⃗ × C) ⃗ = B( ⃗ A ⃗ · C) ⃗ − C( ⃗ A ⃗ · B). ⃗ A a) We start with the equation ⃗v = ω ⃗ × ⃗r

(1.26)

and take its time derivative. We have: ω × ⃗r) d⃗v d(⃗ = dt dt d⃗ ω d⃗r ⃗× . ⇒ ⃗atot = × ⃗r + ω dt dt

(1.27)

Now let’s carefully study each term on the right-hand side of this result. By ω ⃗ = d⃗ definition, we know that α dt is the angular acceleration. Therefore, the first term on the right-hand side gives d⃗ ω ⃗ × ⃗r = ⃗at , × ⃗r = α dt

(1.28)

Solutions to Chapter 1 Student Exercises

⃗ ×⃗r, as seen in Figure which is the tangential acceleration (the cross product of α 1.6, gives a vector in the tangential direction). For the second term on the right-hand side we know that, by definition, r ⃗v = d⃗ ⃗ × ⃗r. So, this term becomes: v=ω dt . In addition, ⃗ d⃗r ⃗ × ⃗v = ω ⃗ × (⃗ =ω ω × ⃗r) = ⃗ac , (1.29) dt which is the centripetal acceleration. ⃗ × (⃗ b) But how do we know that this last term ω ω × ⃗r) corresponds to the centripetal acceleration? First, let’s think about the direction. We saw that ⃗ × (⃗ ⃗ × ⃗v. Therefore, the cross product of ω ⃗ × ⃗v is a vector that ω ω × ⃗r) = ω ⃗ is along is perpendicular to both these vectors and to the plane they form. ω the axis of rotation and ⃗v is tangent to the circle the rotating object executes. By applying the right-hand rule for this cross product, we see that the resulting vector will be in the radial direction and, more specifically, it will be pointing towards the center of the circle, as seen in Figure 1.6. Therefore, the direction of this term is consistent with the direction of the centripetal acceleration. ⃗× ω

ω α ac r

Figure 1.6 A mass rotating in a circle of radius r. The mass has tangential velocity, tangential acceleration, angular velocity, angular acceleration, and centripetal ⃗, α ⃗ , and ⃗ac , respectively. acceleration ⃗v, ⃗at , ω

⃗ × (⃗ Now, let’s focus on the magnitude. To evaluate ω ω × ⃗r) let’s use the ⃗= B ⃗ =ω ⃗ and C⃗ = ⃗r. Then we identity given in the problem, where we set A get: ⃗ × (⃗ ⃗ (⃗ ⃗ ). ω ω × ⃗r) = ω ω · ⃗r) − ⃗r(⃗ ω·ω

(1.30)

⃗ is perpendicular to ⃗r, since ⃗r is along the plane of motion and ω ⃗ is However, ω perpendicular to it, as seen in Figure 1.6. So the dot product between them is

Solutions to Chapter 1 Student Exercises

⃗ · ⃗r = 0). In addition, ω ⃗ is always parallel to itself, so ω ⃗ ·ω ⃗ = ω2 . zero (i.e., ω Therefore, we have: ⃗ × (⃗ ω ω × ⃗r) = −ω2⃗r.

(1.31)

The negative sign indicates that the direction of the resulting vector is opposite to ⃗r (i.e., towards the circle’s center), as already explained. With regards to the magnitude, since v = ωr ⇒ ω = v/r, we have: v2 v2 ∥⃗ ω × (⃗ ω × ⃗r)∥ = rω2 = r 2 = , r r which is the magnitude of the centripetal acceleration, as we know it!

(1.32)

2 Solutions to Chapter 2 Student Exercises

2.1 The location of the center of mass of an extended object can be found by computing the location of the axis about which the torque due to the gravitational force is zero. a) To show this, first consider a one-dimensional rod of length L and mass M placed along the x axis on this page. Let’s pick the origin of the coordinate system to be on the left end of the rod and split the rod into infinitesimal masses dm each with its own position x with respect to the origin and its own gravitational force F⃗g . Each of these gravitational forces will result in a torque about an axis going through the origin of your coordinate system and perpendicular to the plane of the page. If we were to replace the extended object with a point mass M, where should this mass be located with respect to the origin so that we will get the same torque about the axis through the origin due to the gravitational force on this mass F⃗gM ? This location is defined as the center of mass. b) Now that we found the location of the center of mass with respect to the origin, show that the net torque due to the gravitational forces on the infinitesimal masses dm about an axis that goes through the center of mass of the rod is zero. This exercise leads to an important point that is worth exploring. If you recall from introductory mechanics, the x coordinate for the center of mass xCM is given by:

xCM =

N Σi=1 x i mi N Σi=1 mi

(2.1)

for discrete masses m1 , m2 , . . . mi , . . . mN with positions x1 , x2 , . . . xi , . . . xN , re13

Solutions to Chapter 2 Student Exercises

spectively or R xCM = R

xdm dm

(2.2)

for a continuous mass distribution. But how did Equations (2.1) and (2.2) come about? These equations have the form of a weighted average, but one can derive the location of the center of mass by assuming that this is the position on a continuous object about which the gravitational force will not produce torque (assuming a uniform gravitational field, since then the center of mass coincides with the center of gravity). Let’s see why this is the case in the example of a one-dimensional rod shown in Figure 2.1. The rod is an extended object which can be thought of as a collection of many infinitesimal masses dm, each one with its own gravitational force F⃗g . Let’s place the rod along the x axis on this page, as seen in Figure 2.1, with its left end at the origin O. If we pick an axis of rotation perpendicular to the plane of the page going through O, each of these gravitational forces will produce a torque about this axis. Then, the net torque on the rod will be the sum of these torques. Now let’s replace this R rod with a single point object of the same mass M as the rod (i.e., M = dm). Where should this point object be located along the x axis so that the torque due to the gravitational force on the point object is equal to the net torque on the rod about the same axis of rotation? If these torques are equal for the long rod and the point object, then the point object can be considered equivalent to the rod with regard to the effects of the gravitational force. In other words, we can take the gravitation force on the rod F⃗g,tot to be exerted at a point on the rod that has the same location as that of the point object (see Figure 2.1). Now, let’s be more quantitative. Since each dm is located at position x along the rod, each F⃗g has a position vector of magnitude r = x from the axis of rotation. Now, let’s find the net torque about this axis due to the gravitational forces. We have

Σ⃗τO = ⃗τFg,1 + . . . ⃗τFg,i + · · · + ⃗τFg,N .

(2.3)

Since all torques produce clockwise rotation, their sign will be negative. Thus,

Solutions to Chapter 2 Student Exercises

O Fg,1 . . .

Fg,i . . .

Fg,N

+x +

Fg,tot

Figure 2.1 A thin rod of length L and mass M is placed on the x axis so that its left end coincides with the origin O. a) If we split the rod into infinitesimal masses dm1 , ... dmi , ... dmN each of length dx, then the gravitational force Fg on each mass will produced a torque about an axis that goes through the origin and is perpendicular to the plane of the page. b) If we were to replace this rod with a single point object which also has mass M, where should this object be located so that the torque due to its gravitational force about the axis through point O and perpendicular to the page is equal to the net torque on the rod?

for the magnitude of the net torque we have: ΣτO = τFg,1 + . . . + τFg,i + . . . + τFg,N π π π = Fg,1 x1 sin − . . . + Fg,i xi sin + . . . + Fg,N xN sin 2 2 2 = dm1 gx1 + . . . + dmi gxi + . . . + dmN gxN

(2.4)

= g(dm1 x1 + . . . dmi xi + . . . + dmN xN ) Z = g xdm. Now let’s look at the point mass M. Let’s assume that this point mass is located at position xCM to the right of the origin, as shown in Figure 2.1. We know that it has to be to the right of the origin so that its gravitational force also produces a clockwise rotation, just like the rod. So, for the magnitude of this torque we have: ! Z π dm gxCM . (2.5) Στ′O = τFgM = τFgM = FgM xCM sin = MgxCM = 2 Since the point mass is equivalent to the rod, it will produce the same torque about the same axis of rotation. Thus, we have for the magnitudes:

Z ⇒g

ΣτO = Στ′O ! Z xdm = dm gxCM .

(2.6)

Solutions to Chapter 2 Student Exercises

We see that g cancels out so we are left with: R xCM = R

xdm dm

(2.7)

which is the position of the center of mass xCM as defined in Equation 2.2! In other words, the center of mass is the position that the point object needs to have so that the torque due to its gravitational force is the same as the torque of the gravitational force on the rod it represents. b) The equivalency between the point object and the rod is true regardless of the axis of rotation. Therefore, the gravitational force on the point object and on the rod will produce the same torque about any axis. So, let’s assume that the axis of rotation goes through the center of mass and let’s make that point the origin of our coordinate system. Then for the point mass M, the position vector of its gravitational force F⃗gM is xcm = 0. Therefore, π (2.8) Στ⃗′ O = ⃗τFgM ⇒ FgM x sin = 0, 2 and since this torque has to be equal to the torque due to the gravitational force on the rod (i.e., Στ⃗′ O = Σ⃗τO ), we see that the net torque due to the gravitational forces on the infinitesimal masses dm about an axis that goes through the center of mass of the rod is zero.

2.2 Imagine we have a horizontal rod supported either directly below or directly above its center of mass (CM) as shown in Figure 2.2a and 2.2b, respectively. Now, lets tilt the rod slightly and then release it. In the case where the support is below the rod, the rod will continue to tilt and will eventually fall. However, if the rod is supported from directly above its center of mass, it will always return to its original, horizontal position. Explain why this is true. This problem raises the issue of unstable vs. stable equilibrium, illustrated by Figure 2.2a and b, respectively. In these scenarios, the axis of rotation does not go through the center of mass. When the rod is horizontal, in Figure 2.2a the angle between the gravitational force F⃗g and its position vector is ϕ = 180◦ and, therefore, F⃗g does not produce torque. However, if the rod gets tilted a little (say in the counterclockwise direction, as shown), the angle between the gravitational force F⃗g and its position vector is ϕ , 180◦ and, thus, there is a torque due to the gravitational force on the rod. By applying the right-hand

Solutions to Chapter 2 Student Exercises a)

τ φ

Fg Figure 2.2 A rod is supported from directly below or directly above its center of mass (CM). a) A small tilt in the counterclockwise direction will cause the rod to fall, because the torque of the gravitational force will cause further counterclockwise rotation. b) The rod will return to its horizontal position after a tilt in the counterclockwise direction because the torque of the gravitational force will cause an opposite clockwise rotation.

rule, we find that this torque points out of the page, which implies that it results in a counterclockwise rotation. In other words, tilting the rod counterclockwise by the slightest amount will result in F⃗g producing further counterclockwise rotation, so the rod will fall! We would have the same result if the initial disturbance was in the clockwise direction - the torque due to F⃗g would result in further clockwise rotation and the rod would again fall. Therefore, the rod’s initial equilibrium is unstable when it is supported below the center of mass any small disturbance (i.e., deviation from its equilibrium position) will cause the rod to move further away from the initial equilibrium state. Now let’s look at what happens when the rod is supported directly above the center of mass. When the rod is horizontal, as shown in Figure 2.2b, the

Solutions to Chapter 2 Student Exercises

angle between the gravitational force F⃗g and its position vector is ϕ = 0 and, therefore, again F⃗g does not produce torque. However, if the rod gets tilted a little (say in the counterclockwise direction, as shown), the angle between the gravitational force F⃗g and its position vector is ϕ , 0 and, thus, there is torque. By applying the right-hand rule, we find that this torque points into the page, which implies that it results in a clockwise rotation. In other words, tilting the rod counterclockwise by the slightest amount will result in F⃗g producing clockwise rotation, counteracting the counterclockwise initial deviation. Thus, the rod will return to its initial horizontal position! We would have the same result if the initial disturbance was in the clockwise direction - the torque due to F⃗g would result in a counterclockwise rotation and the rod would again become horizontal. Therefore, supporting the rod above its center of mass results in a stable equilibrium - any small disturbance (i.e., deviation from its equilibrium position) will cause the rod to move back to this equilibrium position!

2.3 One end of a rod of length L = 1.0 m is attached to a wall via a junction that is free to rotate. The rod is kept horizontally balanced via a massless spring attached to its other end, as shown in Figure 2.3. You are given that Fg = 100 N and that the center of mass is x = 40 cm away from its right end. Find the force exerted by the spring and the reaction force of the junction.

+y Ry R θ R

+x +

Figure 2.3 A rod is kept horizontal via a junction and a spring. The gravitational force F⃗g is exerted at a distance x = 40 cm from the end of the rod where the ⃗ of unknown direction, so spring is attached. The junction exerts a reaction force R we assume that both its x and y components are positive.

As always, we draw the forces exerted on the rod, the object of interest. ⃗ from the junction, whose magnitude and These forces are the reaction force R

Solutions to Chapter 2 Student Exercises

direction are unknown, the gravitational force F⃗g , which is exerted at the center of mass, and the spring force F⃗ s . The spring force has an upward direction because the rod’s gravitational force is extending the spring downwards. Because ⃗ direction is not known, we will assume that its components are positive and R’s that it forms and angle θ with the +x axis. Of course, we also draw our coordinate system. The rod is in static equilibrium, which implies that all forces and torques have to add up to zero. ΣF⃗ x = 0 ⇒ Rx = 0

(2.9)

and ΣF⃗y = 0 ⇒ Ry − Fg + F s = 0

(2.10)

⇒ Ry = Fg − F s For the rotational equilibrium, we pick our axis of rotation to be perpendicular to the plane of the page and go through point A. We pick point A since the ⃗ is exerted at that point. Based on our discussion in Example unknown force R 4 of Chapter 2, another convenient choice would be an axis of rotation that goes through the rod’s other end, since the spring force is also unknown. We assume that L is the length of the rod, and thus for the net torque about an axis through point A we have: Σ⃗τA = ⃗τRx + ⃗τRy +⃗τFg + ⃗τF s = 0 |{z} |{z} 0, (r=0)

0, (r=0)

⇒ −τFg + τF s = 0 π π = LF s sin ⇒ (L − x)Fg sin 2 2 (L − x)Fg (1 − 0.4) m · 100 N ⇒ Fs = = = 60 N. L 1m

(2.11)

Since now we know the magnitude of the spring force, we can find the value of Ry from Equation (2.10): Ry = Fg − F s = 100 N − 60 N = 40 N,

(2.12)

⃗ is along the +y axis and Since R x = 0 then R = Ry . That is, the reaction force R has a magnitude of 40 N.

Solutions to Chapter 2 Student Exercises

2.4 A rod is attached to a wall via a junction that is free to rotate and is kept balanced by a massless, horizontal rope that is 9.0 m long, as shown in Figure 2.4. You are given that Fg1 = 750 N, Fg2 = 300 N (these weights include the weight of the rod), AB = 6.0 m, AC = 12.0 m and that the length of the rod is L = 15.0 m. Find the tension in the rope and the magnitude and direction of the reaction force of the junction.

9m E rope φ A

C rod B

FT Ry R θ

φ φ

+y Fg2 -

Fg1

Figure 2.4 A rod is kept balanced at an angle ϕ relative to the wall via a junction and a massless, horizontal rope. The gravitational force on the rod is given as two separate vectors, F⃗g1 and F⃗g2 . The force due to tension F⃗T is horizontal but the ⃗ from the junction has an unknown magnitude and direction, so reaction force R we assume that its components are positive.

As always, we draw the forces exerted on the rod, the object of interest. ⃗ from the junction, whose magnitude and These forces are the reaction force R direction are unknown, the gravitational forces F⃗g1 and F⃗g2 , and the tension force F⃗T . The force due to tension is horizontal, pointing to the left. As always, ⃗ direction is not known, we will assume that its components are because R’s positive and that it forms and angle θ with the +x axis. Of course, we also draw our coordinate system. The rod is in static equilibrium, which implies that all forces and torques have to add up to zero. We have: ΣF⃗ x = 0 ⇒ R x − FT = 0 ⇒ R x = FT ,

(2.13)

Solutions to Chapter 2 Student Exercises and ΣF⃗y = 0 ⇒ Ry − Fg1 − Fg2 = 0

(2.14)

⇒ Ry = Fg1 + Fg2 = 750 N + 300 N = 1050 N. For the rotational equilibrium, we pick our axis of rotation to be perpendicular to the plane of the page and go through point A. As we discussed in ⃗ previous problems and examples, we pick point A since the unknown force R is exerted at that point. Another convenient choice would be an axis of rotation that goes through point D, since the force of tension is also unknown. To evaluate the torques, we will need to know the angles between the forces and their respective position vectors. Because we are given the dimensions of the rod and the rope, and we can see that a right triangle is formed by the rod, the rope, and the wall, we can use the Pythagorean theorem to find that the length of the wall segment AE = 12 m. We can also see that the angle ϕ between the rod and the wall is the same as the supplementary angle between the two gravitational forces and their position vectors. Thus, sin ϕ = 9 . As for the force due to tension, the angle it forms with its sin (π − ϕ) = 15 position vector is π2 − ϕ and, thus, sin π2 − ϕ = cos ϕ = 12 15 . For the net torque about an axis through point A and perpendicular to the plane of the page we have: Σ⃗τA = ⃗τRx + ⃗τRy +⃗τFg1 + ⃗τFg2 + ⃗τFT = 0 |{z} |{z} 0, (r=0)

0, (r=0)

⇒ −τFg1 − τFg2 + τFT = 0 ⇒ (AB)Fg1 sin ϕ + (AC)Fg2 sin ϕ = LFT sin ⇒ 6 m · 750 N ·

π 2

−ϕ

(2.15)

9 9 12 + 12 m · 300 N · = 15 m · FT · 15 15 15

⇒ FT = 405 N. Since now we know the magnitude of the force due to tension, we can find the value of R x from Equation (2.13): R x = FT = 405 N.

(2.16)

⃗ are found to be positive, our initial assumption Since both components of R about its components being along the positive directions of their respective axes was correct. ⃗ we have: For the magnitude and direction of R q p (2.17) R = R2x + R2y = (405 N)2 + (1050 N)2 = 1125 N,

Solutions to Chapter 2 Student Exercises tan θ =

Ry 1050 N ⇒ θ = 69◦ . = Rx 405 N

(2.18)

Of course, there is a second solution to Equation (2.18): θ = 249◦ . But, since our results show that both R x , Ry > 0, the angle θ belongs in the first quadrant, and so this second solution is not physically acceptable.

2.5 A uniform wheel with radius α is on a horizontal surface in contact with a step of height h as shown in Figure 2.5. The gravitational force F⃗g is exerted at the wheel’s center O. A force F⃗ is exerted at point Z, which is at a height H above the horizontal surface, which causes the wheel to roll up the step. a) Draw all the forces exerted on the wheel before it rolls up the step. What happens to the ground’s reaction force ⃗nA on the wheel at point A when the wheel starts rolling up the step and loses contact with the ground? b) Find the minimum magnitude of F⃗ needed for the wheel to roll up the step. c) Find the direction ⃗ on the wheel at point B in two different ways: i) of the step’s reaction force R using Newton’s Laws and ii) using geometry and Theorem 3.

+y z

H α h

O R

+x +

Fg Rx θ B A x

Figure 2.5 We exert a horizontal force F⃗ on a uniform wheel of radius α to roll it up a step of height h. As the wheel rolls up the step, point A loses contact with the ground, so the reaction force nA = 0. The wheel is in contact with the step at ⃗ by the step on the wheel at that point. point B, so there is a reaction force R

a) Figure 2.5 shows all the forces exerted on the wheel. There are some important observations that need to be made. Although there is usually a reaction force from the horizontal surface on the wheel exerted at point A, the instant the wheel rolls up the step, it loses contact with the horizontal surface so this force will vanish (i.e., nA = 0). Thus, this force is not shown in Figure 2.5.