CHAPTER 9 ROTATIONAL DYNAMICS

CHAPTER 9 ROTATIONAL DYNAMICS

ANSWERS TO FOCUS ON CONCEPTS QUESTIONS ___________________________________________________________________________________________ 1. (d) A rigid body is in equilibrium if it has zero translational acceleration and zero angular

acceleration. A body, such as a bicycle wheel, can be moving, but the translational and angular accelerations must be zero (a = 0 m/s2 and α = 0 rad/s2).

2. (e) As discussed in Section 9.2, a body is in equilibrium if the sum of the externally applied

forces is zero and the sum of the externally applied torques is zero. 3. (b) The torque τ3 is greater than τ2, because the lever arm for the force F3 is greater than that

for F2. The lines of action for the forces F1 and F4 pass through the axis of rotation. Therefore, the lever arms for these forces are zero, and the forces produce no torque.

4. (b) Since the counterclockwise direction is the positive direction for torque, the torque

produced by the force F1 is τ1 = −(20.0 N)(0.500 m) and that produced by F2 is τ2 = +(35.0 N)[(1.10 m)(cos 30.0°)]. The sum of these torques is the net torque.

5. (e) The clockwise torque produced by F2 is balanced by the counterclockwise torque produced

by F. The torque produced by F2 is (remembering that the counterclockwise direction is positive) τ2 = +F2[(80.0 cm − 20.0 cm)(sin 55.0°)], and the torque produced by F is τ = −(175 N)(20.0 cm). Setting the sum of these torques equal to zero and solving for F2 gives the answer.

6. (d) The sum of the forces (F − 2F + F) equals zero. Select an axis that passes through the center

of the puck and is perpendicular to the screen. The sum of the torques [−FR + 2F(0) + FR ] equals zero, where R is the radius of the puck. Thus, the puck is in equilibrium.

7. Magnitude of F1 = 12.0 N, Magnitude of F2 = 24.0 N

8. (c) The horizontal component of F3 is balanced by F1, and the vertical component of F3 is balanced by F2. Thus, the net force and, hence, the translational acceleration of the box, is zero. For an axis of rotation at the center of the box and perpendicular to the screen, the forces F2 and F3 produce no torque, because their lines of action pass through the axis. The force F1 does produce a torque about the axis, so the net torque is not zero and the box will have an angular acceleration.

9. Distance of center of gravity from support = 0.60 m.

420 ROTATIONAL DYNAMICS

10. (b) The moment of inertia of each particle is given by Equation 9.6 as 2I mr= , where m is its mass and r is the perpendicular distance of the particle from the axis. Using this equation, the moment of inertia of each particle is: A: 2

0 010m r , B: 20 08m r , C: 2

0 09m r .

11. I = 1.7 kg⋅m2

12. Magnitude α of the angular acceleration = 1.3 rad/s2.

13. (a) According to Newton’s second law for rotational motion, Equation 9.7, the angular acceleration is equal to the torque exerted on the wheel (the product of the force magnitude and the lever arm) divided by the moment of inertia. Thus, the angular acceleration of the smaller wheel is α = FR/(MR2) = F/(MR), while that of the larger wheel is

( )( )

( )122

2/

2

F RF MR

M Rα = = , so the smaller wheel has twice the angular acceleration.

14. Magnitude α of the angular acceleration = 12.0 rad/s2 15. (c) The translational kinetic energy is 21

2Mv , where v is the speed of the center of mass of the wheel. The rotational kinetic energy is 21

2 Iω , where I is the moment of inertia and ω is

the angular speed about the axis of rotation. Since I = MR2 and ω = v/R for rolling motion

(See Equation 8.12), the rotational kinetic energy is ( )2

2 2 21 1 12 2 2 ,vI MR Mv

Rω ⎛ ⎞= =⎜ ⎟⎝ ⎠

which

is the same as the translational kinetic energy. Thus, the ratio of the two energies is 1. 16. (c) As each hoop rolls down the incline, the total mechanical energy is conserved. Thus, the

loss in potential energy is equal to the gain in the total kinetic energy (translational plus rotational). Because the hoops have the same mass and fall through the same vertical distance, they lose the same amount of potential energy. Moreover, both start from rest. Therefore, their total kinetic energies at the bottom are the same.

17. (d) As discussed in Section 9.6, the angular momentum a system is conserved (remains

constant) if the net external torque acting on the system is zero. 18. (b) The rotational kinetic energy of a rotating body is 21

2RKE Iω= (see Equation 9.9), where I is the moment of inertia and ω is the angular speed. We also know that her angular momentum, L = Iω (Equation 9.10), is conserved. Solving the last equation for I and

substituting the result into the first equation gives 2 21 1 12 2 2RKE LI Lω ω ω

ω⎛ ⎞= = =⎜ ⎟⎝ ⎠

. Since L

is constant, the final rotational kinetic energy increases as ω increases. 19. I = 0.60 kg⋅m2

Chapter 9 Problems 421 CHAPTER 9 ROTATIONAL DYNAMICS

PROBLEMS

1. SSM REASONING The drawing shows the wheel as it rolls

to the right, so the torque applied by the engine is assumed to be clockwise about the axis of rotation. The force of static friction that the ground applies to the wheel is labeled as fs. This force produces a counterclockwise torque τ about the axis of rotation, which is given by Equation 9.1 as s = fτ l , where l is the lever arm. Using this relation we can find the magnitude fs of the static frictional force.

SOLUTION The countertorque is given as s = fτ l , where fs is the magnitude of the static

frictional force and l is the lever arm. The lever arm is the distance between the line of action of the force and the axis of rotation; in this case the lever arm is just the radius r of the tire. Solving for fs gives

s295 N m 843 N0.350 m

f τ ⋅= = =l

2. REASONING The torque on either wheel is given by Fτ = l (Equation 9.1), where F is the magnitude of the force and l is the lever arm. Regardless of how the force is applied, the lever arm will be proportional to the radius of the wheel.

SOLUTION The ratio of the torque produced by the force in the truck to the torque

produced in the car is

truck truck truck truck

car car car car

0.25 m 1.30.19 m

F Fr rF Fr r

ττ

= = = = =

ll

3. SSM REASONING According to Equation 9.1, we have

Magnitude of torque = Fl

where F is the magnitude of the applied force and l is the lever arm. From the figure in the text, the lever arm is given by (0.28 m) sin 50.0= °l . Since both the magnitude of the torque and l are known, Equation 9.1 can be solved for F.

SOLUTION Solving Equation 9.1 for F, we have

Axis of rotation

r fs


2Magnitude of torque 45 N m 2.1 10 N

(0.28 m) sin 50.0F ⋅= = = ×

°l

4. REASONING The net torque on the branch is the sum of

the torques exerted by the children. Each individual torque τ is given by Fτ = l (Equation 9.1), where F is the magnitude of the force exerted on the branch by a child, and l is the lever arm (see the diagram). The branch supports each child’s weight, so, by Newton’s third law, the magnitude F of the force exerted on the branch by each child has the same magnitude as the child’s weight: F = mg. Both forces are directed downwards. The lever arm for each force is the perpendicular distance between the axis and the force’s line of action, so we have cosd θ=l (see the diagram).

SOLUTION The mass of the first child is m1 = 44.0 kg. This child is a distance d1 = 1.30 m from the tree trunk. The mass of the second child, hanging d2 = 2.10 m from the tree trunk, is m2 = 35.0 kg. Both children produce positive (counterclockwise) torques. The net torque exerted on the branch by the two children is then

Στ = τ1 +τ2 = F11 + F22 = m1gF1

d1cosθ

1

+ m2g

F2

d2 cosθ

2

= g cosθ m1d1 + m2d1( )

Substituting the given values, we obtain

( )( ) ( )( ) ( )( )29.80 m/s cos 27.0 44.0 kg 1.30 m 35.0 kg 2.10 m 1140 N mτΣ = + = ⋅⎡ ⎤⎣ ⎦

o

5. SSM REASONING To calculate the torques, we need to determine the lever arms for

each of the forces. These lever arms are shown in the following drawings:

d

l θ

F Tree

Axis of rotation

2.50 m

Axis 32.0°

T = (2.50 m) cos 32.0°

T

2.50 m

Axis

32.0°

W W = (2.50 m) sin 32.0°

Chapter 9 Problems 423

SOLUTION a. Using Equation 9.1, we find that the magnitude of the torque due to the weight W is

( )( )WMagnitude of torque 10 200 N 2.5 m sin32 13 500 N mW= = ° = ⋅l b. Using Equation 9.1, we find that the magnitude of the torque due to the thrust T is

( )( )TMagnitude of torque 62 300 N 2.5 m cos32 132 000 N mT= = ° = ⋅l 6. REASONING The maximum torque will occur when the force is

applied perpendicular to the diagonal of the square as shown. The lever arm l is half the length of the diagonal. From the Pythagorean theorem, the lever arm is, therefore,

l = 12 (0.40 m)2 + (0.40 m)2 = 0.28 m

Since the lever arm is now known, we can use Equation 9.1 to

obtain the desired result directly. SOLUTION Equation 9.1 gives

τ = Fl = (15 N)(0.28 m) = 4.2 N⋅m

7. SSM REASONING AND SOLUTION The torque produced by each force of magnitude

F is given by Equation 9.1, Fτ = l , where l is the lever arm and the torque is positive since each force causes a counterclockwise rotation. In each case, the torque produced by the couple is equal to the sum of the individual torques produced by each member of the couple.

a. When the axis passes through point A, the torque due to the force at A is zero. The lever arm for the force at C is L. Therefore, taking counterclockwise as the positive direction, we have

0A C FL FLτ τ τ= + = + =

b. Each force produces a counterclockwise rotation. The magnitude of each force is F and each force has a lever arm of / 2L . Taking counterclockwise as the positive direction, we have

2 2A CL LF F FLτ τ τ ⎛ ⎞ ⎛ ⎞= + = + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

c. When the axis passes through point C, the torque due to the force at C is zero. The lever arm for the force at A is L. Therefore, taking counterclockwise as the positive direction, we have


0A C FL FLτ τ τ= + = + =

Note that the value of the torque produced by the couple is the same in all three cases; in other words, when the couple acts on the tire wrench, the couple produces a torque that does not depend on the location of the axis.

8. REASONING We know that the torques generated by the two applied forces sum to zero.

In order for this to be true, one of the forces must tend to produce a clockwise rotation about the pinned end of the meter stick, and the other must tend to produce a counterclockwise rotation about the pinned end. We will assume that the first force, applied perpendicular to the length of the meter stick at its free end, tends to produce a counterclockwise rotation (see the drawing). Counterclockwise is the positive direction.

With this assumption, we obtain

1 1 2 2 1 1 2 20 or F F F F− = =l l l l (1)

The drawing shows that the lever arm 1l of the force F1 is equal to the length of the meter stick: 1l = 1.00 m. The force F2 is applied a distance d from the pinned end of the meter stick. The lever arm 2l of the second force is

2 sind θ=l (2)

SOLUTION Substituting Equation (2) into Equation (1) and solving for the distance d, we obtain

( ) ( )( )( )

1 11 1 2

2

2.00 N 1.00 msin or 0.667 m

sin 6.00 N sin 30.0F

F F d dF

θθ

= = = =ol

l

9. REASONING Each of the two forces produces a torque about the axis of rotation, one

clockwise and the other counterclockwise. By setting the sum of the torques equal to zero, we will be able to determine the angle θ in the drawing.

1l

2l

d θ

θ F1

TOP VIEW

pinned end

F2


SOLUTION The two forces act on the rod at a distance x from the hinge. The torque τ1 produced by the force F1 is given by τ1 = +F1 1l (see Equation 9.1), where F1 is the magnitude of the force and 1l is the lever arm. It is a positive torque, since it tends to produce a counterclockwise rotation. Since F1 is applied perpendicular to the rod, 1 .x=l The torque τ2 produced by F2 is τ2 = −F2 2l , where 2 sinx θ=l (see the drawing). It is a negative torque, since it tends to produce a clockwise rotation. Setting the sum of the torques equal to zero, we have

+F11 + −F22( ) = 0 or +F1x − F2 x sin θ( )2

= 0

The distance x in this relation can be eliminated algebraically. Solving for the angle θ gives

1 1

2 2

38.0 Nsin or sin sin 43.755.0 N

F FF F

θ θ −1 −1⎛ ⎞ ⎛ ⎞= = = = °⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

10. REASONING AND SOLUTION The net torque about the axis in text drawing (a) is

Στ = τ1 + τ2 = F1b − F2a = 0 Considering that F2 = 3F1, we have b – 3a = 0. The net torque in drawing (b) is then

Στ = F1(1.00 m − a) − F2b = 0 or 1.00 m − a − 3b = 0 Solving the first equation for b, substituting into the second equation and rearranging, gives

a = 0.100 m and b = 0.300 m

Table (Top view)

Hinge (axis of rotation)

Rod

θ

F2 = 55.0 N

F1 = 38.0 N

x

θ

2l

90º


11. REASONING Although this arrangement of body parts is vertical, we can apply Equation

9.3 to locate the overall center of gravity by simply replacing the horizontal position x by the vertical position y, as measured relative to the floor.

SOLUTION Using Equation 9.3, we have

ycg =W1y1 +W2 y2 +W3y3

W1 +W2 +W3

=438 N( ) 1.28 m( ) + 144 N( ) 0.760 m( ) + 87 N( ) 0.250 m( )

438 N +144 N +87 N= 1.03 m

12. REASONING The drawing shows the

forces acting on the person. It also shows the lever arms for a rotational axis perpendicular to the plane of the paper at the place where the person’s toes touch the floor. Since the person is in equilibrium, the sum of the forces must be zero. Likewise, we know that the sum of the torques must be zero.

SOLUTION Taking upward to be the

positive direction, we have

F F WFEET HANDS+ − = 0 Remembering that counterclockwise torques are positive and using the axis and the lever

arms shown in the drawing, we find

WW − FHANDSHANDS = 0

FHANDS =WWHANDS

=584 N( ) 0.840 m( )

1.250 m= 392 N

Substituting this value into the balance-of-forces equation, we find

F W FFEET HANDS N N == − = −584 392 192 N The force on each hand is half the value calculated above, or 196 N . Likewise, the force

on each foot is half the value calculated above, or 96 N . 13. SSM REASONING The drawing shows the bridge and the four forces that act on it: the

upward force F1 exerted on the left end by the support, the force due to the weight Wh of the

HANDS = 1.250 m

Axis

FFEET FHANDS

W W = 0.840 m


hiker, the weight Wb of the bridge, and the upward force F2 exerted on the right side by the support. Since the bridge is in equilibrium, the sum of the torques about any axis of rotation must be zero ( )0τΣ = , and the sum of the forces in the vertical direction must be

zero ( )0yFΣ = . These two conditions will allow us to determine the magnitudes of F1 and

F2.

SOLUTION a. We will begin by taking the axis of rotation about the right end of the bridge. The torque produced by F2 is zero, since its lever arm is zero. When we set the sum of the torques equal to zero, the resulting equation will have only one unknown, F1, in it. Setting the sum of the torques produced by the three forces equal to zero gives

( ) ( )4 1

1 h b5 2 0F L W L W LτΣ = − + + =

Algebraically eliminating the length L of the bridge from this equation and solving for F1 gives

( ) ( )4 1 4 11 h b5 2 5 2985 N 3610 N 2590 NF W W= + = + =

b. Since the bridge is in equilibrium, the sum of the forces in the vertical direction must be zero:

1 h b 2 0yF F W W FΣ = − − + =

Solving for F2 gives

2 1 h b 2590 N + 985 N + 3610 N = 2010 NF F W W= − + + = −

Axis F2

F1

Wh

Wb

15 L 1

2 L

+y

+x

+τ


14. REASONING The truck is subject to three vertical forces only (see the free-body diagram), and is in equilibrium. Therefore, the conditions 0yFΣ = (Equation 4.9b) and

0τΣ = (Equation 9.2) apply to the forces and torques acting on it. The ground exerts upward forces FR on the rear wheels and FF on the front wheels, and the earth exerts a downward weight force W on the truck’s center of gravity. The net vertical force on the truck must be zero, so we find

R F 0F F W+ − = (1)

where we have assumed upward to be the positive direction. To apply the zero net torque condition (Equation 9.2), we choose a rotation axis located on the ground, directly below the truck’s center of gravity (see the diagram). The weight force W has no lever arm about this axis, and so it generates no torque. The force FF exerts a counterclockwise torque about this axis, and the force FR exerts a clockwise torque about this axis. Since counterclockwise is the positive direction, we find from Equation 9.2 that F F R R F F R R0 or F F F FτΣ = − = =l l l l (2)

We will use Equations (1) and (2) to find the two unknown forces. SOLUTION a. In order to find the force magnitude FF, we must eliminate the unknown force magnitude FR from Equations (1) and (2). Solving Equation (1) for FR yields R F FF W F mg F= − = − .

Substituting this expression into Equation (2), we obtain

( ) ( )F F F R R F R F F R R or F mg F mg F F mg= − = − + =l l l l l l l (3)

Solving Equation (3) for the force magnitude FF and noting from the drawing in the text that

F R2.30 m and = 0.63 m,=l l we find that

( )( )( )24R

FF R

7460 kg 9.80 m/s 0.63 m1.57 10 N

2.30 m 0.63 mmg

F = = = ×+ +l

l l

b. Returning to Equation (1), we obtain the magnitude of the force on the rear wheels:

( )( )2 4 4R F F 7460 kg 9.80 m/s 1.57 10 N 5.74 10 NF W F mg F= − = − = − × = ×

FF FR

W

Rl Fl

Rotation axis

Free-body diagram of the truck


15. REASONING Since the forearm is in equilibrium, the sum of the torques about any axis of

rotation must be zero ( )0τΣ = . For convenience, we will take the elbow joint to be the axis of rotation.

SOLUTION Let M and F be the magnitudes of the forces that the flexor muscle and test apparatus, respectively, exert on the forearm, and let Ml and Fl be the respective lever arms about the elbow joint. Setting the sum of the torques about the elbow joint equal to zero (with counterclockwise torques being taken as positive), we have

M F 0M FτΣ = − =l l

Solving for M yields ( )( )F

M

190 N 0.34 m1200 N

0.054 mF

M = = =ll

The direction of the force is to the left .

16. REASONING The net torque is the sum of the torques produced by the three forces:

A B Dτ τ τ τΣ = + + . The magnitude of a torque is the magnitude of the force times the lever arm of the force, according to Equation 9.1. The lever arm is the perpendicular distance between the line of action of the force and the axis. A torque that tends to produce a counterclockwise rotation about the axis is a positive torque. Since the piece of wood is at equilibrium, the net torque is equal to zero.

SOLUTION Let L be the length of the short side of the rectangle, so that the length of the

long side is 2L. The counterclockwise torque produced by the force at corner B is BF+ l , and the clockwise torque produced by the force at corner D is DF− l . Assuming that the force at A (directed along the short side of the rectangle) points toward corner B, the counterclockwise torque produced by this force is A AF+ l . Setting the net torque equal to zero gives:

( )( ) ( )A A B D

1A 2

0

12 N 12 N 0

F F F

F L L L

τΣ = + − =

+ − =

l l l

The length L can be eliminated algebraically from this result, which can then be solved for

FA:

( )( )1A 212 N 12 N 6.0 N (pointing toward corner B)F = − + =

Since the value calculated for FA is positive, our assumption that FA points toward corner B

must have been correct. Otherwise, the result for FA would have been negative.


17. SSM REASONING Multiple-Concept Example 8 discusses the static stability factor (SSF) and rollover. In that example, it is determined that the maximum speed v at which a vehicle can negotiate a curve of radius r is related to the SSF according to ( )SSFv rg= . No value is given for the radius of the turn. However, by applying this result separately to the sport utility vehicle (SUV) and to the sports car (SC), we will be able to eliminate r algebraically and determine the maximum speed at which the sports car can negotiate the curve without rolling over.

SOLUTION Applying ( )SSFv rg= to each vehicle, we obtain

( ) ( )SUV SCSUV SCSSF and SSFv rg v rg= = Dividing these two expressions gives

( )( )

( )( ) ( )SCSC SC

SC SUVSUV SUVSUV

SSF SSF 1.4 or 18 m/s 24 m/sSSF 0.80SSF

rgvv v

v rg= = = =

18. REASONING Since the wheelbarrow is in equilibrium, the net torque acting on it must be

zero: 0τΣ = (Equation 9.2). The magnitude of a torque is the magnitude of the force times the lever arm of the force (see Equation 9.1). The lever arm is the perpendicular distance between the line of action of the force and the axis. A torque that tends to produce a counterclockwise rotation about the axis is a positive torque.

SOLUTION The lever arms for the forces can be obtained from the distances shown in the

text drawing for each design. Equation 9.1 can be used to obtain the magnitude of each torque. We will then write an expression for the zero net torque for each design. These expressions can be solved for the magnitude F of the man’s force in each case:

Left design Στ = − 525 N( ) 0.400 m( )− 60.0 N( ) 0.600 m( ) + F 1.300 m( ) = 0

F =525 N( ) 0.400 m( ) + 60.0 N( ) 0.600 m( )

1.300 m= 189 N

Right design Στ = − 60.0 N( ) 0.600 m( ) + F 1.300 m( ) = 0

F =60.0 N( ) 0.600 m( )

1.300 m= 27.7 N

19. REASONING The jet is in equilibrium, so the sum of the external forces is zero, and the

sum of the external torques is zero. We can use these two conditions to evaluate the forces exerted on the wheels.


SOLUTION a. Let Ff be the magnitude of the normal force that the ground exerts on the front wheel.

Since the net torque acting on the plane is zero, we have (using an axis through the points of contact between the rear wheels and the ground)

Στ = −W w + Ff f = 0 where W is the weight of the plane, and w and f are the lever arms for the forces W and Ff,

respectively. Thus,

Στ = −(1.00 × 106 N)(15.0 m − 12.6 m) + Ff (15.0 m) = 0 Solving for Ff gives Ff = 1 60 105. × N . b. Setting the sum of the vertical forces equal to zero yields

ΣFy = Ff + 2Fr − W = 0 where the factor of 2 arises because there are two rear wheels. Substituting the data gives

ΣFy = 1.60 × 105 N + 2Fr − 1.00 × 106 N = 0

Fr = 54.20 10 N×

20. REASONING The minimum value for the coefficient of static friction between the ladder and the ground, so that the ladder does not slip, is given by Equation 4.7:

MAXs s Nf Fµ=

SOLUTION From Example 4, the magnitude of the force of static friction is Gx = 727 N.

The magnitude of the normal force applied to the ladder by the ground is Gy = 1230 N. The minimum value for the coefficient of static friction between the ladder and the ground is

MAXs

sN

727 N 0 5911230 N

x

y

f GF G

µ = = = = .

______________________________________________________________________________

21. REASONING Since the beam is in equilibrium, the sum of the horizontal and vertical

forces must be zero: 0 and 0x yF FΣ = Σ = (Equations 9.4a and b). In addition, the net torque about any axis of rotation must also be zero: 0τΣ = (Equation 9.2).


SOLUTION The drawing shows the beam, as well as its weight W, the force P that the pin exerts on the right end of the beam, and the horizontal and vertical forces, H and V, applied to the left end of the beam by the hinge. Assuming that upward and to the right are the positive directions, we obtain the following expressions by setting the sum of the vertical and the sum of the horizontal forces equal to zero:

Horizontal forces

Vertical forces

P H

P V W

cos ( )

sin ( )

θ

θ

− =

+ − =

0 1

0 2

Using a rotational axis perpendicular to the plane of the paper and passing through the pin,

and remembering that counterclockwise torques are positive, we also set the sum of the torques equal to zero. In doing so, we use L to denote the length of the beam and note that the lever arms for W and V are 12 L and L, respectively. The forces P and H create no torques relative to this axis, because their lines of action pass directly through it.

Torques W 1

2 L( )−VL = 0 (3)

Since L can be eliminated algebraically, Equation (3) may be solved immediately for V:

V = 1

2W = 12 340 N( ) = 170 N

Substituting this result into Equation (2) gives

P W W

P W

sin

sin sin

θ

θ

+ − =

= =°=

12 0

2340

2 39270 N N

Substituting this result into Equation (1) yields

W2sinθ

⎛⎝⎜

⎞⎠⎟

cosθ − H = 0

H = W2 tanθ

= 340 N2tan39°

= 210 N

Beam

θ

Brace

P

W

V H

Pin (rotational axis)

θ


22. REASONING Since the man holds the ball motionless, the ball and the arm are in equilibrium. Therefore, the net force, as well as the net torque about any axis, must be zero.

SOLUTION Using Equation 9.1, the net torque about an axis through the elbow joint is

Στ = M(0.0510 m) – (22.0 N)(0.140 m) – (178 N)(0.330 m) = 0

Solving this expression for M gives M = 31.21 10 N× . The net torque about an axis through the center of gravity is

Στ = – (1210 N)(0.0890 m) + F(0.140 m) – (178 N)(0.190 m) = 0

Solving this expression for F gives F = 31.01 10 N× . Since the forces must add to give a net force of zero, we know that the direction of F is downward .

23. SSM REASONING The drawing shows the forces acting on the

board, which has a length L. The ground exerts the vertical normal force V on the lower end of the board. The maximum force of static friction has a magnitude of µsV and acts horizontally on the lower end of the board. The weight W acts downward at the board's center. The vertical wall applies a force P to the upper end of the board, this force being perpendicular to the wall since the wall is smooth (i.e., there is no friction along the wall). We take upward and to the right as our positive directions. Then, since the horizontal forces balance to zero, we have

µsV – P = 0 (1)

The vertical forces also balance to zero giving

V – W = 0 (2) Using an axis through the lower end of the board, we express the fact that the torques

balance to zero as

PL sin θ −W L2

⎛ ⎝

⎞ ⎠ cos θ = 0 (3)

Equations (1), (2), and (3) may then be combined to yield an expression for θ . SOLUTION Rearranging Equation (3) gives

tan θ = W2P

(4)


But, P = µsV according to Equation (1), and W = V according to Equation (2). Substituting these results into Equation (4) gives

tan θ = V

2µsV=

12µs

Therefore,

θ = tan−1 12µs

⎛ ⎝ ⎜ ⎞

⎠ ⎟ = tan−1 1

2(0.650)⎡ ⎣ ⎢

⎤ ⎦ ⎥ = 37.6°

24. REASONING When the wheel is resting on the ground it is in equilibrium, so the sum of

the torques about any axis of rotation is zero ( )0τΣ = . This equilibrium condition will provide us with a relation between the magnitude of F and the normal force that the ground exerts on the wheel. When F is large enough, the wheel will rise up off the ground, and the normal force will become zero. From our relation, we can determine the magnitude of F when this happens.

SOLUTION The free body diagram shows the forces acting on the wheel: its weight W, the normal force FN, the horizontal force F, and the force FE that the edge of the step exerts on the wheel. We select the axis of rotation to be at the edge of the step, so that the torque produced by FE is zero. Letting

N W F, , and l l l represent the lever arms for the forces FN, W, and F, the sum of the torques is

Στ = −FN r2 − r − h( )2

N

+ W r2 − r − h( )2

W

− F r − h( )

F

= 0

Solving this equation for F gives

( ) ( )22NW F r r h

Fr h

− − −=

−

When the bicycle wheel just begins to lift off the ground the normal force becomes zero (FN = 0 N). When this happens, the magnitude of F is

( ) ( ) ( ) ( ) ( )2 2 22N 25.0 N 0 N 0.340 m 0.340 m 0.120 m

29 N0.340 m 0.120 m

W F r r hF

r h− − − − − −

= = =− −

W

r

F

h

FN

FE

Axis of Rotation


25. SSM REASONING The following drawing shows the beam and the five forces that act

on it: the horizontal and vertical components Sx and Sy that the wall exerts on the left end of the beam, the weight Wb of the beam, the force due to the weight Wc of the crate, and the tension T in the cable. The beam is uniform, so its center of gravity is at the center of the beam, which is where its weight can be assumed to act. Since the beam is in equilibrium, the sum of the torques about any axis of rotation must be zero ( )0τΣ = , and the sum of the

forces in the horizontal and vertical directions must be zero ( )0, 0x yF FΣ = Σ = . These three

conditions will allow us to determine the magnitudes of Sx, Sy, and T.

SOLUTION a. We will begin by taking the axis of rotation to be at the left end of the beam. Then the torques produced by Sx and Sy are zero, since their lever arms are zero. When we set the sum of the torques equal to zero, the resulting equation will have only one unknown, T, in it. Setting the sum of the torques produced by the three forces equal to zero gives (with L equal to the length of the beam)

( ) ( ) ( )1b c2 cos30.0 cos30.0 sin80.0 0W L W L T LτΣ = − ° − ° + ° =

Algebraically eliminating L from this equation and solving for T gives

( ) ( )

( )( ) ( )( )

1b c2

12

cos30.0 cos30.0

sin80.0

1220 N cos30.0 1960 N cos30.02260 N

sin80.0

W WT

° + °=

°

° + °= =

°

Axis

T

Wc

+y

+x

+τ

Wb

Sx

Sy

50.0°

30.0°

30.0°


b. Since the beam is in equilibrium, the sum of the forces in the vertical direction is zero:

b c sin50.0 0y yF S W W TΣ = + − − + ° =

Solving for Sy gives

( )b c sin50.0 N + 1960 N 2260 N sin50.0 1450 NyS W W T= + − ° =1220 − ° =

The sum of the forces in the horizontal direction must also be zero:

cos50.0 0x xF S TΣ = + − ° = so that

( )cos50.0 2260 N cos50.0 NxS T= ° = ° = 1450

26. REASONING Since the outstretched leg is stationary, it is in equilibrium. Thus, the net external torque acting on the leg must equal zero. This net torque arises because of the quadriceps force M and the weight W of the leg. The magnitude τ of each torque can be expressed as Fτ = l (Equation 9.1), where F is the magnitude of the force and l is the lever arm for the force.

SOLUTION At equilibrium the net torque τΣ about an axis through the knee joint is zero,

so that muscle weight 0τ τ τΣ = + = (1)

Using Equation 9.1 to express each torque, we have

muscle muscle weight weight and M Wτ τ= =l l

The torque muscleτ due to M acts counterclockwise and, therefore, is positive. The torque

weightτ due to W acts clockwise and, therefore, is negative. Thus, Equation (1) becomes

( ) ( ) weightmuscle weight

muscle0 or M W M Wτ

⎛ ⎞Σ = + − = = ⎜ ⎟⎜ ⎟⎝ ⎠

ll l

l

The lever arm l is the distance between the line of action of the force and the axis of

rotation, measured on a line that is perpendicular to both. With this in mind, we choose an axis for rotation that is at the knee joint (see the drawing that accompanies the problem statement) and can write the expression for M as follows:

( ) ( )( )

weight

muscle

0.250 m cos30.044.5 N 228 N

0.100 m sin 25.0M W

⎛ ⎞ °= = =⎜ ⎟⎜ ⎟ °⎝ ⎠

ll

______________________________________________________________________________


27. REASONING If we assume that the system is in equilibrium, we know that the vector sum

of all the forces, as well as the vector sum of all the torques, that act on the system must be zero.

The figure below shows a free body diagram for the boom. Since the boom is assumed to be

uniform, its weight BW is located at its center of gravity, which coincides with its geometrical center. There is a tension T in the cable that acts at an angle θ to the horizontal, as shown. At the hinge pin P, there are two forces acting. The vertical force V that acts on the end of the boom prevents the boom from falling down. The horizontal force H that also acts at the hinge pin prevents the boom from sliding to the left. The weight LW of the wrecking ball (the "load") acts at the end of the boom.

By applying the equilibrium conditions to the boom, we can determine the desired forces. SOLUTION The directions upward and to the right will be taken as the positive directions.

In the x direction we have cos 0xF H T θ= − =∑ (1)

while in the y direction we have

L Bsin 0yF V T W Wθ= − − − =∑ (2) Equations (1) and (2) give us two equations in three unknown. We must, therefore, find a

third equation that can be used to determine one of the unknowns. We can get the third equation from the torque equation.

In order to write the torque equation, we must first pick an axis of rotation and determine the

lever arms for the forces involved. Since both V and H are unknown, we can eliminate them from the torque equation by picking the rotation axis through the point P (then both V and H have zero lever arms). If we let the boom have a length L, then the lever arm for LW is cosL φ , while the lever arm for BW is ( / 2)cosL φ . From the figure, we see that the lever


arm for T is sin( – )L φ θ . If we take counterclockwise torques as positive, then the torque equation is

( )B Lcos cos sin 02

LW W L TLφτ φ φ θ⎛ ⎞= − − + − =∑ ⎜ ⎟⎝ ⎠

Solving for T, we have

1

B L2 cossin( – )W W

T φφ θ+

= (3)

a. From Equation (3) the tension in the support cable is

1

42 (3600 N) 4800 Ncos 48 1.6 10 N

sin(48 – 32 )T

+= ° = ×

° °

b. The force exerted on the lower end of the hinge at the point P is the vector sum of the

forces H and V. According to Equation (1), ( )4 4cos 1.6 10 N cos 32 1.4 10 NH T θ= = × ° = ×

and, from Equation (2)

( )4 4sin 4800 N 3600 N 1.6 10 N sin 32 1.7 10 NL BV W W T θ= + + = + + × ° = × Since the forces H and V are at right angles to each other, the magnitude of their vector sum

can be found from the Pythagorean theorem:

2 2 4 2 4 2 4

P (1.4 10 N) (1.7 10 N) 2.2 10 NF H V= + = × + × = ×

28. REASONING Although the crate is in

translational motion, it undergoes no angular acceleration. Therefore, the net torque acting on the crate must be zero: 0τΣ = (Equation 9.2). The four forces acting on the crate appear in the free-body diagram: its weight W, the kinetic friction force fk, the normal force FN, and the tension T in the strap. We will take the edge of the crate sliding along the floor as the rotation axis for applying Equation 9.2. Both the friction force and the normal force act at this point. These two forces, therefore, generate no torque about the axis, so the clockwise torque of the weight W must balance the counterclockwise torque of the tension T in the strap:

T

FN fk

W

L/2

H/2 d

θ

Free-body diagram of the crate


W WT W

T T or

W mgT W T= = =

l ll l

l l (1)

We will apply trigonometry to determine the lever arms Wl and Tl for the weight and the tension, respectively,

and then calculate the magnitude T of the tension in the strap.

SOLUTION The lever arm Wl of the crate’s weight is shown in the diagram at the right, and is given by

( )W cos 25d θ= + ol , where d is the distance between

the axis of rotation (lower edge of the crate) and the crate’s center of gravity, and θ is the angle between that line and the bottom of the crate. The right triangle in the free-body diagram of the crate (drawing on the left) shows how we can use trigonometry to determine both the length d and the angle θ. The length d is the hypotenuse of that right triangle, and the other sides are the half-height (H/2 = 0.20 m) and half-length (L/2 = 0.45 m) of the crate, so by the Pythagorean theorem (Equation 1.7) we find that

( ) ( )2 2

2 20.20 m 0.45 m 0.49 m2 2H Ld ⎛ ⎞ ⎛ ⎞= + = + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

We can find the angle θ from the inverse tangent function:

1 2tan Hθ −=2L

1 0.40 mtan 240.90 m

−⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠o

The lever arm Tl of the tension force is illustrated in the drawing at the right, where we see that ( )T sin 61 25 sin36L L= − =o o ol .

Therefore, from Equation (1), the magnitude of the tension in the strap is

( )

( )( )( )( )

W

T

2

cos 24 25

sin 36

72 kg 9.80 m/s 0.49 m cos49

0.90 m sin 36

430 N

mgdmgT

L

+= =

=

=

o o

o

o

o

ll

25°

T

TlT

L

61°

36°

Lever arm of the tension force

d

wlT

W θ 25°

Lever arm of the crate’s weight


29. SSM REASONING The drawing shows the forces acting on the board, which has a length L. Wall 2 exerts a normal force P2 on the lower end of the board. The maximum force of static friction that wall 2 can apply to the lower end of the board is µsP2 and is directed upward in the drawing. The weight W acts downward at the board's center. Wall 1 applies a normal force P1 to the upper end of the board. We take upward and to the right as our positive directions.

SOLUTION Then, since the horizontal forces balance to zero, we have P1 – P2 = 0 (1) The vertical forces also balance to zero:

µsP2 – W = 0 (2) Using an axis through the lower end of the board, we now balance the torques to zero:

( ) ( )1cos sin 02LW PLθ θ⎛ ⎞ − =⎜ ⎟⎝ ⎠

(3) Rearranging Equation (3) gives

1tan

2WP

θ = (4)

But W = µsP2 according to Equation (2), and P2 = P1 according to Equation (1). Therefore,

W = µsP1, which can be substituted in Equation (4) to show that

s 1 s

1

0.98tan2 2 2PP

µ µθ = = =

or θ = tan–1(0.49) = 26°

From the drawing at the right,

cos dL

θ =

Therefore, the longest board that can be propped between

the two walls is

1.5 m 1.7 mcos cos 26dLθ

= = =°

d

θ

L

θ

P 1

P 2

P 2 µ s W

Wall 1 Wall 2 1


30. REASONING AND SOLUTION The weight W of the left side of the ladder, the normal

force FN of the floor on the left leg of the ladder, the tension T in the crossbar, and the reaction force R due to the right-hand side of the ladder, are shown in the following figure. In the vertical direction −W + FN = 0, so that

FN = W = mg = (10.0 kg)(9.80 m/s2) = 98.0 N

In the horizontal direction it is clear that R = T. The net

torque about the base of the ladder is Στ = – T [(1.00 m) sin 75.0o] – W [(2.00 m) cos 75.0o] + R [(4.00 m) sin 75.0o] = 0 Substituting for W and using R = T, we obtain

( )( )

( )9.80 N 2.00 m cos75.0 17.5 N

3.00 m sin 75.0T °= =

°

31. REASONING The net torque Στ acting on the CD is given by Newton’s second law for

rotational motion (Equation 9.7) as Στ = Ι α, where I is the moment of inertia of the CD and α is its angular acceleration. The moment of inertia can be obtained directly from Table 9.1, and the angular acceleration can be found from its definition (Equation 8.4) as the change in the CD’s angular velocity divided by the elapsed time.

SOLUTION The net torque is Στ = Ι α. Assuming that the CD is a solid disk, its moment of

inertia can be found from Table 9.1 as 212 ,I MR= where M and R are the mass and radius

of the CD. Thus, the net torque is ( )212I MRτ α αΣ = =

The angular acceleration is given by Equation 8.4 as ( )0 / tα ω ω= − , where ω and ω0 are the final and initial angular velocities, respectively, and t is the elapsed time. Substituting this expression for α into Newton’s second law yields

( ) ( )

( )( )

2 2 01 12 2

23 2 412

21 rad/s 0 rad/s17 10 kg 6.0 10 m 8.0 10 N m0.80 s

MR MRt

ω ωτ α

− − −

−⎛ ⎞Σ = = ⎜ ⎟

⎝ ⎠

−⎛ ⎞⎡ ⎤= × × = × ⋅⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠

W

R

T F N

75.0º


32. REASONING For a rigid body rotating about a fixed axis, Newton's second law for rotational motion is given as Iτ α=∑ (Equation 9.7), where I is the moment of inertia of the body and α is the angular acceleration. In using this expression, we note that α must be expressed in rad/s2.

SOLUTION Equation 9.7 gives

22

10.0 N m 1.25 kg m8.00 rad/s

I τα

⋅= = = ⋅∑

______________________________________________________________________________ 33. REASONING The moment of inertia of the stool is the sum of the individual moments of

inertia of its parts. According to Table 9.1, a circular disk of radius R has a moment of inertia of 21

disk disk2I M R= with respect to an axis perpendicular to the disk center. Each thin rod is attached perpendicular to the disk at its outer edge. Therefore, each particle in a rod is located at a perpendicular distance from the axis that is equal to the radius of the disk. This means that each of the rods has a moment of inertia of Irod = Mrod R

2. SOLUTION Remembering that the stool has three legs, we find that the its moment of

inertia is

Istool = Idisk + 3Irod = 12 MdiskR2 + 3Mrod R2

= 12 1.2 kg( ) 0.16 m( )2 + 3 0.15 kg( ) 0.16 m( )2 = 0.027 kg ⋅m2

34. REASONING According to Newton’s second law for rotational motion, Στ = Iα, the

angular acceleration α of the blades is equal to the net torque Στ applied to the blades divided by their total moment of inertia I, both of which are known.

SOLUTION The angular acceleration of the fan blades is

22

1.8 N m 8.2 rad/s0.22 kg mI

τα Σ ⋅= = =⋅

(9.7)

35. REASONING Newton’s second law for rotational motion (Equation 9.2) indicates that the

net external torque is equal to the moment of inertia times the angular acceleration. To determine the angular acceleration, we will use Equation 8.7 from the equations of rotational kinematics. This equation indicates that the angular displacement θ is given by

210 2t tθ ω α= + , where ω0 is the initial angular velocity, t is the time, and α is the angular

acceleration. Since both wheels start from rest, ω0 = 0 rad/s for each. Furthermore, each


wheel makes the same number of revolutions in the same time, so θ and t are also the same for each. Therefore, the angular acceleration α must be the same for each.

SOLUTION Using Equation 9.2, the net torque τΣ that acts on each wheel is given by Iτ αΣ = , where I is the moment of inertia and α is the angular acceleration. Solving

Equation 8.7 for the angular acceleration ( )0

22 t

tθ ω

α⎡ ⎤−

=⎢ ⎥⎢ ⎥⎣ ⎦

and substituting the result into

Equation 9.2 gives ( )0

22 t

I It

θ ωτ α

⎡ ⎤−Σ = = ⎢ ⎥

⎢ ⎥⎣ ⎦

Table 9.1 indicates that the moment of inertia of a hoop is I MRhoop =

2 , while the moment

of inertia of a disk is I MRdisk =12

2 . The net external torques acting on the hoop and the disk are:

( )

( )( ) ( ) ( )( )( )

02hoop 2

22

2

2 13 rad 2 0 rad/s 8.0 s4.0 kg 0.35 m 0.20 N m

8.0 s

tI MR

tθ ω

τ α⎡ ⎤−

Σ = = ⎢ ⎥⎢ ⎥⎣ ⎦

⎡ ⎤−⎢ ⎥= = ⋅⎢ ⎥⎣ ⎦

Hoop

( )

( )( ) ( ) ( )( )( )

021disk 2 2

212 2

2

2 13 rad 2 0 rad/s 8.0 s4.0 kg 0.35 m 0.10 N m

8.0 s

tI MR

tθ ω

τ α⎡ ⎤−

Σ = = ⎢ ⎥⎢ ⎥⎣ ⎦

⎡ ⎤−⎢ ⎥= = ⋅⎢ ⎥⎣ ⎦

Disk

36. REASONING The ladder is subject to

three vertical forces: the upward pull P of the painter on the top end of the ladder, the upward normal force FN that the ground exerts on the bottom end of the ladder, and the downward force W of the ladder’s weight, which acts at the ladder’s center of gravity, halfway between the ends (see the free-body

P

W

FN

Free-body diagram of the ladder

Axis of rotation


diagram of the ladder). The bottom end serves as the axis of rotation. The normal force FN is applied at the axis, so it has no lever arm. The net torque acting on the ladder can be obtained with the aid of Equation 9.1:

Στ = PP

Torque due to P

− WW

Torque due to W

+ FN 0( )

Torque due to FN

= PL−W 1

2L( ) = PL− 1

2mgL (1)

where L is the length of the ladder and m is its mass.

Once we know the net torque τΣ acting on the ladder, we will use Newton’s second law for rotation, Iτ αΣ = (Equation 9.7) to determine the moment of inertia I of the ladder.

SOLUTION a. From Equation (1), the net torque acting on the ladder is

( )( ) ( )( )( )21 12 2245 N 9.75 m 23.2 kg 9.80 m/s 9.75 m 1280 N mPL mgLτΣ = − = − = ⋅

b. The ladder’s moment of inertia is found from Iα τ= Σ (Equation 9.7). Using the result found in part a, we obtain

22

1280 N m 711 kg m1.80 rad/s

I ταΣ ⋅= = = ⋅

37. SSM REASONING a. The angular acceleration α is defined as the change, ω − ω0, in the angular velocity

divided by the elapsed time t (see Equation 8.4). Since all these variables are known, we can determine the angular acceleration directly from this definition.

b. The magnitude τ of the torque is defined by Equation 9.1 as the product of the magnitude

F of the force and the lever arm l . The lever arm is the radius of the cylinder, which is known. Since there is only one torque acting on the cylinder, the magnitude of the force can be obtained by using Newton’s second law for rotational motion, .F Iτ αΣ = =l

SOLUTION a. From Equation 8.4 we have that α = (ω – ω0)/t. We are given that ω0 = 76.0 rad/s,

102 38.0 rad/sω ω= = , and t = 6.40 s, so

20 38.0 rad/s 76.0 rad/s = 5.94 rad/s

6.40 stω ω

α− −= = −

The magnitude of the angular acceleration is 25.94 rad/s .


b. Using Newton’s second law for rotational motion, we have that .F Iτ αΣ = =l Thus, the magnitude of the force is

( )( )2 20.615 kg m 5.94 rad/s44.0 N

0.0830 mIF α ⋅

= = =l

38. REASONING The angular acceleration α that results from the application of a net external

torque Στ to a rigid object with a moment of inertia I is given by Newton’s second law for

rotational motion: Iτα Σ= (Equation 9.7). By applying this equation to each of the two

situations described in the problem statement, we will be able to obtain the unknown angular acceleration.

SOLUTION The drawings at the right illustrate the

two types of rotational motion of the object. In applying Newton’s second law for rotational motion, we need to keep in mind that the moment of inertia depends on where the axis is. For axis 1 (see top drawing), piece B is rotating about one of its ends, so according to Table 9.1, the moment of inertia is

1 2for axis 1 B B3I M L= , where MB and LB are the mass and length of piece B, respectively. For axis 2 (see bottom drawing), piece A is rotating about an axis through its midpoint. According to Table 9.1 the moment of inertia is 1 2

for axis 2 A A12I M L= . Applying

the second law for each of the axes, we obtain

for axis 1 for axis 2for axis 1 for axis 2

and I I

τ τα αΣ Σ= =

As given, the net torque Στ is the same in both expressions. Dividing the expression for axis 2 by the expression for axis 1 gives

for axis 2 for axis 2 for axis 1

for axis 1 for axis 2

for axis 1

I II

I

τα

τα

Σ

= =Σ

Using the expressions from Table 9.1 for the moments of inertia, this result becomes

1 2 2B Bfor axis 2 for axis 1 3 B B

1 22for axis 1 for axis 2 A AA A12

4M LI M LI M LM L

αα

= = =

Noting that A B A B2 and 2M M L L= = , we find that

Axis 1

B

A

Axis 2


( )2 2

for axis 2 B B B B2 2

for axis 1 A A B B

22

for axis 2 for axis 1

4 4 122 2

1 4.6 rad/s 2.3 rad/s2 2

M L M LM L M L

αα

α α

= = =

= = =

39. SSM REASONING The figure below shows eight particles, each one located at a different

corner of an imaginary cube. As shown, if we consider an axis that lies along one edge of the cube, two of the particles lie on the axis, and for these particles r = 0. The next four particles closest to the axis have r = l , where l is the length of one edge of the cube. The remaining two particles have r d= , where d is the length of the diagonal along any one of the faces. From the Pythagorean theorem, d = + = 2 2 2 .

According to Equation 9.6, the moment of inertia of a system of particles is given by 2I mr=∑ .

SOLUTION Direct application of Equation 9.6 gives

2 2 2 2 2 24( ) 2( ) 4( ) 2(2 ) 8I mr m md m m m= = + = + =∑ l l l l or

2 28(0.12 kg)(0.25 m) 0.060 kg mI = = ⋅ 40. REASONING AND SOLUTION The final angular speed of the arm is ω = vT/r, where

r = 0.28 m. The angular acceleration needed to produce this angular speed is α = (ω − ω0)/t. The net torque required is Στ = Iα. This torque is due solely to the force M, so that Στ = ML. Thus,

0It

ML L

ω ωτ

−⎛ ⎞⎜ ⎟∑ ⎝ ⎠= =


Setting ω0 = 0 rad/s and ω = vT/r, the force becomes

( )( )( )( )( )

T

T

20.065 kg m 5.0 m/s460 N

0.025 m 0.28 m 0.10 s

vI

I vr tM

L Lrt

⎛ ⎞⎜ ⎟ ⋅⎝ ⎠= = = =

41. SSM REASONING The drawing shows the two identical sheets and the axis of rotation

for each.

The time t it takes for each sheet to reach its final angular velocity depends on the angular acceleration α of the sheet. This relation is given by Equation 8.4 as ( )0 /t ω ω α= − , where

ω and ω0 are the final and initial angular velocities, respectively. We know that ω0 = 0 rad/s in each case and that the final angular velocities are the same. The angular acceleration can be determined by using Newton’s second law for rotational motion, Equation 9.7, as

/ Iα τ= , where τ is the torque applied to a sheet and I is its moment of inertia. SOLUTION Substituting the relation / Iα τ= into ( )0 /t ω ω α= − gives

( ) ( ) ( )0 0 0It

I

ω ω ω ω ω ωτα τ

− − −= = =

⎛ ⎞⎜ ⎟⎝ ⎠

The time it takes for each sheet to reach its final angular velocity is:

( ) ( )Right 0Left 0

Left Right and II

t tω ωω ω

τ τ−−

= =

The moments of inertia I of the left and right sheets about the axes of rotation are given by

the following relations, where M is the mass of each sheet (see Table 9.1 and the drawings above): 21

Left 13I ML= and 21Right 23 .I ML= Note that the variables M, ω, ω0, and τ are the

same for both sheets. Dividing the time-expression for the right sheet by that for the left sheet gives

( )

( )

Right 0 2 21Right Right 23 2

2 21Left LeftLeft 0 1 13

It I ML Lt II ML L

ω ωτω ωτ

−

= = = =−

Axis of Rotation

L2 = 0.20 m L1 = 0.40 m


Solving this expression for tRight yields

( ) ( )( )

222

Right Left 2 21

0.20 m8.0 s 2.0 s

0.40 m

Lt t

L

⎛ ⎞= = =⎜ ⎟⎜ ⎟⎝ ⎠

42. REASONING The time t it takes to completely unwind the hose from the reel is related to

the reel’s angular displacement θ and its angular acceleration α by 210 2t tθ ω α= +

(Equation 8.7). The reel is initially at rest, so ω0 = 0 rad/s. Substituting this into 21

0 2t tθ ω α= + , and solving for the elapsed time t, we obtain

( ) 2 21 12 2

20 rad/s or t t t t θθ α αα

= + = =

(1)

The hose unwinds from the reel without slipping, so the total arc length s traversed by a point on the reel’s rim is equal to the total length L of the hose. The reel’s angular displacement θ, therefore, is related to the length of the hose and the radius R of the reel by s L Rθ= = (Equation 8.1). Thus, the reel’s angular displacement is given by θ = L/R, so that Equation (1) becomes

2 LRtα

⎛ ⎞⎜ ⎟⎝ ⎠=

(2)

The angular acceleration α of the reel depends upon the net torque and the reel’s moment inertia I via Iα τ= Σ (Equation 9.7).

SOLUTION Assuming that the hose unwinds in the counterclockwise direction, the torque exerted on the reel by the tension in the hose is positive. Using Equation 9.1, we have

TRτ = , where T is the magnitude of the tension in the hose and R is the radius of the reel. This torque is opposed by the frictional torque τf , so the net torque on the reel is

fTRτ τΣ = − . Thus, the angular acceleration of the reel is

fTRI I

ττα−Σ= = (9.7)

Substituting Equation 9.7 into Equation (2) now yields the elapsed time:

( )( )( )

( ) ( )( )

2

f f

2 2 15.0 m 0.44 kg m2 12 s0.160 m 25.0 N 0.160 m 3.40 N m

LLIRt TR R TR

Iτ τ

⎛ ⎞⎜ ⎟ ⋅⎝ ⎠= = = =− − − ⋅⎡ ⎤⎣ ⎦

43. REASONING


a. The moment of inertia for the three-ball system is 2 2 21 1 2 2 3 3I m r m r m r= + + (Equation 9.6),

where m1, m2, and m3 are the masses of the balls and r1, r2, and r3 are the distances from the axis. In system A, the ball whose mass is m1 does not contribute to the moment of inertia, because the ball is located on the axis and r1 = 0 m. In system B, the ball whose mass is m3 does not contribute to the moment of inertia, because it is located on the axis and r3 = 0 m.

b. The magnitude of the torque is equal to the magnitude F of the force times the lever

arm l (see Equation 9.1). In system A the lever arm is 3.00 m.=l In B the lever arm is 0 m,=l since the line of action of the force passes through the axis of rotation.

c. According to Newton’s second law for rotational motion, Equation 9.7, the angular

acceleration α is given by ( ) / ,Iα τ= Σ where τΣ is the net torque and I is the moment of inertia. The angular velocity ω is given by Equation 8.4 as ω = ω0 + α t, where ω0 is the initial angular velocity and t is the time.

SOLUTION

a. The moment of inertia for each system is System A

( )( ) ( )( ) ( )( )

2 2 21 1 2 2 3 3

2 2 2 29.00 kg 0 m + 6.00 kg 3.00 m + 7.00 kg 5.00 m = 229 kg m

I m r m r m r= + +

= ⋅

System B

( )( ) ( )( ) ( )( )

2 2 21 1 2 2 3 3

2 2 2 29.00 kg 5.00 m + 6.00 kg 4.00 m + 7.00 kg 0 m = 321 kg m

I m r m r m r= + +

= ⋅

b. A torque that tends to produce a counterclockwise rotation about the axis is a positive torque. The torque produced by the force has a magnitude that is equal to the product of the force magnitude and the lever arm:

The torque is negative because it produces a clockwise rotation about the axis.

c. The final angular velocity ω is related to the initial angular velocity ω0, the angular acceleration α, and the time t by 0 tω ω α= + (Equation 8.4). The angular acceleration is

given by Newton’s second law for rotational motion as ( ) / Iα τ= Σ (Equation 9.7), where τΣ is the net torque and I is the moment of inertia. Since there is only one torque acting on

System A ( )( )424 N 3.00 m 1270 N mFτ = − = − = − ⋅l

System B ( )( )424 N 0 m 0 N mFτ = = = ⋅l


each system, it is the net torque, so .τ τΣ = Substituting this expression for α into Equation 8.4 yields

0 0t tIτω ω α ω ⎛ ⎞= + = + ⎜ ⎟⎝ ⎠

In both cases the initial angular velocity is ω0 = 0 rad/s, since the systems start from rest.

The final angular velocities after 5.00 s are:

44. REASONING Each door starts from rest, so that it is initial angular velocity is 0 0.00 rad / sω = . The angle θ through which each door turns in a time t is the same. Due to

the torque created by the applied force, each door has an angular acceleration α. The variables ω0, θ, t, and α are related according to 1 2

0 2t tθ ω α= + (Equation 8.7), which is one

of the equations of rotational kinematics. The angular acceleration α is related to the torque τ created by the applied force via Newton’s second law for rotation about a fixed axis

Iτ τ αΣ = = (Equation 9.7), where I is the moment of inertia of the door. In addition, the magnitude τ of the torque can be expressed as Fτ = l (Equation 9.1), where F is the magnitude of the applied force and l is the lever arm for the force.

SOLUTION Since 0 0.00 rad / sω = for each door, Equation 8.7 becomes

1 12 20 2 2t t tθ ω α α= + = (8.7)

Using Fτ = l in Newton’s second law for rotation, we obtain

or FF II

τ α α= = = ll

Substituting this result for α into Equation 8.7, we have

1 12 22 2

2 or F It t tI F

θθ α ⎛ ⎞= = =⎜ ⎟⎝ ⎠l

l

We can now apply this result for the time t to both doors:

System A ( ) ( )0 21270 N m0 rad/s + 5.00 s 27.7 rad/s

229 kg mt

Iτω ω

⎛ ⎞− ⋅⎛ ⎞= + = = −⎜ ⎟⎜ ⎟ ⋅⎝ ⎠ ⎝ ⎠

System B ( ) ( )0 20 N m0 rad/s + 5.00 s 0 rad/s

321 kg mt

Iτω ω

⎛ ⎞⋅⎛ ⎞= + = =⎜ ⎟⎜ ⎟ ⋅⎝ ⎠ ⎝ ⎠


A BA B

A B

2 2 and

I It t

F Fθ θ

= =l l

In these expressions for tA and tB, the angle θ and the force magnitude F are the same, so

dividing tB by tA gives

B

BB B A

A B AA

A

2

2

IFt I

t IIF

θ

θ= =

l ll

l

(1)

According to Table 9.1 in the text, 1 2

A 3I ML= and 1 2

B 12I ML= , where M and L are the

mass and width of each door, respectively. In addition, the lever arms are A L=l and 1

B 2L=l . Substituting these values into Equation (1) gives

( )( )

( )( )1 2

12B B A AB1 1 2A B A 2 3

1 3.00 s or 2.12 s2 2 2

ML Lt I tt

t I L ML= = = = = =

ll

______________________________________________________________________________ 45. SSM REASONING The angular acceleration of the bicycle wheel can be calculated from

Equation 8.4. Once the angular acceleration is known, Equation 9.7 can be used to find the net torque caused by the brake pads. The normal force can be calculated from the torque using Equation 9.1.

SOLUTION The angular acceleration of the wheel is, according to Equation 8.4,

20 3.7 rad/s 13.1 rad/s 3.1 rad/s3.0 st

ω ωα

− −= = = − If we assume that all the mass of the wheel is concentrated in the rim, we may treat the

wheel as a hollow cylinder. From Table 9.1, we know that the moment of inertia of a hollow cylinder of mass m and radius r about an axis through its center is 2I mr= . The net torque that acts on the wheel due to the brake pads is, therefore,

2( )I mrτ α α= =∑ (1)

From Equation 9.1, the net torque that acts on the wheel due to the action of the two brake

pads is k–2 fτ =∑ l (2)


where fk is the kinetic frictional force applied to the wheel by each brake pad, and 0.33 m=l is the lever arm between the axle of the wheel and the brake pad (see the

drawing in the text). The factor of 2 accounts for the fact that there are two brake pads. The minus sign arises because the net torque must have the same sign as the angular acceleration. The kinetic frictional force can be written as (see Equation 4.8)

k k Nf Fµ= (3)

where µk is the coefficient of kinetic friction and FN is the magnitude of the normal force

applied to the wheel by each brake pad. Combining Equations (1), (2), and (3) gives

2k N–2( ) ( )F mrµ α=l

2 2 2

Nk

– –(1.3 kg)(0.33 m) (–3.1 rad/s ) 0.78 N2 2(0.85)(0.33 m)mrF αµ

= = =l

46. REASONING The parallel axis theorem states that the moment of inertia I about an

arbitrary axis is given by 2

cmI I Mh= + (1) where Icm is the moment of inertia relative to an axis that passes through the center of mass

and is parallel to the axis of interest, M is the total mass of the object (the solid cylinder in this case), and h is the distance between the two axes. The axis of interest here is the axis that lies on the surface of the cylinder and is perpendicular to its circular ends. Thus, Icm is the moment of inertia relative to an axis that passes through the center of mass and is perpendicular to the cylinder’s circular ends, and referring to Table 9.1 in the text, we see that 1 2

cm 2I MR= , where R is the radius of the cylinder. Note that in Equation (1) the

distance h between these two axes is just R, the radius of the cylinder. SOLUTION According to the parallel axis theorem as stated in Equation (1), the moment

of inertia of the cylinder relative to an axis that lies on the surface of the cylinder and is perpendicular to its circular ends is

1 32 2 2 2

cm 2 2I I Mh MR MR MR= + = + =

47. REASONING The following drawing shows the drum, pulley, and the crate, as well as the

tensions in the cord


r1

r2

Drum m1 = 150 kg r1 = 0.76 m

Pulley m2 = 130 kg

Crate m3 = 180 kg

REASONI

T1

T2 T1

T2

Let T1 represent the magnitude of the tension in the cord between the drum and the pulley.

Then, the net torque exerted on the drum must be, according to Equation 9.7, Στ = I1α1, where I1 is the moment of inertia of the drum, and α1 is its angular acceleration. If we assume that the cable does not slip, then Equation 9.7 can be written as

−T1r1 + τ

τ∑

= m1r12( )

I1

ar1

⎛

⎝⎜

⎞

⎠⎟

α1

(1)

where τ is the counterclockwise torque provided by the motor, and a is the acceleration of

the cord (a = 1.2 m/s2). This equation cannot be solved for τ directly, because the tension T1 is not known.

We next apply Newton’s second law for rotational motion to the pulley in the drawing:

+T1r2 − T2r2τ∑

= 1

2 m2r22( )

I2

ar2

⎛

⎝⎜

⎞

⎠⎟

α2

(2)

where T2 is the magnitude of the tension in the cord between the pulley and the crate, and I2

is the moment of inertial of the pulley. Finally, Newton’s second law for translational motion (ΣFy = m a) is applied to the crate,

yielding + −

∑

=T m g

F

m a

y

2 3 3 (3)

SOLUTION Solving Equation (1) for T1 and substituting the result into Equation (2), then


solving Equation (2) for T2 and substituting the result into Equation (3), results in the following value for the torque

( )

( )

11 1 2 3 32

2 212(0.76 m) (1.2 m/s ) 150 kg + 130 kg +180 kg (180 kg)(9.80 m/s ) 1700 N m

r a m m m m gτ ⎡ ⎤= + + +⎣ ⎦

⎡ ⎤= + = ⋅⎣ ⎦

48. REASONING a. The kinetic energy is given by Equation 9.9 as KER = 1

22Iω . Assuming the earth to be a

uniform solid sphere, we find from Table 9.1 that the moment of inertia is I MR= 25

2 . The

mass and radius of the earth are M = 5.98 × 1024 kg and R = 6.38 × 106 m (see the inside of the text’s front cover). The angular speed ω must be expressed in rad/s, and we note that the earth turns once around its axis each day, which corresponds to 2π rad/day.

b. The kinetic energy for the earth’s motion around the sun can be obtained from

Equation 9.9 as KER = 12

2Iω . Since the earth’s radius is small compared to the radius of

the earth’s orbit (Rorbit = 1.50 × 1011 m, see the inside of the text’s front cover), the moment of inertia in this case is just I MR= orbit

2 . The angular speed ω of the earth as it goes around the sun can be obtained from the fact that it makes one revolution each year, which corresponds to 2π rad/year.

SOLUTION a. According to Equation 9.9, we have

KER = 12 Iω 2 = 1

225 MR2( )ω 2

= 12

25 5.98×1024 kg( ) 6.38×106 m( )2⎡

⎣⎢

⎤⎦⎥

2π rad1 day

⎛⎝⎜

⎞⎠⎟

1 day24 h

⎛⎝⎜

⎞⎠⎟

1 h3600 s

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

2

= 2.57 ×1029 J

b. According to Equation 9.9, we have


KER = 12 Iω

2 = 12 MRorbit

2( )ω 2

= 12 5.98 ×1024 kg( ) 1.50 ×1011 m( )2 2π rad

1 yr⎛⎝⎜

⎞⎠⎟

1 yr365 day

⎛⎝⎜

⎞⎠⎟

1 day24 h

⎛⎝⎜

⎞⎠⎟

1 h3600 s

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

2

= 2.67 ×1033 J

49. SSM REASONING AND SOLUTION a. The tangential speed of each object is given by Equation 8.9, v rT = ω . Therefore, For object 1: vT1 = (2.00 m)(6.00 rad/s) = 12.0 m / s For object 2: vT2 = (1.50 m)(6.00 rad/s) = 9.00 m / s For object 3: vT3 = (3.00 m)(6.00 rad/s) = 18.0 m / s b. The total kinetic energy of this system can be calculated by computing the sum of the

kinetic energies of each object in the system. Therefore,

KE = + +12 1 1

2 12 2 2

2 12 3 3

2m v m v m v KE kg)(12.0 m / s) kg)(9.00 m / s) kg)(18.0 m / s) 1.08 2 2 2= + + = ×1

2 6 00 4 00 3 00 10( . ( . ( . 3 J

c. The total moment of inertia of this system can be calculated by computing the sum of the moments of inertia of each object in the system. Therefore,

I mr m r m r m r= = + +∑ 2

1 12

2 22

3 32

I = + + = ⋅( . ( . ( . .6 00 4 00 3 00 60 0 kg)(2.00 m) kg)(1.50 m) kg)(3.00 m) kg m2 2 2 2

d. The rotational kinetic energy of the system is, according to Equation 9.9,

KE (60.0 kg m )(6.00 rad / s) 1.08R2 2= = ⋅ = ×1

22 1

2 10Iω 3 J This agrees, as it should, with the result for part (b).


50. REASONING The kinetic energy of a rotating object is expressed as KER = 12

2Iω (Equation 9.9), where I is the object’s moment of inertia and ω is its angular speed. According to Equation 9.6, the moment of inertia for rod A is just that of the attached particle, since the rod itself is massless. For rod A with its attached particle, then, the moment of inertia is I MLA = 2 . According to Table 9.1, the moment of inertia for rod B is I MLB = 1

32 .

SOLUTION Using Equation 9.9 to calculate the kinetic energy, we find that

Rod A KER = 12 IAω

2 = 12 ML2( )ω 2

= 12 0.66 kg( ) 0.75 m( )2 4.2 rad/s( )2 = 3.3 J

Rod B KER = 12 IBω

2 = 12

13 ML2( )ω 2

= 16 0.66 kg( ) 0.75 m( )2 4.2 rad/s( )2 = 1.1 J

51. SSM REASONING The kinetic energy of the flywheel is given by Equation 9.9. The

moment of inertia of the flywheel is the same as that of a solid disk, and, according to Table 9.1 in the text, is given by I MR= 1

22 . Once the moment of inertia of the flywheel is

known, Equation 9.9 can be solved for the angular speed ω in rad/s. This quantity can then be converted to rev/min.

SOLUTION Solving Equation 9.9 for ω, we obtain,

ω = = = × = ×2 2

12

2

( ) ( )KE KE 4(1.2 10 J)(13 kg)(0.30 m)

6.4 10 rad / sR R9

24

I MR

Converting this answer into rev/min, we find that

ω = 6.4 × 104 rad/s( ) 1 rev

2π rad⎛⎝⎜

⎞⎠⎟

60 s1 min

⎛⎝⎜

⎞⎠⎟= 6.1×105 rev/min


52. REASONING Each blade can be approximated as a thin rod rotating about an axis perpendicular to the rod and passing through one end. The moment of inertia of a blade is given in Table 9.1 as 21

3ML , where M is the mass of the blade and L is its length. The total moment of inertia I of the two blades is just twice that of a single blade. The rotational kinetic energy KER of the blades is given by Equation 9.9 as 21

R 2KE Iω= , where ω is the angular speed of the blades.

SOLUTION

a. The total moment of inertia of the two blades is

( )( )22 2 2 21 1 2 23 3 3 3 240 kg 6.7 m 7200 kg mI ML ML ML= + = = = ⋅

b. The rotational kinetic energy is

( )( )22 2 61 1R 2 2KE 7200 kg m 44 rad/s 7.0 10 JIω= = ⋅ = ×

53. REASONING The rotational kinetic energy of a solid sphere is given by Equation 9.9

as 21R 2KE Iω= , where I is its moment of inertia and ω its angular speed. The sphere has

translational motion in addition to rotational motion, and its translational kinetic energy is 21

T 2KE mv= (Equation 6.2), where m is the mass of the sphere and v is the speed of its center of mass. The fraction of the sphere’s total kinetic energy that is in the form of rotational kinetic energy is KER/(KER + KET).

SOLUTION The moment of inertia of a solid sphere about its center of mass is 22

5I mR= , where R is the radius of the sphere (see Table 9.1). The fraction of the sphere’s total kinetic energy that is in the form of rotational kinetic energy is

( )( )

2 21 2 2 22 212 5 5R 2

2 2 2 2 21 1 2 2 2 21 2 1R T 2 2 52 5 2

KE=

KE KE

mR RII mv R vmR mv

ω ωωω ωω

= =+ + ++

Since the sphere is rolling without slipping on the surface, the translational speed v of the center of mass is related to the angular speed ω about the center of mass by v = Rω (see Equation 8.12). Substituting v = Rω into the equation above gives

( )

2 2 2 22 25 5R2 2 2 22 2 22R T 5 5

KE 2KE KE 7

R RR v R R

ω ωω ω ω

= = =+ + +

54. REASONING Only the conservative force of gravity does work on the objects, so the total mechanical energy is conserved as they move down the ramp. The total mechanical energy E at any height h above the zero level is


E = 12

mv2

Translationalkinetic energy

+ 1

2Iω 2

Rotationalkinetic energy

+ mgh

Gravitationalpotentialenergy

As the objects move down the ramp, potential energy is converted into kinetic energy.

However, the kinetic energy for the marble is shared between the translational form ( )1 22mv

and the rotational form ( )1 22Iω , whereas the kinetic energy for the cube is all translational.

Therefore, at the bottom of the ramp, the marble will have the smaller center-of-mass speed, because it will arrive there with less translational kinetic energy than the cube has. We expect, then, that the ratio of the center-of-mass speeds vcube/vmarble will be greater than one.

SOLUTION When applying energy conservation, we will assume that the zero level for

measuring the height h is located at the bottom of the ramp. As applied to the marble, energy conservation gives

12

mvmarble2 + 1

2Iωmarble

2

Total mechanical energyat bottom of ramp

= mgh

Total mechanical energyat top of ramp

(1)

In Equation (1) we have used the fact that the marble starts at rest at the top of the ramp.

Since the marble rolls without slipping, we know that marble marble /v rω = (Equation 8.12), where r is the radius of the marble. Referring to Table 9.1, we also know that the marble’s moment of inertia is 2 2

5I mr= . Substituting these two expressions into Equation (1) gives

( )2

1 1 22 2 marblemarble marble2 2 5

10 or 7

v ghmv mr mgh vr

⎛ ⎞+ = =⎜ ⎟⎝ ⎠

As applied to the cube, energy conservation gives

12

mvcube2

Total mechanical energyat bottom of ramp

= mgh

Total mechanical energyat top of ramp

or vcube = 2gh

where we have used the fact that the cube starts at rest at the top of the ramp and does not

rotate. The desired ratio of the center-of-mass speeds is cube

marble

2 14 1.181010 / 7

v ghv gh

= = =

______________________________________________________________________________ 55. REASONING Because we are ignoring frictional losses, the total mechanical energy

2 21 12 2E mv I mghω= + + of both objects is conserved as they roll down the hill. We will


apply this conservation principle twice: first, to determine the height h0 − hf of the hill, and second, to determine the translational speed vf of the frozen juice can at the bottom of the hill. Both the basketball and the frozen juice can roll without slipping, so the translational speed v of either one is related to its radius r and angular speed ω by v rω= (Equation 8.12).

SOLUTION a. Applying the energy conservation principle to the basketball, we obtain

12 mvf

2 + 12 Iω f

2 + mghf

Ef

= 1

2 mv02 + 1

2 Iω02 + mgh0

E0

(1)

The basketball starts from rest, so we have v0 = 0 m/s and ω0 = 0 rad/s. The basketball is a

thin-walled spherical shell, so its moment of inertia is given by 223I mr= (see Table 9.1 in

the text). Substituting these values along with f fv rω = (Equation 8.12) into Equation (1) yields

12 m

2 1 2f 2 3v m+ 2r( ) fv

r

2

m⎛ ⎞ +⎜ ⎟⎝ ⎠ f 0 0gh m= + + ( )2 2 251 1

0 f f f 0 f2 3 6 or gh v v v g h h+ = = −

Solving for the height h0 − hf of the hill, we obtain

( )( )

22f

0 f 2

5 5 6.6 m/s3.7 m

6 6 9.80 m/s

vh h

g− = = =

b. In Equation (1), we again substitute v0 = 0 m/s and ω0 = 0 rad/s, but this time we use

212I mr= for the moment of inertia (see Table 9.1 in the text), because the frozen juice can

is a solid cylinder:

12 m

2 1 1f 2 2v m+ 2r( ) fv

r

2

m⎛ ⎞ +⎜ ⎟⎝ ⎠ f 0 0gh m= + + ( )2 2 231 1

0 f f f 0 f2 4 4 or gh v v v g h h+ = = −

The final translational speed of the frozen juice can is

( ) ( ) ( )( )20 f 0 f2

f f

4 9.80 m/s 3.7 m4 4 or 7.0 m/s

3 3 3g h h g h h

v v− −

= = = = 56. REASONING The drawing shows the rod in its initial

(dashed lines) and final (solid lines) orientations. Since both friction and air resistance are absent, the total mechanical energy is conserved. In this case, the total

Center of mass

Center of mass

Pivot

hf


mechanical energy is

E = 12

Iω 2

Rotationalkinetic energy

+ mgh

Gravitationalpotential energy

In this expression, m is the rod’s mass, I is its moment of

inertia, ω is its angular speed, and h is the height of its center of mass above a reference level that we take to be the ground. Since the rod is uniform, its center of mass is located at the center of the rod. The energy-conservation principle will guide our solution.

SOLUTION Conservation of the total mechanical energy dictates that

12

Iω f2 + mghf

Final totalmechanical energy

= 1

2Iω0

2 + mgh0

Initial totalmechanical energy

(1)

Since the rod is momentarily at rest in its final orientation, we know that ωf = 0 rad/s, so that

Equation (1) becomes

( )1 12 2f 0 0 0 f 02 2

or mgh I mgh I mg h hω ω= + = − (2) We can relate the initial angular speed ω0 to the initial linear speed v0 by using

Equation 8.9: 00

vL

ω = . With this substitution, the fact that f 0h h L− = (see the drawing),

and the fact that 1 23

I mL= (see Table 9.1 in the text), Equation (2) indicates that

( ) ( )2

1 1 1 12 2 200 f 0 02 2 3 6

or or v

I mg h h mL mgL v gLL

ω ⎛ ⎞= − = =⎜ ⎟

⎝ ⎠

Solving for v0 gives

( )( )20 6 6 9.80 m/s 0.80 m 6.9 m/sv gL= = =

57. SSM REASONING AND SOLUTION The conservation of energy gives

mgh + (1/2) mv2 + (1/2) Iω2 = (1/2) mv02 + (1/2) Iω0

2 If the ball rolls without slipping, ω = v/R and ω0 = v0/R. We also know that I = (2/5) mR2.

Substitution of the last two equations into the first and rearrangement gives


( ) ( )( )22 210 100 7 73.50 m/s 9.80 m/s 0.760 m 1.3 m/sv v gh= − = − =

58. REASONING We first find the speed 0v of the ball when it becomes airborne using the

conservation of mechanical energy. Once 0v is known, we can use the equations of kinematics to find its range x.

SOLUTION When the tennis ball starts from rest, its total mechanical energy is in the form

of gravitational potential energy. The gravitational potential energy is equal to mgh if we take h = 0 m at the height where the ball becomes airborne. Just before the ball becomes airborne, its mechanical energy is in the form of rotational kinetic energy and translational kinetic energy. At this instant its total energy is 2 21 1

02 2mv Iω+ . If we treat the tennis ball as a thin-walled spherical shell of mass m and radius r, and take into account that the ball rolls down the hill without slipping, its total kinetic energy can be written as

2

1 1 1 1 2 52 2 2 2 200 0 02 2 2 2 3 6

( )v

mv I mv mr mvr

ω ⎛ ⎞+ = + =⎜ ⎟

⎝ ⎠

Therefore, from conservation of mechanical energy, he have

5 20 06

6 or 5ghmgh mv v= =

The range of the tennis ball is given by 0 0(cos ) xx v t v tθ= = , where t is the flight time of

the ball. From Equation 3.3b, we find that the flight time t is given by

0 0 0 0( ) 2 sin

–y y y

y y y

v v v v vt

a a aθ− − −

= = =

Therefore, the range of the tennis ball is

00 0

2 sin (cos ) –x

y

vx v t v

aθ

θ⎛ ⎞⎜ ⎟= =⎜ ⎟⎝ ⎠

If we take upward as the positive direction, then using the fact that –ya g= and the

expression for 0v given above, we find

220

2cos sin 2cos sin 6 12 cos sin 5 5

12 (1.8 m) (cos 35 )(sin 35 )= 2.0 m5

ghx v hg gθ θ θ θ θ θ

⎛ ⎞⎛ ⎞ ⎛ ⎞= = =⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= ° °


59. SSM REASONING Let the two disks constitute the system. Since there are no external

torques acting on the system, the principle of conservation of angular momentum applies. Therefore we have L Linitial final= , or

I I I IA A B B A B finalω ω ω+ = +( )

This expression can be solved for the moment of inertia of disk B. SOLUTION Solving the above expression for I B , we obtain

IB = IA

ω final –ωAωB –ω final

⎛

⎝⎜

⎞

⎠⎟ = (3.4 kg ⋅m2) –2.4 rad/s–7.2 rad/s

–9.8 rad/s–(–2.4 rad/s)⎡

⎣⎢

⎤

⎦⎥ = 4.4 kg ⋅m2

60. REASONING The supernova explosion proceeds entirely under the influence of internal

forces and torques. External forces and torques play no role. Therefore, the star’s angular momentum is conserved during the supernova and its subsequent transformation from a solid sphere into an expanding spherical shell: f 0L L= . The star’s initial and final angular momenta are given by L Iω= (Equation 9.10), where I is the star’s moment of inertia, and ω is its angular velocity. Initially, the star is a uniform solid sphere with a moment of inertia given by 22

5I MR= (see Table 9.1 in the text). Following the supernova, the moment of

inertia of the expanding spherical shell is 223I MR= (see Table 9.1 in the text). We will use

the angular-momentum-conservation principle to calculate the final angular velocity ωf of the expanding supernova shell.

SOLUTION Applying the angular-momentum-conservation principle yields

Ifω fLf

= I0ω0

L0

or 2

3 M Rf2ω f =

25 M R0

2ω0 or ω f =3R0

2ω0

5Rf2

Substituting R0 = R, Rf = 4.0R, and ω0 = 2.0 rev/d, we obtain the final angular velocity of the expanding shell:

2

f3R

ω = 0

5 4.0 R

ω

( )( )0

23 3 2.0 rev/d

0.075 rev/d80 80ω

= = =

61. REASONING The carousel rotates on frictionless bearings and without air resistance, so

no net external torque acts on the system comprised of the carousel and the person on it. Therefore, the total angular momentum L Iω= (Equation 9.10) of the system is conserved


(Lf = L0). As the person moves closer to the center of the carousel, the person’s distance r from the rotation axis decreases from r0 = 1.50 m to rf = 0.750 m. Therefore, the moment of

inertia 2PI mr= (Equation 9.6) of the person also decreases, from 2

P0 0I mr= to 2Pf fI mr= .

Consequently, the angular speed of the system increases from ω0 to ωf , preserving the system’s total angular momentum. We will use the angular-momentum-conservation principle to find the mass m of the person.

SOLUTION The person and the carousel have the same initial angular velocity ω0, and the same final angular velocity ωf. Therefore, the conservation of angular momentum principle can be expressed as

IPfω f + ICω f

Lf

= IP0ω0 + ICω0

L0

(1)

where IC is the moment of inertia of the carousel (without the person). Substituting

2P0 0I mr= and 2

Pf fI mr= (Equation 9.6) for the person’s initial and final moments of inertia into Equation (1), we obtain

mrf2

IPf

ω f + ICω f = mr0

2

IP0

ω0 + ICω0 or IC ω f −ω0( ) = m r0

2ω0 − rf2ω f( )

Solving for the mass m of the person yields

( ) ( )( )( ) ( ) ( ) ( )

2C f 02 2 2 20 0 f f

125 kg m 0.800 rad/s 0.600 rad/s28 kg

1.50 m 0.600 rad/s 0.750 m 0.800 rad/s

Im

r r

ω ωω ω

⋅ −−= = =

− − 62. REASONING Once the motorcycle is in the air, it is subject to no net external torque, since

gravity and air resistance are being ignored. Therefore, its total angular momentum is conserved: f 0L L= . The total angular momentum of the motorcycle is the sum of the angular momentum E E EL I ω= (Equation 9.10) of the engine and the angular momentum

M M ML I ω= of the rest of the motorcycle (including the rider):

IEωEf + IMωMf

Lf

= IEωE0 + IMωM0

L0

(1)

We will use Equation (1) to find the ratio IE/IM of the moments of inertia of the engine and the rest of the motorcycle. We note that, so long as all angular velocities are expressed in rev/min, there is no need to convert to SI units (rad/s).

SOLUTION Initially, only the engine is rotating, so the rest of the motorcycle has no angular velocity: ωM0 = 0 rad/s. Solving Equation (1) for the ratio IE/IM, we obtain


( ) ( ) E Mf

E Ef M Mf E E0 M E Ef E0 M MfM Ef E0

0 or or I

I I I I I II

ωω ω ω ω ω ω

ω ω−

+ = + − = − =−

As usual, clockwise rotation is negative, and counterclockwise is positive. The ratio of the moments of inertia is, then,

( )( ) ( )

4E Mf

M Ef E0

3.8 rev/min7.9 10

12 500 rev/min 7700 rev/minII

ωω ω

−− − −= = = ×

− + − +

63. REASONING The rod and bug are taken to be the system of objects under consideration,

and we note that there are no external torques acting on the system. As the bug crawls out to the end of the rod, each exerts a torque on the other. However, these torques are internal torques. The conservation of angular momentum states that the total angular momentum of a system remains constant (is conserved) if the net average external torque acting on the system is zero. We will use this principle to find the final angular velocity of the system.

SOLUTION The angular momentum L of the system (rod plus bug) is given by

Equation 9.10 as the product of the system’s moment of inertia I and angular velocity ω, or L = Iω. The conservation of angular momentum can be written as

Iω Final angularmomentum

= I0ω0

Initial angularmomentum

where ω and ω0 are the final and initial angular velocities, respectively, and I and I0 are the final and initial moments of inertia. The initial moment of inertia is given. The initial moment of inertia of the bug is zero, because it is located at the axis of rotation. The final moment of inertia is the sum of the moment of inertia of the bug and that of the rod; I = Ibug + I0. When the bug has reached the end of the rod, its moment of inertia is

Ibug = mL2, where m is its mass and L is the length of the rod. The final angular velocity of the system is, then,

( ) ( )( ) ( )

0 0 00 0 0 2

bug 0 0

3 2

23 3 21.1 10 kg m0.32 rad/s 0.26 rad/s

4.2 10 kg 0.25 m + 1.1 10 kg m

I I II I I mL I

ω ω ω ω

−

− −

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎜ ⎟= = = ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟+ +⎝ ⎠ ⎝ ⎠⎝ ⎠

⎡ ⎤× ⋅⎢ ⎥= =⎢ ⎥× × ⋅⎢ ⎥⎣ ⎦

64. REASONING We consider a system consisting of the person and the carousel. Since the

carousel rotates on frictionless bearings, no net external torque acts on this system. Therefore, its angular momentum is conserved, and we can set the total angular momentum of the system before the person hops on equal to the total angular momentum afterwards.


SOLUTION The initial angular momentum of the carousel is carousel 0I ω (Equation 9.10) ,

where Icarousel is the moment of inertia of the carousel and ω0 is its initial angular velocity. After the person climbs aboard, the total angular momentum of the carousel and person is carousel f person fI Iω ω+ , where ωf is the final angular velocity. According to Equation 9.6, the

person’s moment of inertial is I MRperson =2 , since he is at the outer edge of the carousel,

which has a radius R. Applying the conservation of angular momentum, we have

Icarouselω f + Ipersonω f

Final totalangular momentum

= Icarouselω0

Initial totalangular momentum

ω f =Icarouselω0

Icarousel + Iperson

=Icarouselω0

Icarousel + MR2

=125 kg ⋅m2( ) 3.14 rad/s( )

125 kg ⋅m2 + 40.0 kg( ) 1.50 m( )2= 1.83 rad/s

65. REASONING Let the space station and the people within it constitute the system. Then as

the people move radially from the outer surface of the cylinder toward the axis, any torques that occur are internal torques. Since there are no external torques acting on the system, the principle of conservation of angular momentum can be employed.

SOLUTION Since angular momentum is conserved,

I Ifinal final 0ω ω= 0 Before the people move from the outer rim, the moment of inertia is

I I m r0 500= +station person person2

or I 0

93 00 10 10= × ⋅ + = × ⋅. ( kg m kg m2 9 2500)(70.0 kg)(82.5 m) 3.242 If the people all move to the center of the space station, the total moment of inertia is

I Ifinal station= = × ⋅3 00 10. 9 2 kg m Therefore,

ωω

final

0

0

final

3.24= =

× ⋅× ⋅

=II

1010

1 089

9 2

2

kg m 3.00 kg m

.

This fraction represents a percentage increase of 8 percent .


66. REASONING Since the rod (length = d, mass = m) changes shape without the aid of

external torques, the principle of conservation of angular momentum applies. This principle states that the rod’s total angular momentum remains constant. Thus, the total angular momentum is the same in its final straight shape as it is in its initial bent shape.

SOLUTION Angular momentum is the momentum of inertia I of a rotating object times the

object’s angular velocity ω, according to Equation 9.10. Conservation of the total angular momentum indicates that

Istraightωstraight

Final angularmomentum

= Ibentωbent


or ωstraight =

IbentωbentIstraight

(1)

In its bent shape, the rod’s horizontal half (length = d/2, mass = m/2) rotates about the axis

perpendicular to its left end (see text drawing), so that its mass is located at varying distances from the axis. According to Table 9.1, its moment of inertia is

( )( )21 1 1 1 23 2 2 24m d md= . As the vertical half rotates around the axis, all of its mass is

located at the same distance of d/2 from the axis. Therefore, its moment of inertia is

( )( )21 1 1 22 2 8m d md= , according to Equation 9.4. The total moment of inertia of the bent

rod, then is the sum of these two contributions: 1 1 12 2 2

bent 24 8 6I md md md= + = (2)

According to Table 9.1 the momentum of inertia of the straight rod is

1 2straight 3I md= (3)

Substituting Equations (2) and (3) into Equation (1) gives

( )( )1 26bent bent

straight 1 2straight 3

9.0 rad / s4.5 rad / s

mdII md

ωω = = =

67. REASONING AND SOLUTION After the mass has moved inward to its final path the

centripetal force acting on it is T = 105 N.


105 N

r

Its centripetal acceleration is

ac = v2/R = T/m Now

v = ωR so R = T/(ω2m) The centripetal force is parallel to the line of action (the string), so the force produces no

torque on the object. Hence, angular momentum is conserved.

Iω = I0ω0 so that ω = (I0/I)ω0 = (R02/R2)ω0

Substituting and simplifying, we obtain

( )( ) ( )4 24 2 4 23 0 0 0 0 33 0.500 kg 1.00 m 6.28 rad / s

or 0.573 m105 N

mR mRR R

T Tω ω

= = = = 68. REASONING AND SOLUTION The block will just start to move when the centripetal

force on the block just exceeds maxsf . Thus, if rf is the smallest distance from the axis at

which the block stays at rest when the angular speed of the block is ωf, then µs FN = mrf ωf

2, or µs mg = mrf ωf2. Thus,

µs g = rf ωf

2 (1) Since there are no external torques acting on the system, angular momentum will be

conserved. I0 ω0 = If ωf where I0 = mr0

2, and If = mrf2. Making these substitutions yields

r0

2 ω0 = rf2 ωf (2)

Solving Equation (2) for ωf and substituting into Equation (1) yields:


42 0

s f 0 4f

rg r

rµ ω=

Solving for rf gives 1/3 1/32 4 2 4

0 0f 2

s

(2.2 rad/s) (0.30 m) 0.17 m(0.75)(9.80 m/s )

rr

gωµ

⎛ ⎞ ⎡ ⎤= = =⎜ ⎟ ⎢ ⎥⎜ ⎟ ⎣ ⎦⎝ ⎠

69. SSM REASONING In both parts of the problem, the

magnitude of the torque is given by Equation 9.1 as the magnitude F of the force times the lever arm l . In part (a), the lever arm is just the distance of 0.55 m given in the drawing. However, in part (b), the lever arm is less than the given distance and must be expressed using trigonometry as ( )0.55 m sinθ=l . See the drawing at the right.

SOLUTION a. Using Equation 9.1, we find that

( )( )Magnitude of torque 49 N 0.55 m 27 N mF= = = ⋅l b. Again using Equation 9.1, this time with a lever arm of ( )0.55 m sinθ=l , we obtain

( )( )

( )( ) ( )( )1

Magnitude of torque 15 N m 49 N 0.55 m sin

15 N m 15 N msin or sin 3449 N 0.55 m 49 N 0.55 m

F θ

θ θ −

= ⋅ = =

⎡ ⎤⋅ ⋅= = = °⎢ ⎥⎢ ⎥⎣ ⎦

l

70. REASONING Before any sand strikes the disk, only the disk is rotating. After the sand has

landed on the disk, both the sand and the disk are rotating. If the sand and disk are taken to be the system of objects under consideration, we note that there are no external torques acting on the system. As the sand strikes the disk, each exerts a torque on the other. However, these torques are exerted by members of the system, and, as such, are internal torques. The conservation of angular momentum states that the total angular momentum of a system remains constant (is conserved) if the net average external torque acting on the system is zero. We will use this principle to find the final angular velocity of the system.

SOLUTION The angular momentum of the system (sand plus disk) is given by

Equation 9.10 as the product of the system’s moment of inertia I and angular velocity ω, or L = Iω. The conservation of angular momentum can be written as

0.55 m

49 N

θ

lever arm

axis


Iω Final angularmomentum

= I0ω0


where ω and ω0 are the final and initial angular velocities, respectively, and I and I0 are final

and initial moments of inertia. The initial moment of inertia is given, while the final moment of inertia is the sum of the values for the rotating sand and disk, I = Isand + I0. We note that the sand forms a thin ring, so its moment of inertia is given by (see Table 9.1) 2

sand sand sand = I M R , where Msand is the mass of the sand and Rsand is the radius of the ring. Thus, the final angular velocity of the system is, then,

( )( )( )

0 0 00 0 0 2

sand 0 sand sand 0

2

2 20.10 kg m0.067 rad/s 0.037 rad/s

0.50 kg 0.40 m + 0.10 kg m

I I II I I M R I

ω ω ω ω⎛ ⎞⎛ ⎞⎛ ⎞

= = = ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎡ ⎤⋅⎢ ⎥= =⎢ ⎥⋅⎣ ⎦

71. REASONING According to Table 9.1 in the text, the moment of inertia I of the disk is

1 22

I MR= , where M is the disk’s mass and R is its radius. Thus, we can determine the mass from this expression, provided we can obtain a value for I. To obtain the value for I, we will use Newton’s second law for rotational motion, Iτ αΣ = (Equation 9.7), where Στ is the net torque and α is the angular acceleration.

SOLUTION From the moment of inertia of the disk, we have

1 222

2 or II MR MR

= = (1) Using Newton’s second law for rotational motion, we find for I that

or I I ττ ααΣΣ = =

Substituting this expression for I into Equation (1) gives

2 22 2IMR R

ταΣ= = (2)

The net torque Στ is due to the 45-N force alone. According to Equation 9.1, the magnitude

of the torque that a force of magnitude F produces is Fl , where l is the lever arm of the force. In this case, we have R=l , since the force is applied tangentially to the disk and perpendicular to the radius. Substituting FRτΣ = into Equation (2) gives

( )

( )( )2 2 2

2 45 N2 2 2 5.0 kg0.15 m 120 rad/s

FR FMRR R

ταα α

Σ= = = = =


72. REASONING At every instant before the plank begins to tip, it is in equilibrium, and the

net torque on it is zero: 0τΣ = (Equation 9.2). When the person reaches the maximum distance x along the overhanging part, the plank is just about to rotate about the right support. At that instant, the plank loses contact with the left support, which consequently exerts no force on it. This leaves only three vertical forces acting on the plank: the weight W of the plank, the force FR due to the right support, and the force P due to the person (see the free-body diagram of the plank). The force FR acts at the right support, which we take as the axis, so its lever arm is zero. The lever arm for the force P is the distance x. Since counterclockwise is the positive direction, Equation 9.2 gives

WW 0 or

WW Px x

PτΣ = − = =

ll (1)

SOLUTION The weight W = 225 N of the plank is known, and the force P due to the person is equal to the person’s weight: P = 450 N. This is because the plank supports the person against the pull of gravity, and Newton’s third law tells us that the person and the plank exert forces of equal magnitude on each other. The plank’s weight W acts at the center of the uniform plank, so we have (see the drawing)

1 1W W2 2 or d L L d+ = = −l l (2)

where d = 1.1 m is the length of the overhanging part of the plank, and L = 5.0 m is the length of the entire plank. Substituting Equation (2) into Equation (1), we obtain

( ) ( ) ( )1 12 2225 N 5.0 m 1.1 m

0.70 m450 N

W L dx

P

⎡ ⎤− −⎣ ⎦= = =

x

L/2

FR

W

P

d

Wl

Free-body diagram of the plank

Axis


73. SSM REASONING The rotational analog of Newton's second law is given by Equation 9.7, Iτ α=∑ . Since the person pushes on the outer edge of one section of the door with a force F that is directed perpendicular to the section, the torque exerted on the door has a magnitude of FL, where the lever arm L is equal to the width of one section. Once the moment of inertia is known, Equation 9.7 can be solved for the angular acceleration α.

The moment of inertia of the door relative to the rotation axis is I = 4IP, where IP is the

moment of inertia for one section. According to Table 9.1, we find 21P 3I ML= , so that the

rotational inertia of the door is 243I ML= .

SOLUTION Solving Equation 9.7 for α, and using the expression for I determined above,

we have

22 4 44

3 33

68 N 0.50 rad/s(85 kg)(1.2 m)

FL FMLML

α = = = =

74. REASONING To determine the angular

acceleration of the pulley and the tension in the cord attached to the block, we will apply Newton’s second law to the pulley and the block separately. Only one external forces acts on the pulley, as its free-body diagram at the right shows. This force T is due to the tension in the cord. The torque that results from this force is the net torque acting on the pulley and obeys Newton’s second law for rotational motion (Equation 9.7). Two external forces act on the block, as its free-body diagram at the right indicates. These are (1) the force T′ due to the tension in the cord and (2) the weight mg of the block. The net force that results from these forces obeys Newton’s second law for translational motion (Equation 4.2b).

SOLUTION Applying Newton’s second law for rotational motion to the pulley gives

TR Iτ αΣ = + = (1) where we have written the torque as the magnitude of the tension force times the lever arm

(the radius) as specified by Equation 9.1. In addition, we have assigned this torque a positive value, since it causes a counterclockwise rotation of the pulley. To obtain a value for T, we note that the tension has the same magnitude everywhere in the massless cord, so

Free-body diagram for pulley

T

R

Free-body diagram for block

mg

T′

+y

+x

+τ


that T T ′= . Thus, by applying Newton’s second law for translation (Equation 4.2b), we obtain

or F T mg ma T T mg ma′ ′Σ = − = − = = − where we have used a to denote the magnitude of the vertical acceleration of the block and

included the minus sign to account for the fact that the block is accelerating downward. Substituting this result for T into Equation (1) gives

( )mg ma R Iα− = (2)

To proceed further, we must deal with a. Note that the pulley rolls without slipping against

the cord, so a and α are related according to a Rα= (Equation 8.13). With this substitution, Equation (2) becomes

( ) 2 or mg m R R I mgR mR Iα α α α⎡ ⎤− = = +⎣ ⎦

Solving for α, we find that

( )( )( )( )( )

22

2 2 3 2

2.0 kg 9.80 m/s 0.040 m180 rad/s

2.0 kg 0.040 m 1.1 10 kg m

mgRmR I

α−

= = =+ + × ⋅

The value for the tension can be now obtained by substituting this value for α into

Equation (1):

( )( )3 2 21.1 10 kg m 180 rad/s or 5.0 N

0.040 mITR I TRαα

−× ⋅= = = =

98 N

47 N

M

0.61 m

0.28 m

0.069 m

29°

(0.069 m) sin 29°

cg Axis


75. REASONING The arm, being stationary, is in equilibrium, since it has no translational or angular acceleration. Therefore, the net external force and the net external torque acting on the arm are zero. Using the fact that the net external torque is zero will allow us to determine the magnitude of the force M. The drawing at the right shows three forces: M, the 47-N weight of the arm acting at the arm’s center of gravity (cg), and the 98-N force that acts upward on the right end of the arm. The 98-N force is applied to the arm by the ring. It is the reaction force that arises in

response to the arm pulling downward on the ring. Its magnitude is 98 N, because it supports the 98-N weight hanging from the pulley system. Other forces also act on the arm at the shoulder joint, but we can ignore them. The reason is that their lines of action pass directly through the axis at the shoulder joint, which is the axis that we will use to determine torques. Thus, these forces have zero lever arms and contribute no torque.

SOLUTION The magnitude of each individual torque is the magnitude of the force times

the corresponding lever arm. The forces and their lever arms are as follows: Force Lever Arm

98 N 0.61 m 47 N 0.28

M (0.069 m) sin 29° Each torque is positive if it causes a counterclockwise rotation and negative if it causes a

clockwise rotation about the axis. Thus, since the net torque must be zero, we see that

( )( ) ( )( ) ( )98 N 0.61 m 47 N 0.28 m 0.069 m sin 29 0M ⎡ ⎤− − ° =⎣ ⎦ Solving for M gives

( )( ) ( )( )( )

98 N 0.61 m 47 N 0.28 m1400 N

0.069 m sin 29M

−= =

°

76. REASONING The moment of inertia of a point particle is the mass m of the particle times the square of the perpendicular distance r of the particle from the axis about which the particle is rotating, according to Equation 9.4. For an object that is not a point particle, its mass is spread out over a region of space, and the range of distances where the mass is located must be taken into account. The moments of inertia given in Table 9.1 in the text illustrate how the shape of an object can influence its moment of inertia.


SOLUTION a. In Table 9.1 the moment of inertia of a rod relative to an axis that is perpendicular to the

rod at one end is given by 2 2 21 1

rod 3 3 (2.00 kg)(2.00 m) 2.67 kg mI ML= = = ⋅ b. According to Equation 9.4, the moment of inertia of a point particle of mass m relative to

a rotation axis located a perpendicular distance r from the particle is 2I mr= . Suppose that all the mass of the rod were located at a single point located at perpendicular distance r from the axis in part (a). If this point particle has the same moment of inertia as the rod in part (a), then the distance r, which is called the radius of gyration, is given by

22.67 kg m 1.16 m2.00 kg

Irm

⋅= = =

______________________________________________________________________________ 77. SSM REASONING When the modules pull together, they do so by means of forces that

are internal. These pulling forces, therefore, do not create a net external torque, and the angular momentum of the system is conserved. In other words, it remains constant. We will use the conservation of angular momentum to obtain a relationship between the initial and final angular speeds. Then, we will use Equation 8.9 (v = rω) to relate the angular speeds ω0 and ωf to the tangential speeds v0 and vf.

SOLUTION Let L be the initial length of the cable between the modules and ω0 be the

initial angular speed. Relative to the center-of-mass axis, the initial momentum of inertia of the two-module system is I0 = 2M(L/2)2, according to Equation 9.6. After the modules pull together, the length of the cable is L/2, the final angular speed is ωf, and the momentum of

inertia is If = 2M(L/4)2. The conservation of angular momentum indicates that

Ifω f

Final angularmomentum

= I0ω0


2M L4

⎛⎝⎜

⎞⎠⎟

2⎡

⎣⎢⎢

⎤

⎦⎥⎥ω f = 2M L

2⎛⎝⎜

⎞⎠⎟

2⎡

⎣⎢⎢

⎤

⎦⎥⎥ω0

ω f = 4ω0

According to Equation 8.9, ωf = vf/(L/4) and ω0 = v0/(L/2). With these substitutions, the

result that ωf = 4ω0 becomes


vfL / 4

= 4v0

L / 2

⎛

⎝⎜⎞

⎠⎟ or vf = 2v0 = 2 17 m/s( ) = 34 m/s

78. REASONING

a. The kinetic energy of the rolling wheel is the sum of its translational ( )212 mv and

rotational ( )212 Iω kinetic energies. In these expressions m and I are, respectively, the mass

and moment of inertia of the wheel, and v and ω are, respectively, its linear and angular speeds. The sliding wheel only has translational kinetic energy, since it does not rotate.

b. As the wheels move up the incline plane, the total mechanical energy is conserved, since only the conservative force of gravity does work on each wheel. Thus, the initial kinetic energy at the bottom of the incline is converted entirely into potential energy when the wheels come to a momentary halt. The potential energy PE is given by PE = mgh (Equation 6.5), where h is the height of the wheel above an arbitrary zero level.

SOLUTION

a. Since the rolling wheel is a disk, its moment of inertia is 212 = I mR (see Table 9.1),

where R is the radius of the disk. Furthermore, its angular speed ω is related to the linear speed v of its center of mass by Equation 8.12 as ω = v/R. Thus, the total kinetic energy of the rolling wheel is

The kinetic energy of the sliding wheel is

b. As each wheel rolls up the incline, its total mechanical energy is conserved. The initial kinetic energy KE at the bottom of the incline is converted entirely into potential energy PE when the wheels come to a momentary halt. Thus, the potential energies of the wheels have the values calculated in part a for the total kinetic energies.

The potential energy of a wheel is given by Equation 6.5 as PE = mgh, where g is the

acceleration due to gravity and h is the height relative to an arbitrary zero level. Therefore, the height reached by each wheel is as follows:

Rolling Wheel

( )

( )( )

22 2 2 21 1 1 1 1

2 2 2 2 2

223 34 4

KE = + +

2.0 kg 6.0 m/s 54 J

vmv I mv mRR

mv

ω ⎛ ⎞= ⎜ ⎟⎝ ⎠

= = =

Sliding Wheel ( )( )221 1

2 2KE = 2.0 kg 6.0 m/s 36 Jmv = =


79. REASONING AND SOLUTION Consider the left board, which has a length L and a weight

of (356 N)/2 = 178 N. Let Fv be the upward normal force exerted by the ground on the board. This force balances the weight, so Fv = 178 N. Let fs be the force of static friction, which acts horizontally on the end of the board in contact with the ground. fs points to the right. Since the board is in equilibrium, the net torque acting on the board through any axis must be zero. Measuring the torques with respect to an axis through the apex of the triangle formed by the boards, we have

+ (178 N)(sin 30.0°)2L⎛ ⎞

⎜ ⎟⎝ ⎠ + fs(L cos 30.0°) – Fv (L sin 30.0°) = 0

or s V44.5 N + cos 30.0 – sin 30.0 = 0f F° °

so that

s(178 N)(sin 30.0 ) – 44.5 N 51.4 N

cos 30.0f °= =

°

80. REASONING AND SOLUTION Newton's law applied to the 11.0-kg object gives

T2 − (11.0 kg)(9.80 m/s2) = (11.0 kg)(4.90 m/s2) or T2 = 162 N A similar treatment for the 44.0-kg object yields

T1 − (44.0 kg)(9.80 m/s2) = (44.0 kg)(−4.90 m/s2) or T1 = 216 N For an axis about the center of the pulley

T2r − T1r = I(−α) = (1/2) Mr2 (−a/r) Solving for the mass M we obtain

M = (−2/a)(T2 − T1) = [−2/(4.90 m/s2)](162 N − 216 N) = 22.0 kg

Rolling Wheel ( )( )2

PE 54 J = = = 2.8 m2.0 kg 9.80 m/s

hmg

Sliding Wheel ( )( )2

PE 36 J = = = 1.8 m2.0 kg 9.80 m/s

hmg


81. SSM REASONING

To determine the angular acceleration of the dual pulley and the tension in the cable attached to the crate, we will apply Newton's second law to the pulley and the crate separately. Four external forces act on the dual pulley, as its free-body diagram in the figure shows. These are (1) the tension

T1 in

the cable connected to the motor, (2) the tension

T2 in the cable

attached to the crate, (3) the pulley's weight

WP , and (4) the

reaction force P exerted on the

dual pulley by the axle. The force

P arises because the two cables and the pulley's weight pull the pulley down and to

the left into the axle, and the axle pushes back, thus keeping the pulley in place. The net torque that results from these forces obeys Newton's second law for rotational motion (Equation 9.7). Two external forces act on the crate, as its free-body diagram in figure indicates. These are (1) the cable tension

T2′ and (2) the weight m

g of the crate. The net force that results from these forces obeys Newton's second law for translational motion (Equation 4.2b).

Using the lever arms 1 and 2 shown in the figure, we can apply the second law to the rotational motion of the dual pulley: τ = T11 −T22 = Iα∑ (1) Note that the forces

P and

WP have zero lever arms, since their lines of action pass through

the axis of rotation. Thus, these forces contribute nothing to the net torque. Applying Newton’s second law to the upward translational motion of the crate gives: Fy = T2′∑ −mg = may Note that the magnitude of the tension in the cable between the crate and the pulley is T2′ = T2 , so that we can solve the above equation for the tension and obtain:


T2′ = T2 = mg +may Because the cable attached to the crate rolls on the pulley without slipping, the linear acceleration ay of the crate is related to the angular acceleration α of the pulley via ay = rα (Equation 8.13), where r = 2. Thus, we have ay = 2α . We can now substitute our expressions for T2 and ay into Equation (1): Iα = T11 −T22 = T11 − (mg +may )2 = T11 − (mg + 2α )2 Solving for α gives:

α = T11 −mg2

I +m22 = (2150 N)(0.600 m)− (451 kg)(9.80 m/s2 )(0.200 m)

46.0 kg ⋅m2 + (451 kg)(0.200 m)2 = 6.3 rad/s2

Using this value for α and Equation (1), we can now solve for T2 :

T2 =

T11 − Iα2

= (2150 N)(0.600 m)− (46.0 kg ⋅m2 )(6.3 rad/s2 )0.200 m

= 5.00 ×103 N

_____________________________________________________________________________ 82. CONCEPTS (i) Tipping is a rotational or angular motion.

Since the crate starts from rest, an angular acceleration is needed, which can only be created by a net external torque. Thus, the torque created by the force

P causes the tipping.

(ii) The magnitude of the torque is the product of the magnitude of the force and the lever arm. Thus, for a minimum force, the lever arm should be a maximum, and the force

P should be applied at the upper left corner of the

crate, as shown in the figure (lower right). CALCULATIONS Only the pushing force

P and the weight

W of the crate produce

external torques with respect to the rotational axis through the lower right corner of the crate. The obstruction also applies a force to the crate, but it creates no torque since its line of action passes through the axis. Since the sum of the external torques is zero, we refer to the figure for the lever arms, and write

τ = −PP∑ +W W = 0


The lever arm for the force P is P = L , where L is the length of the side of the crate. The

lever arm for the weight is W = L 2 , because the crate is uniform, and the center of gravity is at the center of the crate. Substituting these lever arms into the torque equation, we obtain

_____________________________________________________________________________ 83. SSM CONCEPTS (i) The solid sphere has a moment of inertia of 25MR

2 , while the shell has a moment of inertia of 23MR

2 . Since the masses and radii of the spheres are the same, it follows that the shell has the greater moment of inertia. The reason is that more of the mass of the shell is located farther from the rotational axis than is the case for the solid sphere. In the solid sphere, some of the mass is located close to the axis and, therefore, does not contribute as much to the moment of inertia. (ii) Newton’s second law for rotation specifies that the angular acceleration is α = Στ( ) I , whereΣτ is the net external torque and I is the moment of inertia. Since the moment of inertia is in the denominator, the angular acceleration is smaller when I is greater. Because it has the greater moment of inertia, the shell has the angular acceleration with the smaller magnitude. (iii) Since the angular acceleration of the shell has the smaller magnitude, the shell requires a longer time for the deceleration to reduce its angular velocity to zero. CALCULATIONS According to the equations of rotational kinematics, whereω and ω 0 are the final and initial angular velocities, respectively. Taking the initial time to be t0 = 0 , we write From Newton’s second law for rotation, the angular acceleration is given byα = Στ( ) I , and substituting this into the previous equation results in In applying this result, we arbitrarily choose the direction of the initial rotation to be positive. With this choice, the torque must be negative, since it causes a deceleration. Using the proper moments of inertia, we find the following times for the spheres to come to a halt:

−PL +W 12 L( ) = 0 or P = 1

2W = 12 580 N( ) = 290 N

α = ω −ω 0

t − t0

t = ω −ω 0

α

t = ω −ω 0

Στ( ) I =I ω −ω 0( )

Στ


Solid sphere:

Spherical

_____________________________________________________________________________

t =I ω −ω 0( )

Στ=

25MR2 ω −ω 0( )

Στ

=25 1.5 kg( ) 0.20 m( )2 0 rad/s( )− 24 rad/s( )⎡⎣ ⎤⎦

−0.12 N ⋅m= 4.8 s

t =I ω −ω 0( )

Στ=

23MR2 ω −ω 0( )

Στ

=23 1.5 kg( ) 0.20 m( )2 0 rad/s( )− 24 rad/s( )⎡⎣ ⎤⎦

−0.12 N ⋅m= 8.0 s

CHAPTER 9 ROTATIONAL DYNAMICS

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