Download PSE 3e Chapter 12 EOC Conceptual Questions Larry Smith

12
ROTATION OF A RIGID BODY
Conceptual Questions
12.1. As suggested by the figure, we will assume that the larger sphere is more massive. Then the center of gravity
would be at point a because if we suspend the dumbbell from point a then the counterclockwise torque due to the
large sphere (large weight times small lever arm) will be equal to the clockwise torque due to the small sphere (small
weight times large lever arm).
Look at the figure and mentally balance the dumbbell on your finger; your finger would have to be at point a. The
sun-earth system is similar to this except that the sun’s mass is so much greater than the earth’s that the center of
mass (called the barycenter for astronomical objects orbiting each other) is only 450 km from the center of the sun.
12.2. To double the rotational energy without changing ω requires doubling the moment of inertia. The moment of
inertia is proportional to R 2 so R must increase by 2.
12.3. The rotational kinetic energy is =
K rot
1
1 2
=
I
MR 2 . Since the mass is the same for all three
I ω . For a disk,
2
2
disks, the quantity R 2ω 2 determines the ranking. Thus K a = K b > K c .
12.4. No. The moment of inertia does not have any dependence on a quantity that indicates an object is rotating,
such as ω or α , so an object does not have to be rotating to have a moment of inertia.
12.5. Mass that is farther away from the axis of rotation contributes more to the moment of inertia
=
I
∫r
2
dm.
2
Here, r is the distance from the axis of rotation to the mass element dm. Note r is always positive. For a rod, there is
more mass farther away from an axis through the rod’s end than one through its middle.
12.6. Because sphere 2 has twice the radius, its mass is greater by a factor of 23 = 8, =
since m
4 3
π r rsteel . The
3
added mass is also distributed farther from the center, so, I ∝ mr 2 leads to I 2 ∝ (8m1 )(2r1 ) 2 =
32 I1.
12.7. It will be easier to rotate the solid sphere because the hollow sphere’s mass is generally distributed farther
from its center. If you roll both simultaneously down an incline, the solid sphere will win.
12.8. τ e > τ a = τ b > τ c = τ d > τ f The torque
=
τ rF sin θ . We must calculate each torque:
L
 
τa =   F
2
L
 
τb =  F
4
L
=
τc  =
 F sin 45°
2
L
F=
sin 45°
2
τ e = L(2 F )
=
τd
2
LF
4
=
τ f LF=
sin 0° 0
2
LF
4
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12-1
12-2
Chapter 12
12.9. (a) Negative, since it causes the ball to rotate clockwise.
(b) The angular velocity holds steady (iii) after the push has ended, since the torque has ended, and there is no more
angular acceleration.
(c) The torque is zero after the push has ended because there is no longer an applied force.
12.10. Since τ = I α, α=
τ
aa =
I
. Also, τ = Fr and I = mr 2 ⇒ α =
F0
m0r0
2 F0
=
aa
=
b
a
2m0r0
F
. Calculate α for each case:
mr
F0
1
=
aa
=
c
a
m0 (2r0 ) 2
2 F0
1
=
aa
=
d
a
(2m0 )(2r0 ) 2
So aaaa
a = b > c = d.
12.11. The block attached to the solid cylinder hits first. The solid cylinder has a smaller moment of inertia since
more of its mass is closer to the rotation axis, so it has less resistance to a change in its rotational motion. The torque
applied by the string attached to the block makes the solid cylinder change its rotation and unwind the string faster.
12.12. The moment of inertia for the tuck position is smaller than that of the pike position. Since the angular
momentum of the diver is conserved, any initial angular velocity is increased more when the diver moves to the tuck
position relative to the pike position.
12.13. The angular momentum L of disk b is larger than the angular momentum of disk a. Calculate L for each:
1 2
mra ωa
2
1
Lb I=
m 2ra2
=
bωb
2
La I=
=
aωa
( )  12 ω=
2
a
1

2  mra2ω
=
a  2 La
2

Exercises and Problems
Section 12.1 Rotational Motion
12.1. Model: A spinning skater, whose arms are outstretched, is a rigid rotating body.
Visualize:
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Rotation of a Rigid Body
12-3
Solve: The speed v = rω , where r =
140 cm/2 =
0.70 m. Also, 180
rpm (180)2pp
/60
rad/s 6 rad/s. Thus,
=
=
v =.
(0 70 m)(6π rad/s) =.
13 2 m/s.
Assess: A speed of 13.2 m/s ≈ 26 mph for the hands is a little high, but reasonable.
12.2. Model: Assume constant angular acceleration.
 2p rad  min 
Solve: (a) The final angular velocity is ωf =
(2000 rpm) 
209.4 rad/s. The definition of angular

=
 rev  60 s 
acceleration gives us
∆ω ω − ωi 209.4 rad/s − 0 rad/s
=
a = f =
= 419 rad/s 2
∆t
∆t
0.50 s
The angular acceleration of the drill is 4.2 × 102 rad/s 2 .
1
1
(b) θ f = θi + ωi ∆t + a (∆t ) 2 = 0 rad + 0 rad + (419 rad/s 2 )(0.50 s) 2 = 52.4 rad
2
2
 rev 
The drill makes (52.4 rad) 
8.3 revolutions.
=
 2π rad 
12.3. Model: Assume constant angular acceleration.
Visualize:
 2p rad  min 
Solve: The initial angular velocity
is ωi (60 rpm) 
=
=
 2p rad/s.
 rev  60 s 
The angular acceleration is
ω − ωi 0 rad/s − 2π rad/s
a
= f =
= 2 0.251 rad/s 2
∆t
25 s
The angular velocity of the fan blade after 10 s is
ωf = ωi + a (t − t0 ) = 2π rad/s + (−0.251 rad/s 2 )(10 s − 0 s) = 3.77 rad/s
The tangential speed of the tip of the fan blade is
vt =
rω =
(0.40 m)(3.77 rad/s) =
1.5 m/s
1
1
(b) θ f = θi + ωi ∆t + a (∆t ) 2 = 0 rad + (2π rad/s)(25 s) + (−0.251 rad/s 2 )(25 s) 2 = 78.64 rad
2
2
The fan turns 78.64 rad =.
12 5 rev ≈ 13 rev while coming to a stop.
12.4. Model: Assume constant angular acceleration.
Visualize:
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12-4
Chapter 12
Solve: (a) Since at = ra , find α first. With 90 rpm = 9.43 rad/s and 60 rpm = 6.28 rad/s,
∆ω 9.43 rad − 6.28 rad/s
=
= 0.314 rad/s 2
∆t
10 s
The angular acceleration of the sprocket and pedal are the same. So
a=
at =
ra =
(0.18 m)(0.314 rad/s 2 ) =
0.057 m/s 2
(b) The length of chain that passes over the sprocket during this time is L = r ∆θ . Find ∆θ :
1
2
θ f = θi + ωi ∆t + a (∆t ) 2
1
2
The length of chain which has passed over the top of the sprocket is
θf − θi =∆θ =(6.28 rad/s)(10 s) + (0.314 rad/s 2 )(10 s)2 =78.5 rad
L=
(0.10 m)(78.5 rad) =
7.9 m
Section 12.2 Rotation About the Center of Mass
12.5. Model: The earth and moon are particles.
Visualize:
Choosing xE = 0 m sets the coordinate origin at the center of the earth so that the center of mass location is the
distance from the center of the earth.
Solve:
mE xE + mM xM (5.98 × 1024 kg)(0 m) + (7.36 × 1022 kg)(3.84 × 108 m)
=
xcm =
mE + mM
5.98 × 1024 kg + 7.36 × 1022 kg
= 4.67 × 106 m ≈ 4.7 × 106 m
Assess: The center of mass of the earth-moon system is called the barycenter and is located beneath the surface of
the earth. Even though xE = 0 m the earth influences the center of mass location because mE is in the denominator of
the expression for xcm .
12.6. Visualize: The coordinates of the three masses mA , mB , and mC are (0 cm, 0 cm), (0 cm, 10 cm), and (10 cm, 0 cm),
respectively.
Solve: The coordinates of the center of mass are
mA xA + mB xB + mC xC (100 g)(0 cm) + (200 g)(0 cm) + (300 g)(10 cm)
=
xcm =
= 5.0 cm
mA + mB + mC
(100 g + 200 g + 300 g)
=
ycm
mA yA + mB yB + mC yC (100 g)(0 cm) + (200 g)(10 cm) + (300 g)(0 cm)
=
= 3.3 cm
mA + mB + mC
(100 g + 200 g + 300 g)
12.7. The coordinates of the three masses mA , mB , and mC are (0 cm, 10 cm), (10 cm, 10 cm), and (10 cm, 0 cm),
respectively.
Solve: The coordinates of the center of mass are
mA xA + mB xB + mC xC (200 g)(0 cm) + (300 g)(10 cm) + (100 g)(10 cm)
=
xcm =
= 6.7 cm
mA + mB + mC
(200 g + 300 g + 100 g)
=
ycm
mA yA + mB yB + mC yC (200 g)(0 cm) + (300 g)(10 cm) + (100 g)(0 cm)
=
= 5.0 cm
mA + mB + mC
(200 g + 300 g + 100 g)
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Rotation of a Rigid Body
12-5
12.8. Model: The balls are particles located at the ball’s respective centers.
Visualize:
Solve: The center of mass of the two balls measured from the left hand ball is
(100 g)(0 cm) + (200 g)(30 cm)
=
xcm = 20 cm
100 g + 200 g
The linear speed of the 100 g ball is
 2π rad  min 
v1 =
rω =
xcmω =
(0.20 m)(120 rev/min) 
2.5 m/s

=
 rev  60 s 
Section 12.3 Rotational Energy
12.9. Model: The earth is a rigid, spherical rotating body.
Solve: The rotational kinetic energy of the earth is =
K rot
1 I ω 2 . The
2
moment of inertia of a sphere about its diameter
(see Table 12.2) is I = 52 M earth R and the angular velocity of the earth is
2
2π rad
=7.27 × 10−5 rad/s
24 × 3600 s
ω=
Thus, the rotational kinetic energy is
1 2

K rot =  M earth R 2  ω 2
2 5

1
= (5.98 × 1024 kg)(6.37 × 106 m) 2 (7.27 × 102 5 rad/s) 2 =2.57 × 1029 J
5
12.10. Model: The disk is a rigid body rotating about an axis through its center.
Visualize:
Solve: The speed of the point on the rim is given by vrim
= Rω . The angular velocity ω of the disk can be
1 Iω 2 =
determined from its rotational kinetic energy which is K =
0.15 J. The moment of inertia I of the disk about
2
its center and perpendicular to the plane of the disk is given by
1
1
I = MR 2 = (0.10 kg)(0.040 m) 2 = 8.0 × 102 5 kg m 2
2
2
2(0.15 J)
0.30 J
2
⇒ω =
=
⇒ ω = 61.237 rad/s
I
8.0 × 10−5 kg m 2
Now, we can go back to the first equation to find vrim . We get vrim =
Rω =
(0.040 m)(61.237 rad/s) =
2.4 m/s.
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12-6
Chapter 12
12.11. Model: The triangle is a rigid body rotating about an axis through the center.
Visualize: Please refer to Figure EX12.11. Each 200 g mass is a distance r away from the axis of rotation, where r is
given by
0.20 m
0.20 m
= cos30° ⇒ r =
= 0.2309 m
r
cos30°
Solve: (a) The moment of inertia of the triangle is I =
3 × mr 2 =
3(0.200 kg)(0.2309 m) 2 =
0.032 kg m 2.
(b) The frequency of rotation is given as 5.0 revolutions per s or 10π rad/s. The rotational kinetic energy is
1
1
K rot . = I ω 2 = (0.0320 kg m 2 )(10.0π rad/s) 2 =15.8 J =16 J
2
2
12.12. Model: The baton is a thin rod rotating about a perpendicular axis through its center of mass.
Solve: The moment of inertia of a thin rod rotating about its center
is I
=
1
ML2 . For the baton,
12
1
I=
(0.400 kg)(0.96 m) 2 =
0.031 kg m 2
12
The rotational kinetic energy of the baton is
2
K rot =
1 2 1

 2π rad  min  
I ω = (0.031 kg m 2 )  (100 rev/min) 

  = 1.68 J ≈ 1.7 J
2
2
 rev  60 s  

Section 12.4 Calculating Moment of Inertia
12.13. Model: The moment of inertia of any object depends on the axis of rotation. In the present case, the rotation
axis passes through mass A and is perpendicular to the page.
∑ mi xi mA xA + mB xB + mC xC + mD xD
Solve: (a)=
xcm =
∑ mi
mA + mB + mC + mD
ycm
(100 g)(0 m) + (200 g)(0 m) + (200 g)(0.10 m) + (200 g)(0.10 m)
= 0.057 m
100 g + 200 g + 200 g + 200 g
m y + mB yB + mC yC + mD yD
= A A
mA + mB + mC + mD
(100 g)(0 m) + (200 g)(0.08 m) + (200 g)(0.08 cm) + (200 g )(0 m)
= 0.046 m
700 g
(b) The distance from the axis to mass C is 12.81 cm. The moment of inertia through A and perpendicular to
the page is
=
IA =
∑ mi ri2 =
mA rA2 + mBrB2 + mC rC2 + mD rD2
i
=(0.100 kg)(0 m) 2 + (0.200 kg)(0.08 m) 2 + (0.200 kg)(0.1281 m) 2 + (0.200 kg)(0.10 m) 2 =0.0066 kg m 2
12.14. Model: The moment of inertia of any object depends on the axis of rotation.
Visualize:
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Rotation of a Rigid Body
∑ mi xi mA xA + mB xB + mC xC + mD xD
=
∑ mi
mA + mB + mC + mD
Solve: (a)=
xcm
ycm
12-7
(100 g)(0 m) + (200 g)(0 m) + (200 g)(0.10 m) + (200 g)(0.10 m)
= 0.057 m
100 g + 200 g + 200 g + 200 g
m y + mB yB + mC yC + mD yD
= A A
mA + mB + mC + mD
(100 g)(0 m) + (200 g)(0.08 m) + (200 g)(0.08 cm) + (200 g )(0 m)
= 0.046 m
700 g
(b) The moment of inertia about a diagonal that passes through B and D is
=
=
I BD mA rA2 + mC rC2
where we must compute rA = rC which are the distances from the diagonal. From triangle ABD we see that
8
θ = tan −1   = 38.66°. Now rA = (0.10 m)sin 38.66° = 0.06247 m Thus,
 10 
I BD =
(0.100 kg)(0.06247) 2 + (0.200 kg)(0.06247) 2 =
0.0012 kg m 2
Assess: Note that the masses B and D, being on the axis of rotation, do not contribute to the moment of inertia.
12.15. Model: The three masses connected by massless rigid rods are a rigid body.
Solve: (a) xcm =
∑ mi xi (0.100 kg)(0 m) + (0.200 kg)(0.06 m) + (0.100 kg)(0.12 m)
=
= 0.060 m
∑ mi
0.100 kg + 0.200 kg + 0.100 kg
∑ mi yi
ycm =
=
∑ mi
(0.100 kg)(0 m) + (0.200 kg)
(
)
(0.10 m) 2 − (0.06 m) 2 + (0.100 kg)(0 m)
0.100 kg + 0.200 kg + 0.100 kg
= 0.040 m
(b) The moment of inertia about an axis through A and perpendicular to the page is
IA =
∑ mi ri2 =
mB (0.10 m) 2 + mC (0.10 m) 2 =
(0.100 kg)[(0.10 m) 2 + (0.10 m) 2 ] =
0.0020 kg m 2
(c) The moment of inertia about an axis that passes through B and C is
I BC = mA
(
(0.10 m) 2 − (0.06 m) 2
) = 0.00128 kg m
2
2
≈ 0.0013 kg m 2
Assess: Note that mass mA does not contribute to I A , and the masses mB and mC do not contribute to I BC .
12.16. Model: The door is a slab of uniform density.
Solve: (a) The hinges are at the edge of the door, so from Table 12.2,
1
I=
(25 kg)(0.91 m) 2 =
6.9 kg m 2
3
(b) The distance from the axis through the center of mass along the height of the door is
 0.91 m

d =
− 0.15 m  = 0.305 m. Using the parallel–axis theorem,
2


1
I=
I cm + Md 2 =(25 kg)(0.91 m) 2 + (25 kg)(0.305 cm) 2 =
4.1 kg m 2
12
Assess: The moment of inertia is less for a parallel axis through a point closer to the center of mass.
12.17. Model: The CD is a disk of uniform density.
Solve: (a) The center of the CD is its center of mass. Using Table 12.2,
1
1
I cm = MR 2 = (0.021 kg)(0.060 m) 2 = 3.8 × 10−5 kg m 2
2
2
(b) Using the parallel–axis theorem with d = 0.060 m,
I = I cm + Md 2 = 3.8 × 10−5 kg m 2 + (0.021 kg)(0.060 m) 2 = 1.14 × 10−4 kg m 2
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12-8
Chapter 12
Section 12.5 Torque
12.18. Visualize:


Solve: Torque by a force is defined as τ = Fr sin o where φ is measured counterclockwise from the r vector to the F
vector. The net torque on the pulley about the axle is the torque due to the 30 N force plus the torque due to the 20 N force:
(30 N)r1 sinφ1 + (20 N)r2=
sinφ2 (30 N)(0.02 m) sin ( − 90°) + (20 N)(0.02 m) sin (90°)
= ( − 0.60 N m) + (0.40 N m) =−0.20 N m
Assess: A negative torque causes a clockwise acceleration of the pulley.
12.19. Visualize:
The two equal but opposite 50 N forces, one acting at point P and the other at point Q, make a couple that causes a
net torque.
Solve: The distance between the lines of action
is l d cos30°. The net torque is given by
=
τ =lF =(d cos30°) F =(0.10 m)(0.866)(50 N) =4.3 N m
12.20. Model: The disk is a rotating rigid body.
Visualize:
The radius of the disk is 10 cm and the disk rotates on an axle through its center.
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Rotation of a Rigid Body
12-9
Solve: The net torque on the axle is
τ = FA rA sin φA + FBrB sin φB + FC rC sin φC + FD rD sin φD
= (30 N)(0.10 m)sin(−90°) + (20 N)(0.050 m)sin 90° + (30 N)(0.050 m)sin135° + (20 N)(0.10 m)sin 0°
=
−3 N m + 1 N m + 1.0607 N m =
−0.94 N m
Assess: A negative torque means a clockwise rotation of the disk.
12.21. Model: The beam is a solid rigid body.
Visualize:

The steel beam experiences a torque due to the gravitational force on the construction worker ( FG )C and the

gravitational force on the beam ( FG ) B . The normal force exerts no torque since the net torque is calculated about the
point where the beam is bolted into place.

Solve: The net torque on the steel beam about point O is the sum of the torque due to ( FG )C and the torque due to

( FG ) B . The gravitational force on the beam acts at the center of mass.
=
τ (( FG )C )(4.0 m)sin(−90°) + (( FG )B )(2.0 m)sin(−90°)
=
−(70 kg)(9.80 m/s 2 )(4.0 m) − (500 kg)(9.80 m/s 2 )(2.0 m) =
−12.5 kN m
The negative torque means these forces would cause the beam to rotate clockwise. The magnitude of the torque is
12.5 kN m.
12.22. Model: Model the arm as a uniform rigid rod. Its mass acts at the center of mass.
Visualize:
Solve: (a) The torque is due both to the gravitational force on the ball and the gravitational force on the arm:
=
τ τ ball + τ arm
= (mb g )rb sin 90° + (ma g )ra sin 90°
(3.0
=
kg)(9.8 m/s 2)(0.70 m)+(4.0 kg)(9.8 m/s 2)(0.35 m) 34 N m
(b) The torque is reduced because the moment arms are reduced. Both forces act at φ= 45° from the radial line, so
=
τ τ ball + τ arm
= (mb g )rb sin 45° + (ma g )ra sin 45°
= (3.0 kg)(9.8 m/s 2)(0.70 m)(0.707) + (4.0 kg)(9.8 m/s 2)(0.35 m)(0.707) = 24 N m
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12-10
Chapter 12
Section 12.6 Rotational Dynamics
Section 12.7 Rotation About a Fixed Axis
12.23. Solve: τ = Iα is the rotational analog of Newton’s second law F = ma. We have
=
=
=
τ (2.0 kg m 2 )(4.0
rad/s 2 ) 8.0
kg m 2 /s 2 8.0 N m.
12.24. Visualize: Since α = τ /I , a graph of the angular acceleration looks just like the torque graph with the numerical
values divided by I = 4.0 kg m 2 .
Solve: From the discussion in Chapter 4,
ω=
ωi + area under the angular acceleration α curve between ti and tf
f
The area under the curve between t = 0 s and t = 3 s is 0.50 rad/s. With ω1 = 0 rad/s, we have
ωf =
0 rad/s + 0.50 rad/s =
0.50 rad/s
12.25. Model: Two balls connected by a rigid, massless rod are a rigid body rotating about an axis through the
center of mass. Assume that the size of the balls is small compared to 1 m.
Visualize:
We placed the origin of the coordinate system on the 1.0 kg ball.
Solve: The center of mass and the moment of inertia are
(1.0 kg)(0 m) + (2.0 kg)(1.0 m)
=
= 0.667
=
m and ycm 0 m
xcm
(1.0 kg + 2.0 kg)
∑ mi ri2 =
(1.0 kg)(0.667 m) 2 + (2.0 kg)(0.333 m) 2 =
0.667 kg m 2
I about cm =
We have ωf = 0 rad/s, tf − ti =
−20 rpm =
−20(2pp
rad/60 s) =
− 23 rad/s, so ωf =
5.0 s, and ωi =
ωi + α (tf − ti ) becomes
2π
 2π

0 rad/s = −
rad/s  + aa
(5.0 s) ⇒ = rad/s 2
15
 3

Having found I and α , we can now find the torque τ that will bring the balls to a halt in 5.0 s:
2
 4π
2  2π
=
=
=
I about
rad/s 2  =
N m 0.28 N m
ta
cm
 kg m 
3
 15
 45
The magnitude of the torque is 0.28 N m, applied in the counterclockwise direction.
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Rotation of a Rigid Body
12-11
12.26. Model: The compact disk is a rigid body rotating about its center.
Visualize:
Solve: (a) The rotational kinematic equation ω1 =ω0 + α (t1 − t0 ) gives
200
 2pp

= 0 rad + aa
(2000 rpm) 
(3.0 s − 0 s) ⇒ =
rad/s 2
 rad/s
9
 60 
The torque needed to obtain this operating angular velocity is
 200π

Ia =
(2.5 × 10−5 kg m 2 ) 
rad/s 2  =
1.75 × 10−3 N m
τ=
 9

(b) From the rotational kinematic equation,
1
1  200π

2
θ1 = θ 0 + ω0 (t1 − t0 ) + a (t1 − t0 ) 2 = 0 rad + 0 rad + 
rad/s 2  ( 3.0 s − 0 s )
2
2 9

100π
= 100
=
= 50 rev
revolutions
π rad
2π
Assess: Fifty revolutions in 3 seconds is a reasonable value.
12.27. Model: Model the rod as thin enough to use I = 13 ML2 .
Visualize:
Solve: From Newton’s second law we have ∆L =t∆t. Combine with ∆L = I ∆ω and then solve for ω1.
t∆t = I ∆ω
With ω0 = 0 we have ∆ω = ω1 − ω0 = ω1. Also τ = rF where r = 0.25 m because the rod is hit perpendicular to it.
t∆t rF ∆t (0.25 m)(1000 N)(0.0020 s)
ω1 =∆ω = = 2 =
=
8.0 rad/s
2
I
1 ML
3
1 (0.75 kg)(0.50 m)
3
Assess: The units check out, and 8.0 rad/s seems like a reasonable answer.
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12-12
Chapter 12
Section 12.8 Static Equilibrium
12.28. Model: The rod is in rotational equilibrium, which means that t net = 0.
Visualize:
As the gravitational force on the rod and the hanging mass pull down (the rotation of the rod is exaggerated in the
figure), the rod touches the pin at two points. The piece of the pin at the very end pushes down on the rod; the right
end of the pin pushes up on the rod. To understand this, hold a pen or pencil between your thumb and forefinger, with
your thumb on top (pushing down) and your forefinger underneath (pushing up).
Solve: Calculate the torque about the left end of the rod. The downward force exerted by the pin acts through this

point, so it exerts no torque. To prevent rotation, the pin’s normal force npin exerts a positive torque (ccw about the
left end) to balance the negative torques (cw) of the gravitational force on the mass and rod. The gravitational force
on the rod acts at the center of mass, so
2
2
tt
0 Nm =
net =
pin − (0.40 m)(2.0 kg)(9.8 m/s ) − (0.80 m)(0.50 kg)(9.8 m/s )
⇒ t=
pin 11.8 Nm ≈ 12 N m
12.29. Model: The massless rod is a rigid body.
Visualize:

Solve: To be in equilibrium, the object must be in both translational equilibrium ( Fnet = 0 N) and rotational
equilibrium (t net = 0 N m). We have ( Fnet ) y = (40 N) − (100 N) + (60 N) = 0 N, so the object is in translational
equilibrium. Measuring t net about the left end,
t net =
(60 N)(3.0 m)sin(+90°) + (100 N)(2.0 m)sin(−90°) =
−20 N m
The object is not in equilibrium.
12.30. Model: The object balanced on the pivot is a rigid body.
Visualize:
Since the object is balanced on the pivot, it is in both translational equilibrium and rotational equilibrium.
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Rotation of a Rigid Body
12-13

Solve: There are three forces acting on the object: the gravitational force FG acting through the center of mass of
1

the long rod, the gravitational force FG acting through the center of mass of the short rod, and the normal force
2

P on the object applied by the pivot. The translational equilibrium equation ( Fnet ) y = 0 N is
( )
( )
−( FG )1 − ( FG ) 2 + P= 0 N ⇒ P= ( FG )1 + ( FG ) 2= (1.0 kg)(9.8 m/s 2 ) + (4.0 kg)(9.8 m/s 2 )= 49 N
Measuring torques about the left end, the equation for rotational equilibrium t net = 0 N m is
Pd − w1 (1.0 m) − w2 (1.5 m) =
0Nm
⇒ (49 N)d − (1.0 kg)(9.8 m/s 2 )(1.0 m) − (4.0 kg)(9.8 m/s 2 )(1.5 m)= 0 N ⇒ d = 1.40 m
Thus, the pivot is 1.4 m from the left end.
12.31. Model: The see-saw is a rigid body. The cats and bowl are particles.
Visualize:
Solve: The see-saw is in rotational equilibrium. Calculate the net torque about the pivot point.
t net =
0=
( FG )1 (2.0 m) − ( FG ) 2 (d ) − ( FG ) B (2.0 m)
=
m2 gd m1g (2.0 m) − mB g (2.0 m)
=
d
(m1 − mB )(2.0 m) (5.0 kg − 2.0 kg)(2.0 m)
=
= 1.5 m
m2
4.0 kg
Assess: The smaller cat is close but not all the way to the end by the bowl, which makes sense since the combined
mass of the smaller cat and bowl of tuna is greater than the mass of the larger cat.
Section 12.9 Rolling Motion
12.32. Solve: (a) According to Equation 12.36, the speed of the center of mass of the tire is
vcm = Rω = 20 m/s ⇒ ω =
vcm 20 m/s
 60 
2
=
= 66.67 rad/s = (66.7) 
 rpm = 6.4 × 10 rpm
R
0.30 m
 2p 
(b) The speed at the top edge of the tire relative to the ground is =
vtop 2=
vcm 2(20 m/s)
= 40 m/s.
(c) The speed at the bottom edge of the tire relative to ground is vbottom = 0 m/s.
12.33. Model: The can is a rigid body rolling across the floor.
Solve: The rolling motion of the can is a translation of its center of mass plus a rotation about the center of mass.
The moment of inertia of the can about the center of mass is 12 MR 2 , where R is the radius of the can. Also vcm = Rω ,
where ω is the angular velocity of the can. The total kinetic energy of the can is
K= K cm + K rot=
=
1
1
1
11
 v 
2
2
2
+ I cmω=
+  MR 2  cm 
Mvcm
Mvcm
2
2
2
2 2
 R 
2
3
3
2
=
Mvcm
(0.50 kg)(1.0=
m/s) 2 0.38 J
4
4
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12-14
Chapter 12
12.34. Model: The sphere is a rigid body rolling down the incline without slipping.
Visualize:
The initial gravitational potential energy of the sphere is transformed into kinetic energy as it rolls down.
Solve: (a) If we choose the bottom of the incline as the zero of potential energy, the energy conservation equation
will be K f = U i . The kinetic energy consists of both translational and rotational energy. This means
1
1
1 2
1

2
K f =I cmω 2 + Mvcm
Mgh ⇒  MR 2  ω 2 + M ( Rω ) 2 =
Mgh
=
2
2
2 5
2

7
2 2
⇒ MR=
ω Mg (2.1 m)sin 25°
10
=
⇒ω
g (2.1 m)(sin 25°)
=
R2
10
7
g (2.1 m)(sin 25°)
88 rad/s
=
(0.04 m) 2
10
7
(b) From part (a)
K total =
1
1
7
1
1 2
1

2
=
I cmω 2 + Mvcm
MR 2ω 2 and K rot = I cmω 2 =  MR 2  ω 2 = MR 2ω 2
2
2
10
2
2 5
5

⇒
1 MR 2ω 2
K rot
1 10 2
=5
= × =
K total 7 MR 2ω 2 5 7 7
10
12.35. Model: The mechanical energy of both the hoop (h) and the sphere (s) is conserved. The initial gravitational
potential energy is transformed into kinetic energy as the objects roll down the slope. The kinetic energy is a
combination of translational and rotational kinetic energy. We also assume no slipping of the hoop or of the sphere.
Visualize:
The zero of gravitational potential energy is chosen at the bottom of the slope.
Solve: The energy conservation equation for the sphere or hoop K f + U gf =Ki + U gi is
1
1
1
1
I (ω1 ) 2 + m(v1 ) 2 + mgy=
I (ω0 ) 2 + m(v0 ) 2 + mgy0
1
2
2
2
2
For the sphere, this becomes
2
1 2
1
2  (v1 )s
2
mR

 2 + m(v1 )s + 0 J = 0 J + 0 J + mghs
2 5
2
 R
⇒
7
gh ⇒ (v1 )s =10 gh/7 =10(9.8 m/s 2 )(0.30 m)/7 =
(v1 )s2 =
2.05 m/s
10
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Rotation of a Rigid Body
12-15
For the hoop, this becomes
1
(v ) 2 1
(mR 2 ) 1 2h + m(v1 ) 2h + 0 J = 0 J + 0 J + mghhoop
2
2
R
(v1 ) h2
⇒ hhoop =
g
For the hoop to have the same velocity as that of the sphere,
(v1 )s2 (2.05 m/s) 2
=
= 42.9 cm
g
9.8 m/s 2
The hoop should be released from a height of 43 cm.
h=
hoop
Section 12.10 The Vector Description of Rotational Motion
12.36. Visualize: Please refer to Figure EX12.36. To determine angle α , put the tails of the vectors together.
 
 
Solve: (a) The magnitude of A × B is=
AB sin α (6)(4)sin
=
45° 17. The direction of A × B, using the right-hand
 
rule, is out of the page. Thus, A × B =
(17, out of the page).
  
 
(b) The magnitude of C × D =
0.
is CD sin α (6)(4)sin180
=
° 0. Thus C × D =




12.37. Solve: (a) The magnitude of A × B is=
AB sin α (6)(4)sin
=
45° 21.21. The direction of A × B is given by

 

the right-hand rule. To curl our fingers from A to B, we have to point our thumb into the page. Thus, A × B =
(21, into the page).
 
(b)
=
C × D ((6)(4)sin 90°, out of the
=
page) (24, out of the page).
12.38. Solve: (a) (iˆ × ˆj ) × iˆ = kˆ × iˆ = ˆj
(b) iˆ × ( ˆj × iˆ) =×
iˆ (−kˆ) =
−iˆ × kˆ =
−(− ˆj ) =ˆj
12.39. Solve: (a) iˆ × (iˆ × ˆj ) =×
iˆ kˆ =− ˆj

(b) (iˆ × ˆj ) × kˆ = kˆ × kˆ = 0


12.40. Solve: A × B = (3iˆ + ˆj ) × (3iˆ − 2 ˆj + 2kˆ)
= 9iˆ × iˆ − 6iˆ × ˆj + 6iˆ × kˆ + 3 ˆj × iˆ − 2 ˆj × ˆj + 2 ˆj × kˆ
= 0 − 6kˆ + 6(− ˆj ) + 3(−kˆ) − 0 + 2iˆ = 2iˆ − 6 ˆj − 9kˆ




because iˆ × iˆ =
0. Thus D = niˆ, where n could be any real number.
 


(b) C × E =
6kˆ implies that E must be along the ĵ vector, because iˆ × ˆj =
kˆ. Thus E = 2 ˆj.
 


(c) C × F =
−3 ˆj implies that F must be along the k̂ vector, because iˆ × kˆ =− ˆj. Thus F = 1kˆ.

12.41. Solve: (a) C × D =
0 implies that D must also be in the same or opposite direction as the C vector or zero,



12.42. Solve: τ = r × F = (5iˆ + 5 ˆj ) × (−10 ˆj ) N m

[ 50(iˆ × ˆj ) − 50( ˆj × ˆj )] N m =−
[ 50(+ kˆ) − 0] N m =−50kˆ N m
=−



12.43. Solve: L =×
(3.0iˆ + 2.0 ˆj ) m × (0.1 kg)(4.0ˆj ) m/s
r mv =
= 1.20(iˆ × ˆj ) kg m 2 /s + 0.8( ˆj × ˆj ) kg m 2 /s = 1.20kˆ kg m 2 /s + 0 kg m 2 /s
= 1.20kˆ kg m 2 /s or (1.20 kg m 2 /s, out of page)
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12-16
Chapter 12
12.44. Solve:
 

L = r × mv = (1.0iˆ + 2.0 ˆj ) m × (0.200 kg)(3.0 m/s)(cos 45°iˆ − sin 45° ˆj )
=(0.42iˆ × iˆ − 0.42iˆ × ˆj + 0.85 ˆj × iˆ − 0.85 ˆj × ˆj ) kg m 2 /s =−(1.27 kˆ) kg m 2 /s or (1.27 kg m 2 /s, into page)
Section 12.11 Angular Momentum
12.45. Model: The bar is a rotating rigid body. Assume that the bar is thin.
Solve: The angular velocity
rpm (120)(2pp
)/60
rad/s 4 rad/s. From Table 12.2, the moment of inertia
=
ω 120
=
=
1 ML2 . The angular momentum is
of a rod about its center is I = 12
1
=
= 2.1 kg m 2 /s
L I=
ω   (0.50 kg)(2.0 m) 2 (4π rad/s)
12
 
If we wrap our fingers in the direction of the rod’s rotation, our thumb will point in the z direction or out of the page.
Consequently,

L = (2.1 kg m 2 /s, out of the page)
12.46. Model: The disk is a rotating rigid body.
Solve: From Table 12.2, the moment of inertia of the disk about its center is
1
1
2
2
I
MR=
(2.0 kg)(0.020 m)=
4.0 × 10−4 kg m 2
=
2
2
The angular velocity ω is 600 rpm =
Iω =
(4.0 × 10−4 kg m 2 )(20π rad/s) =
600 × 2pp
/60 rad/s =
20 rad/s. Thus, L =
0.025 kg m 2 /s. If we wrap our right fingers in the direction of the disk’s rotation, our thumb will point in the
− x direction. Consequently,

L=
−0.025 iˆ kg m 2 /s =
(0.025 kg m 2 /s, into page)
12.47. Model: The bowling ball is a solid sphere.
Solve: From Table 12.2, the moment of inertia about a diameter of a solid sphere is
2
2
(5.0 kg)(0.11
m) 2 0.0242 kg m 2
=
I =
MR 2
=
5
5
Require
=
L 0.23 kg m 2=
/s I=
ω (0.0243 kg m 2 )ω
⇒ω =
(9.5 rad/s)
 rev   60 s 
In rpm, this is (9.5 rad/s) 

 = 91 rpm.
 2p rad   min 
12.48. Model: Model the turntable as a rigid disk rotating on frictionless bearings. As the blocks fall from above
and stick on the turntable, the turntable slows down due to increased rotational inertia of the (turntable + blocks)
system. Any torques between the turntable and the blocks are internal to the system, so angular momentum of the
system is conserved.
Visualize: The initial moment of inertia is I1 and the final moment of inertia is I 2 .
Solve: The initial moment of inertia is=
I1 I=
disk
2
1 mR
=
2
1 (2.0
2
2
kg)(0.10 m)
=
0.010 kg m 2 and the final moment
of inertia is
=
I 2 I1 + 2=
mR 2 0.010 kg m 2 + 2(0.500 kg) × (0.10=
m) 2 0.010 kg m 2 + 0.010 kg=
m 2 0.020 kg m 2
Let ω1 and ω2 be the initial and final angular velocities. Then
Lf =Li ⇒ ω2 I 2 =ω1I1 ⇒ ω2 =
I1ω1 (0.010 kg m 2 )(100 rpm)
=
=50 rpm
I2
0.020 kg m 2
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Rotation of a Rigid Body
12-17
12.49. Model: Model all the rest of the body other than the arms as one object (call it the trunk, even though it
includes head and legs), then we can write
+ 2 yarm marm
y
m
ycm = trunk trunk
M
where M = mtrunk + marm = 70 kg (the mass of the whole body).
Visualize: The language “by how much does he raise his center of mass” makes us think of writing ∆ycg.
Since we have modeled the arm as a uniform cylinder 0.75 m long, its own center of gravity is at its geometric center,
0.375 m from the pivot point at the shoulder. So raising the arm from hanging down to straight up would change the
height of the center of gravity of the arm by twice the distance from the pivot to the center of gravity: (∆ycg)arm =
2(0.375 m) = 0.75 m.
Solve:
( ycm ) trunk mtrunk + 2( ycm )arm, up marm
∆ ( ycm )=
body ( ycm ) with arms up − ( ycm ) with arms =
down
M
( ycm ) trunk mtrunk + 2( ycm )arm, down marm
−
M
2marm
2marm
2(3.5 kg)
=
(( ycm )arm, up − ( ycm )arm, down ) =
(∆ycm)arm =
(0.75 m) = 0.075 m = 7.5 cm
M
M
70 kg
Assess: 7.5 cm seems like a reasonable amount, not a lot, but not too little. The trunk term subtracted out, which is
both expected and good because we didn’t know ( ycg ) trunk.
12.50. Model: The structure is a rigid body rotating about its center of mass.
Visualize:
We placed the origin of the coordinate system on the 300 g ball.
Solve: First, we calculate the center of mass:
(300 g)(0 cm) + (600 g)(40 cm)
=
xcm = 26.67 cm
300 g + 600 g
Next, we will calculate the moment of inertia about the structure’s center of mass:
I =(300 g)( xcm ) 2 + (600 g)(40 cm − xcm ) 2
= (0.300 kg)(0.2667 m) 2 + (0.600 kg)(0.1333 m) 2 = 0.032 kg m 2
Finally, we calculate the rotational kinetic energy:
2
=
K rot
1 2 1
 100 × 2π

=
=
Iω
(0.032 kg m 2 ) 
rad/s
 1.75 J ≈ 1.8 J
2
2
 60

12.51. Model: The wheel is a rigid rolling body.
Visualize:
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12-18
Chapter 12
Solve: The front of the disk is moving forward at velocity vcm . Also, because of rotation the point is moving
downward at velocity v=
ω vcm . So, this point has a speed
rel R=
v=
2
2
vcm
+ vcm
=
2vcm =
2(20 m/s) = 28 m/s
Assess: The speed v is independent of the radius of the wheel.
12.52. Visualize:
Solve: We will consider a vertical strip of width dx and of mass dm at a position x from the origin. The formula for
the x component of the center of mass is
1
xcm =
x dm
M∫
The area of the=
steel plate is A
1=
(0.2 m)(0.3 m)
2
0.030 m 2 . Mass dm in the strip is the same fraction of M as dA is
of A. Thus
dm dA
M
 0.800 kg 
2
=
⇒ dm = dA = 
 dA = (26.67 kg/m )l dx
M
A
A
 0.030 m 2 
The relationship between l and x is
l
x
2
=
=
⇒l
x
0.20 m 0.30 m
3
Therefore,
0.3 m
=
xcm
1
(17.78 kg/m 2 ) x3
(17.78 kg/m 2 ) (0.3 m) 2
2 2 2
=
=
= 20 cm
x
dx
(26.67
kg/m
)
 
M∫
M
3
0.8 kg
3
3
0m
Due to symmetry ycm = 0 cm.
12.53. Model: The disk is a rigid rotating body. The axis is perpendicular to the plane of the disk.
Visualize:
Solve: (a) From Table 12.2, the moment of inertia of a disk about its center is
1
1
=
I =
MR 2
(2.0 kg)(0.10
=
m) 2 0.010 kg m 2
2
2
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Rotation of a Rigid Body
12-19
(b) To find the moment of inertia of the disk through the edge, we can make use of the parallel axis theorem:
I = I center + Mh 2 = (0.010 kg m 2 ) + (2.0 kg)(0.10 m) 2 = 0.030 kg m 2
Assess: The larger moment of inertia about the edge means there is more inertia to rotational motion about the edge
than about the center.
12.54. Model: The object is a rigid rotating body. Assume the masses m1 and m2 are small and the rod is thin.
Visualize: Please refer to Figure P12.54.
Solve: The moment of inertia of the object is the sum of the moment of inertia of the rod, mass m1, and
mass m2 . Using Table 12.2 for the moment of inertia of the rod, we get
2
= I rod about center + I m1 + I m=
I object
2
=
1
L
L
ML2 + m1   + m2  
12
2
4
2
1
1
1
L2  M
m2 
ML2 + m1L2 + m2=
L2
 + m1 +

12
4
16
4 3
4 
2
1
Assess: With m
=
=
1 m
2 0 kg, I rod 12 ML , as expected.
12.55. Visualize:
We chose the origin of the coordinate system to be on the axis of rotation, that is, at a distance d from one end of the rod.
Solve: The moment of inertia can be calculated as follows:
x2
I = ∫ x 2 dm
and
x1
⇒ I=
For d = 0 m, I =
1 ML2 ,
3
M
L
L−d
∫
−d
M
x 2dx= 
 L
and for d =
3 L−d
x

 3
=
−d
dm dx
M
= ⇒ dm = dx
M
L
L
1 M

3 L
M

3
3
]
[( L − d )3 + d 3 ]
[( L − d ) − (− d ) =
3L

1 L,
2
3
3
M  L   L   1
ML2
  +   =

3L  2   2   12
Assess: The special cases d = 0 m and d = L/2 of the general formula give the same results that are found in Table 12.2.
=
I
12.56. Visualize:
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12-20
Chapter 12
Solve: We solve this problem by dividing the disk between radii r1 and r2 into narrow rings of mass dm. Let
dA = 2π rdr be the area of a ring of radius r. The mass dm in this ring is the same fraction of the total mass M as dA
is of the total area A.
(a) The moment of inertia can be calculated as follows:
M
M
I disk ∫ r 2dm and=
dm =
dA
=
(2π r )dr
2
A
π (r2 − r12 )
=
⇒ I disk
M
r2
r
2
2M r 4
r 3dr
=
2
2 ∫
2
(r2 − r1 ) r
(r2 − r12 ) 4
2M
2
(2π r ) dr
∫r =
π (r22 − r12 ) r
1
1
=
2M
(r24
2
2
4(r2 − r1 )
− r14=
)
r2
r1
M 2 2
(r2 − r1 )
2
Replacing r1 with r and r2 with R, the moment of inertia of the disk through its center is
=
I disk
1 M (R2
2
+ r 2 ).
(b) For r = 0 m, I disk = 12 MR 2 . This is the moment of inertia for a solid disk or cylinder about the center.
Additionally, for r ≅ R, we have I = MR 2 . This is the expression for the moment of inertia of a cylindrical hoop or
ring about the center.
(c) The initial gravitational potential energy of the disk is transformed into kinetic energy as it rolls down. If we
choose the bottom of the incline as the zero of potential energy, and use vcm = ω R, the energy conservation equation
K f = U i is
2
 2 2 vcm 1
2
Mgyi =
Mg (0.50 m)sin 20°
 ( R + r ) 2 + Mvcm =
2
R

2
2
r2 
2 R +r  1 2
2 1 1
v
v
⇒ vcm
+
=
+
+
= 1.6759 m 2 /s 2



cm
cm
2  2
 2 4 4 R 2 
R
4




2 

3 (0.015 m)
2
1.6759 m 2 /s 2 ⇒ v=
=
vcm
 +
cm 1.37 m/s ≈ 1.4 m/s
2

4
4(0.020
m)


For a sliding particle on a frictionless surface K f = U i , so
1 2 1
1 M
2
I ω + Mvcm
Mgh ⇒ 
=
2
2
2 2
1 2
mv
=
mgyi ⇒=
vf
f
2
=
2 gy
i
vcm
=
° 1.83 m/s ⇒ =
2 g (0.50 m)sin 20
0.75
vf
That is, vcm is 75% of the speed of a particle sliding down a frictionless ramp.
12.57. Model: The plate has uniform density.
Visualize:
Solve: The moment of inertia is
I = ∫ r 2dm.
M
M
=
dA
dx dy.
A
L2
L
L
L
L
x 2 + y 2 . With − ≤ x ≤ and − ≤ y ≤ ,
2
2
2
2
Let the mass of the plate be M. Its area is L2 . A region of area dA located at (x, y) has mass=
dm
The distance from the axis of rotation to the point (x, y) is=
r
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Rotation of a Rigid Body
L
2
L
2
L
2

M  x3
2
2 M 
I=
+ y2x 
∫ ∫ ( x + y )  L2  dx dy =
2 ∫ 

L L  3
L L

−
2
−
−
2
L
2
dy
−
2
L
2
L

 L3 y 2 L  − L3 y 2 L  
M 2  L3
M  L4
y3
2
dy= 2 ∫  + Ly dy= 2  + L
= 2 ∫ +
−
−

 24

2
2  
3
L L  24
L L  12
L  12


−
−

2
2
M
=
12-21
L
2



L
− 
2
L
2
 L3 L3   1 2
M  L4

L
+
 +   = ML

L2  12
 24 24   6
12.58. Solve: From Equation 12.16,
=
I
∫ (x
2
+ y 2 )dm
A small region of area dA has mass dm, and
=
dm
The area of the plate is
1 (0.20
2
M
M
=
dA
dx dy
A
A
m)(0.30 m) = 0.030 m 2 . So
M  0.800 kg 
2
= =
 26.67 kg m
A  0.030 m 2 
1
1
1
The limits for x are 0 ≤ x ≤ 30 cm. For a particular value of x, − x ≤ y ≤ x. Note that ± is the slope of the top
3
3
3
and bottom edges of the triangle. Therefore,
=
I
1
x
30 cm 3
∫
0
∫
( x 2 + y 2 )(26.67 kg/m 2 )dx dy
1
− x
3
1
= (26.67 kg/m )
2
= (26.67 kg/m 2 )
30 cm 
x
0
− x
3
∫
y3  3
2
dx
 x y + 
3  1

30 cm 
∫
0
2 3 2 x3 
 x +
dx
81 
3
30 cm
 56  1 
kg/m 2 )   x 4 
0.037 kg m 2
= (26.67
=
 81  4  0
12.59. Model: Assume the woman is in equilibrium, so ∑ F =
0 and ∑τ =
0.
Visualize: Choose the axis to be the left end of the board.
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12-22
Chapter 12
Solve: Use ∑τ =
0.
L
( mb ) g= 0 N ⋅ m
2
1
L
L(scalereading) g − (mb ) g L[(scalereading) − mb ]
2
2
d
=
= =
mw g
mw
∑τ= L(scalereading) g − d ( mw ) g −
1
(2.5 m)[(25 kg) − (6.1 kg)]
2
= 0.91 m
60 kg
Assess: This is a little more than halfway up the body of a woman of average height.
12.60. Model: The ladder is a rigid rod of length L. To not slip, it must be in both translational equilibrium

( Fnet = 0 N) and rotational equilibrium (t net = 0 N m). We also apply the model of static friction.
Visualize:

Since the wall is frictionless, the only force from the wall on the ladder is the normal force n2 . On the other hand, the



floor exerts both the normal force n1 and the static frictional force fs . The gravitational force FG on the ladder acts
through the center of mass of the ladder.


Solve: The x- and y-components of Fnet = 0 N are
∑ Fx = n2 − fs = 0 N ⇒ fs = n2
∑ Fy = n1 − FG = 0 N ⇒ n1 = FG
The minimum angle occurs when the static friction is at its maximum value fs max = ms n1. Thus we have
n=
f=
2
s ms n=
1 ms mg . We choose the bottom corner of the ladder as a pivot point to obtain t net , because two forces
pass through this point and have no torque about it. The net torque about the bottom corner is
t net = d1mg − d 2n2 = (0.5L cosθ min )mg − ( L sin θ min ) ms mg = 0 N m
0.5 0.5
1.25 ⇒ θ min =
51°
⇒ 0.5cosθ min =
ms sin θ min ⇒ tan θ min = = =
ms 0.4
12.61. Model: The beam is a rigid body of length 3.0 m and the student is a particle.
Visualize:
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Rotation of a Rigid Body
12-23


Solve: To stay in place, the beam must be in both translational equilibrium ( Fnet = 0 N) and rotational equilibrium
=
(t net 0 N m). The first condition is
∑ Fy =
−( FG ) beam − ( FG )student + F1 + F2 =
0N
2
⇒ F1 +=
F2 ( FG ) beam + ( FG )student
= (100 kg + 80 kg)(9.80 m/s
=
) 1764 N
Taking the torques about the left end of the beam, the second condition is
−( FG ) beam (1.5 m) − ( FG )student (2.0 m) + F2 (3.0 m) =
0Nm
−(100 kg)(9.8 m/s 2 )(1.5 m) − (80 kg)(9.8 m/s 2 )(2.0 m) + F2 (3.0 m) =
0Nm
⇒=
F2 1013 N ≈ 1000 N
From F1 + F2 =
1764 N, we get F1 =
1764 N − 1013 N =
0.75 kN.
Assess: To establish rotational equilibrium, the choice for the pivot is arbitrary. We can take torques about any point
on the body of interest.
12.62. Model: The structure is a rigid body.
Visualize:
Solve: We pick the left end of the beam as our pivot point. We don’t need to know the forces Fh and Fv because the
pivot point passes through the line of application of Fh and Fv and therefore these forces do not exert a torque. For the
beam to stay in equilibrium, the net torque about this point is zero. We can write
t about left end =
−( FG ) B (3.0 m) − ( FG ) W (4.0 m) + (T sin150°)(6.0 m) =
0Nm
Using
=
( FG ) B (1450 kg)(9.8 m/s 2 ) and =
( FG ) W (80 kg)(9.8 m/s 2 ), the torque equation can be solved to yield
T = 15,300 N. The tension in the cable is slightly more than the cable rating. The worker should be worried.
12.63. Model: Model the beam as a rigid body. For the beam not to fall over, it must be both in translational


equilibrium ( Fnet = 0 N) and rotational equilibrium =
(t net 0 Nm).
Visualize:
The boy walks along the beam a distance x, measured from the left end of the beam. There are four forces acting on


the beam. F1 and F2 are from the two supports, ( FG ) b is the gravitational force on the beam, and ( FG ) B is the
gravitational force on the boy.
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12-24
Chapter 12
Solve: We pick our pivot point on the left end through the first support. The equation for rotational equilibrium is
−( FG ) b (2.5 m) + F2 (3.0 m) − ( FG ) B x =
0Nm
−(40 kg)(9.80 m/s 2 )(2.5 m) + F2 (3.0 m) − (20 kg)(9.80 m/s 2 ) x =
0Nm
The equation for translation equilibrium is
∑ Fy = 0 N = F1 + F2 − ( FG ) b − ( FG ) B
2
⇒ F1 + F=
) 588 N
2 ( FG ) b + ( FG )=
B (40 kg + 20 kg)(9.8 m/s =
Just when the boy is at the point where the beam tips, =
F1 0 N. Thus F2 = 588 N. With this value of F2 , we can
simplify the torque equation to:
−(40 kg)(9.80 m/s 2 )(2.5 m) + (588 N)(3.0 m) − (20 kg)(9.80 m/s 2 ) x =
0Nm
⇒ x =4.0 m
Thus, the distance from the right end is 5.0 m − 4.0 m = 1.0 m.
12.64. Solve: The bricks are stable when the net gravitational torque on each individual brick or combination of
bricks is zero. This is true as long as the center of gravity of each individual brick and any combination is over a base
of support. To determine the relative positions of the bricks, work from the top down. The top brick can extend past
the second brick by L/2. For maximum extension, their combined center of gravity will be at the edge of the third
brick, and the combined center of gravity of the three upper bricks will be at the edge of the fourth brick. The
combined center of gravity of all four bricks will be over the edge of the table.
Measuring from the left edge of brick 2, the center of gravity of the top two bricks is
L
m   + mL
m1x1 + m2 x2
3
2
=
L
( x12 )com = =
m1 + m2
2m
4
Thus the top two bricks can extend L/4 past the edge of the third brick. The top three bricks have a center of mass
L
 3L 
 5L 
m  + m  + m 
m1x1 + m2 x2 + m3 x3
2
4
  =
 
 4  5L
( x123 )com =
=
m1 + m2 + m3
3m
6
Thus the top three bricks can extend past the edge of the fourth brick by L/6. Finally, the four bricks have a combined
center of mass at
L
 4L 
 11L 
 17L 
m  + m  + m
 + m

2
6 
12 



 12  7 L
=
(x1234 )com =
4m
8
The center of gravity of all four bricks combined is 7 L/8 from the left edge of the bottom brick, so brick 4 can
extend L/8 past the table edge. Thus the maximum distance to the right edge of the top brick from the table edge is
L L L L 25
d max = + + + =
L
8 6 4 2 24
Thus, yes, it is possible that no part of the top brick is directly over the table because d max > L.
Assess: As crazy as this seems, the center of gravity of all four bricks is stably supported, so the net gravitational
torque is zero, and the bricks do not fall over.
12.65. Model: The pole is a uniform rod. The sign is also uniform.
Visualize:
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Rotation of a Rigid Body
12-25
Solve: The geometry of the rod and cable give the angle that the cable makes with the rod.
 250 
 = 51.3°
 200 
θ = tan2 1 
The rod is in rotational equilibrium about its left-hand end.
1
 
1
 
0=
t net =
−(100 cm)(FG ) P − (80 cm)   (FG )S − (200 cm)   (FG )S + (200 cm)T sin 51.3°
2
2
=
−(100 cm)(5.0 kg)(9.8 m/s 2 ) − mS (9.8 m/s 2 )(140 cm) + (156 cm)T
With T =
300 N , mS =
30.6 kg ≈ 31 kg.
Assess: A mass of 30.6 kg is reasonable for a sign.
12.66. Model: The sacs are constrained by the stamens. The gravitational torque is 2000 times less than the
straightening torque, so ignore it in part b.
Visualize: Refer to the diagram below.
Solve: (a) The moment of inertia of the stamen and sac can be calculated using kinematic equations.
2
R
I = mstamen   + msac R 2 = (1.0 × 10−6 kg)(5.0 × 10−4 m) 2 + (1.0 × 10−6 kg)(1.0 × 10−3 m) 2 = 1.25 × 10−12 kg ⋅ m 2
2
An additional significant figure has been kept in this intermediate result.
The “straightening torque” is
τ=
Ia =
(1.25 × 10−12 kg ⋅ m 2 )(2.32 × 107 rad/s 2 ) =
2.91 × 10−5 N ⋅ m ≈ 2.9 × 10−5 N ⋅ m
(b) We can use the kinematic equations to find the angular acceleration of a sac, and then the corresponding
tangential acceleration.
2∆θ
a=
∆t
2
(
2π rad
)
2(60°) 360°
=
=2.32 × 107 rad/s 2
−3 2
(0.30 × 10 s)
The tangential acceleration is
at =a R =(2.32 × 107 rad/s 2 )(1.0 × 10−3 m) =2.32 × 104 m/s 2
Using the kinematic equations with the result above gives
vf =at ∆t =(2.32 × 104 m/s 2 )(0.30 × 10−3 s) =7.0 m/s
Assess: These results seem reasonable. The accelerations are huge, while the torque is relatively small. This is due to
the relatively low moment of inertia of the structure.
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12-26
Chapter 12
12.67. Model: The bar is a solid body rotating through its center.
Visualize:
Solve: (a) The two forces form a couple. The net torque on the bar about its center is
Iα
t net = LF = Iα ⇒ F =
L
where F is the force produced by one of the air jets. We can find I and α as follows:
1
1
I=
ML2 =
(0.50 kg)(0.60 m) 2 =
0.015 kg m 2
12
12
(t1 − t0 ) ⇒ 150 rpm =5.0 rad/s =0 rad + (10 s − 0 s) ⇒ =0.50 rad/s 2
ω1 =ω0 + apaap
(0.015 kg m 2 )(0.5p rad/s 2 )
⇒F=
=0.0393 N
(0.60 m)
The force
=
F 39 mN.
(b) The torque of a couple is the same about any point. It is still t net
= LF . However, the moment of inertia has
changed.
LF
1
1
where I = ML2 = (0.500 kg)(0.6 m) 2 =0.060 kg m 2
taa
net =LF =I ⇒ =
I
3
3
(0.0393 N) × (0.60 m)
⇒a =
=0.393 rad/s 2
0.060 kg m 2
Finally,
ω1 = ω0 + a (t1 − t0 ) = 0 rad/s + (0.393 rad/s 2 )(10 s − 0 s)
(3.93)(60)
3.93 rad/s =
rpm =
37.5 rpm
=
2p
The angular speed is 38 rpm.
Assess: Note that ω ∝ α and α ∝ 1/I . Thus, ω ∝ 1/I . I about the center of the rod is 4 times smaller than I about one
end of the rod. Consequently, ω is 4 times larger.
12.68. Model: The flywheel is a rigid body rotating about its central axis.
Visualize:
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Rotation of a Rigid Body
12-27
Solve: (a) The radius of the flywheel is R = 0.75 m and its mass =
is M 250 kg. The moment of inertia about the
axis of rotation is that of a disk:
1
1
I=
MR 2 =
(250 kg)(0.75 m) 2 =
70.31 kg m 2
2
2
The angular acceleration is calculated as follows:
2
2
taat
net =I ⇒ = net /I =(50 N m)/(70.31 kg m ) =0.711 rad/s
Using the kinematic equation for angular velocity gives
(t1 − t0 ) = 1200 rpm = 40 rad/s = 0 rad/s + 0.711 rad/s 2 (t1 − 0 s)
ω1 = ω0 + ap
177 s
⇒ t1 =
(b) The energy stored in the flywheel is rotational kinetic energy:
1
1
K rot = I ω12 = (70.31 kg m 2 )(40π rad/s) 2 =5.55 × 105 J
2
2
The energy stored is 5.6 × 105 J.
energy delivered (5.55 × 105 J)/2
(c) Average power delivered =
=
=1.39 × 105 W ≈ 140 kW
time interval
2.0 s
 ω full energy − ω half energy 
∆ω
(d) Because=
tat
I , ⇒= I = I 
= ω1 (from part (a))
= 40p rad/s. ωhalf energy
 . ωfull energy
∆t
∆t


can be obtained as:
1 2
1
K
5.55 × 105 J
I ω half energy = K rot ⇒ ωhalf energy = rot =
88.85 rad/s
=
2
2
I
70.31 kg m 2
Thus
 40 π rad/s − 88.85 rad/s 
τ=
(70.31 kg m 2 ) 
1.30 kN m
=

2.0 s

12.69. Model: The pulley is a rigid rotating body. We also assume that the pulley has the mass distribution of a
disk and that the string does not slip.
Visualize:
Because the pulley is not massless and frictionless, tension in the rope on both sides of the pulley is not the same.
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12-28
Chapter 12
Solve: Applying Newton’s second law to m1, m2 , and the pulley yields the three equations:
T1 − ( FG=
)1 m1a1
− ( FG ) 2 +=
T2 m2a2
T2 R − T1R − 0.50 Nm
= Ia
Noting that − a2 = a1 = a, I = 12 mp R 2 , and a = a/R, the above equations simplify to
T1 − m1=
g m1a
0.50 Nm
1
 a  1 0.50 Nm 1
=
T2 − =
T1  mp R 2   +
mp a +
R
2
0.060 m
2
 R  R
=
m2 g − T
2 m2 a
Adding these three equations,
1 

(m2 − m1=
) g a  m1 + m2 + mp  + 8.333 N
2 

(m − m1 ) g − 8.333 N (4.0 kg − 2.0 kg)(9.8 m/s 2 ) − 8.333 N
⇒a= 2
=
=1.610 m/s 2
m1 + m2 + 12 mp
2.0 kg + 4.0 kg + (2.0 kg/2)
We can now use kinematics to find the time taken by the 4.0 kg block to reach the floor:
1
1
y1 = y0 + v0 (t1 − t0 ) + a2 (t1 − t0 ) 2 ⇒ 0 = 1.0 m + 0 + (−1.610 m/s 2 )(t1 − 0 s) 2
2
2
2(1.0 m)
⇒ t1 =
=1.1 s
(1.610 m/s 2 )
12.70. Model: Assume the string does not slip on the pulley.
Visualize:
The free-body diagrams for the two blocks and the pulley are shown. The tension in the string exerts an upward force
on the block m2 , but a downward force on the outer edge of the pulley. Similarly the string exerts a force on block
m1 to the right, but a leftward force on the outer edge of the pulley.
Solve: (a) Newton’s second law for m1 and m2 is T = m1a1 and T − m2 g =
m2a2 . Using the constraint −a2 =
+ a1 =
a,
we have T = m1a and −T + m2 g =
m2a. Adding these equations, we get m2=
g (m1 + m2 )a, or
a=
m2 g
m1m2 g
⇒ T= m1a=
m1 + m2
m1 + m2
(b) When the pulley has mass m, the tensions (T1 and T2 ) in the upper and lower portions of the string are different.
Newton’s second law for m1 and the pulley are:
T1 =
m1a
and
T1R − T2 R =
− Ia
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Rotation of a Rigid Body
12-29
We are using the minus sign with α because the pulley accelerates clockwise. Also, =
a Ra . Thus, T1 = m1a and
I a aI
=
R R R2
T2 − T1=
Adding these two equations gives
I 

=
T2 a  m1 + 2 
R 

Newton’s second law for m2 is T2 − m2 g =
m2a2 =
− m2a. Using the above expression for T2 ,
I 
m2 g

a  m1 + 2  + m2a= m2 g ⇒ a=
R 
m1 + m2 + I/R 2

Since I = 12 mp R 2 for a disk about its center,
a=
m2 g
m1 + m2 + 12 mp
With this value for a we can now find T1 and T2:
=
T1 m=
1a
m1m2 g
m1 + m2 + 12 mp
2
=
T2 a (m1 + I/R=
)
m2 g
1  m2 (m1 + 12 mp ) g

+
m
m=
1
p

(m1 + m2 + 12 mp ) 
2  m1 + m2 + 12 mp
Assess: For m = 0 kg, the equations for a, T1 and T2 of part (b) simplify to
=
a
m2 g
m1m2 g
m1m2 g
=
and T1
=
and T2
m1 + m2
m1 + m2
m1 + m2
These agree with the results of part (a).
12.71. Model: The disk is a rigid spinning body.
Visualize: Please refer to Figure P12.71. The initial angular velocity is 300 rpm or (300)(2
=
π )/60 10 π rad/s. After
3.0 s the disk stops.
Solve: Using the kinematic equation for angular velocity,
ω1 = ω0 + aa
(t1 − t0 ) ⇒ =
ω1 − ω0
t1 − t0
=
(0 rad/s − 10π rad/s) −10π
=
rad/s 2
(3.0 s − 0 s)
3
Thus, the torque due to the force of friction that brings the disk to rest is
( 1 mR 2 )a
Ia
1
1
π


τ =Iaa
=− fR ⇒ f =2
=2 2
=− (mR ) =− (2.0 kg)(0.15 m)  −10 rad/s 2  =1.57 N ≈ 1.6 N
R
R
2
2
3


The minus sign with τ = − fR indicates that the torque due to friction acts clockwise.
12.72. Model: Assume the turbine is a rigid rotating body.
Visualize: Start with Newton’s second law: ∑τ =
Iα . The net torque is just the frictional torque we seek.
Solve: Apply the definition of α .
=
t I=
α I
∆ω
∆t
1ω .
When the turbine has reduced its rotation speed by 50% then ∆ω =
2 i
I
∆t = ωi
2t
This suggests that a graph of ∆t vs. ωi should be a straight line whose slope is I/2τ .
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12-30
Chapter 12
We see that the fit is quite good and that the slope is 0.1188 s 2 , so the frictional torque is
I
2.6 kg ⋅ m 2
=
= 10.94 N ⋅ m ≈ 11 N ⋅ m
2 ⋅ slope 2(0.1188 s 2 )
Assess: Your boss wants the frictional torque to be as small as possible, but 11 N ⋅ m seems reasonable.
=
τ
12.73. Model: Assume that the hollow sphere is a rigid rolling body and that the sphere rolls up the incline without
slipping. We also assume that the coefficient of rolling friction is zero.
Visualize:
The initial kinetic energy, which is a combination of rotational and translational energy, is transformed in
gravitational potential energy. We chose the bottom of the incline as the zero of the gravitational potential energy.
Solve: The conservation of energy equation K f + U gf =Ki + U gi is
1
1
1
1
2
2
M (v1 )cm
+ I cm (ω1 ) 2 + Mgy
=
M (v0 )cm
+ I cm (ω0 ) 2 + Mgy0
1
2
2
2
2
0 J + 0 J +=
Mgy1
1
1 2
1
1
(v ) 2

2
2
2
+  MR 2  (ω0 )cm
+ 0 J ⇒=
+ MR 2 0 2cm
M (v0 )cm
Mgy1
M (v0 )cm
2
2 3
2
3
R

5 (v ) 2
5
5 (5.0 m/s) 2
0 cm
2
⇒
=
= 2.126 m
(v0 )cm
y1 6 =
6
6 9.8 m/s 2
g
The distance traveled along the incline is
y1
2.126 m
s=
=
= 4.3 m
sin 30°
0.5
Assess: This is a reasonable stopping distance for an object rolling up an incline when its speed at the bottom of the
incline is approximately 10 mph.
⇒=
gy1
12.74. Model: The disk is a rigid body rotating on an axle passing through one edge. The gravitational potential
energy is transformed into rotational kinetic energy as the disk is released.
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Rotation of a Rigid Body
12-31
Visualize:
We placed the origin of the coordinate system at a distance R just below the axle. In the initial position, the center of
mass of the disk is at the same level as the axle. The center of mass of the disk in the final position is coincident with
the origin of the coordinate system.
Solve: (a) The torque is due to the gravitational force on the disk acting at the center of mass. Thus
τ=
(mg ) R =
(5.0 kg)(9.8 m/s 2 )(0.30 m) =
14.7 Nm
The moment of inertia about the disk’s edge is obtained using the parallel-axis theorem:
1
3
3
2
I=
I cm + mR 2 =
mR 2 + mR 2 =
mR 2 =
0.675 kg m 2
  (5.0 kg)(0.30 m) =
2
2
2
 
14.7 Nm
τ
⇒a = =
= 22 rad/s 2
I 0.675 kg m 2
(b) The energy conservation equation K f + U gf =Ki =U gi is
1 2
1
1
I ω1 + mgy1 = I ω02 + mgy0 ⇒ I ω12 + 0 J = 0 J + mgR
2
2
2
ω1 =
2mgR
=
I
2(5.0 kg)(9.8 m/s 2 )(0.30 m)
= 6.6 rad/s
0.675 kg m 2
Assess: An angular velocity of 6.6 rad/s (or 1.05 revolutions/s) as the center of mass of the disk reaches below the
axle is reasonable.
12.75. Model: The hoop is a rigid body rotating about an axle at the edge of the hoop. The gravitational torque on
the hoop causes it to rotate, transforming the gravitational potential energy of the hoop’s center of mass into
rotational kinetic energy.
Visualize:
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12-32
Chapter 12
We placed the origin of the coordinate system at the hoop’s edge on the axle. In the initial position, the center of
mass is a distance R above the origin, but it is a distance R below the origin in the final position.
Solve: (a) Applying the parallel-axis theorem, I edge =I cm + mR 2 =mR 2 + mR 2 =2mR 2 . Using this expression in the
energy conservation equation K f + U gf =Ki + U gi yields:
1
1
I edgeω12 + mgy1 = I edgeω02 + mgy0
2
2
(b) The speed of the lowest point on the hoop is
2g
R
1
(2mR 2 )ω12 − mgR = 0 J + mgR ⇒ ω1 =
2
2g
(2 R )
8 gR
=
R
Assess: Note that the speed of the lowest point on the loop involves a distance of 2R instead of R.
v (ω
=
=
1 )(2 R )
12.76. Model: The long, thin rod is a rigid body rotating about a frictionless pivot on the end of the rod. The
gravitational torque on the rod causes it to rotate, transforming the gravitational potential energy of the rod’s center
of mass into rotational kinetic energy.
Visualize:
We placed the origin of the coordinate system at the pivot point. In the initial position, the center of mass is a
distance 12 L above the origin. In the final position, the center of mass is at y = 0 m and thus has zero gravitational
potential energy.
Solve: (a) The energy conservation equation for the rod K f + U gf =Ki + U gi is
11 2 2
J 0 J + mg ( L/2) ⇒ ω=
1
 mL  ω1 + 0=
2 3

1 2
1 2
I ω1 + mgy=
I ω0 + mgy0
1
2
2
(b) The speed at the tip of the rod is=
vtip (ω=
1) L
3 g/L
3gL .
12.77. Model: The sphere attached to a thin rod is a rigid body rotating about the rod. Assume the rod is vertical
and the sphere solid.
Visualize: Please refer to Figure P12.77. The sphere rotates because the string wrapped around the rod exerts a torque τ .
Solve: The torque exerted by the string on the rod is τ= Tr .
From the parallel-axis theorem, the moment of inertia of the sphere about the rod’s axis is
2
2
MR 2 13
R
I off center =I cm + M   = MR 2 +
= MR 2
2
5
4
20
 
From Newton’s second law,
α=
τ
Tr
20Tr
=
=
I (13MR 2 /20) 13MR 2
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Rotation of a Rigid Body
12-33
12.78. Model: The angular momentum of the satellite in the elliptical orbit is a constant.
Visualize:
Solve: (a) Because the gravitational force is always along the same direction as the direction of the moment arm
  
vector, the torque τ = r × Fg is zero at all points on the orbit.
(b) The angular momentum of the satellite at any point on the elliptical trajectory is conserved. The velocity is

perpendicular to r at points a and b, so β= 90° and=
L mvr . Thus
r 
Lb = La ⇒ mvb rb = mva ra ⇒ vb =  a  va
 rb 
30,000 km
30,000 km
− 9000 km= 6000 km and rb=
+ 9000 km= 24,000 km
ra=
2
2
 6000 km 
=
⇒ vb  =
 (8000 m/s) 2000 m/s
 24,000 km 
(c) Using the conservation of angular momentum Lc = La , we get
r 
mvc rc sin β c = mva ra ⇒ vc =  a  va /sin β c
rc = (9000 km) 2 + (12,000 km) 2 = 1.5 × 107 m
r
 c
From the figure, we see that sin β c =12,000/15,000 =0.80. Thus
 6000 km  (8000 m/s)
=
vc =
4000 m/s

0.80
 15,000 km 
12.79. Model: For the (bullet + door) system, the angular momentum is conserved in the collision.
Visualize:
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12-34
Chapter 12


Solve: As the bullet hits the door, its velocity v is perpendicular to r. Thus the initial angular momentum about the
rotation axis, with r = L, is
Li =
mBvB L =
(0.010 kg)(400 m/s)(1.0 m) =
4.0 kg m 2 /s
After the collision, with the bullet in the door, the moment of inertia about the hinges is
1
1
mD L2 + mB L2 = (10.0 kg)(1.0 m) 2 + (0.010 kg)(1.0 m) 2 =3.343 kg m 2
=
I I door +=
I bullet
3
3
Therefore, Lf =
Iω =
(3.343 kg m 2 )ω . Using the angular momentum conservation equation Lf =
Li (3.343 kg m 2 )ω =
1.2 rad/s.
4.0 kg m 2 /s and thus ω =
12.80. Model: Model the turntable as a rigid disk rotating on frictionless bearings. For the (turntable + block)
system, no external torques act as the block moves outward towards the outer edge. Angular momentum is thus
conserved.
Visualize: The initial moment of inertia of the turntable is I1 and the final moment of inertia is I 2 .
1 mR 2 =
1 (0.2 kg)(0.2 m) 2 =
Solve: The initial moment of inertia is I1 =
I disk =
0.0040 kg m 2 . As the block reaches
2
2
the outer edge, the final moment of inertia is
I 2 =I1 + mB R 2 =0.0040 kg m 2 + (0.020 kg)(0.20 m) 2
= 0.0040 kg m 2 + 0.0008 kg m 2 = 0.0048 kg m 2
Let ω1 and ω2 be the initial and final angular velocities, then the conservation of angular momentum equation is
Lf =Li ⇒ ω2 I 2 =ω1I1 ⇒ ω2 =
I1ω1 (0.0040 kg m 2 )(60 rpm)
=
=50 rpm
I2
(0.0048 kg m 2 )
Assess: A change of angular velocity from 60 rpm to 50 rpm with an increase in the value of the moment of inertia
is reasonable.
12.81. Model: Model the merry-go-round as a rigid disk rotating on frictionless bearings about an axle in the center
and John as a particle. For the (merry-go-round + John) system, no external torques act as John jumps on the merrygo-round. Angular momentum is thus conserved.
Visualize: The initial angular momentum is the sum of the angular momentum of the merry-go-round and the
angular momentum of John. The final angular momentum as John jumps on the merry-go-round is equal to
I finalωfinal .
Solve: John’s initial angular momentum is that of a particle:
=
LJ mJ vJ R sin
=
β mJ vJ R. The angle β= 90° since
John runs tangent to the disk. The conservation of angular momentum equation Lf = Li is
1

I finalωfinal= Ldisk + LJ=  MR 2  ωi + mJ vJ R
2

2p  rad 
1
=   (250 kg)(1.5 m) 2 (20 rpm) 
=
.5 m) 814 kg m 2 /s
 + (30 kg)(5.0 m/s)(1
2
60
rpm
 


814 kg m 2 /s
⇒ ωfinal =
I final
I final= I disk + I J=
1
1
MR 2 + mJ R 2=
(250 kg)(1.5 m) 2 + (30 kg)(1.5 m) 2= 349 kg m 2
2
2
814 kg m 2 /s
ωfinal =
349 kg m 2
2.33 rad/s =
22 rpm
=
12.82. Model: Model the skater as a cylindrical torso with two rod-like arms that are perpendicular to the axis of
the torso in the initial position and collapse into the torso in the final position.
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Rotation of a Rigid Body
12-35
Visualize:
Solve: For the initial position, the moment of inertia is I1 = I Torso + 2 I Arm . The moment of inertia of each arm is that
of a 66-cm-long rod rotating about a point 10 cm from its end, and can be found using the parallel-axis theorem. In
the final position, the moment of inertia =
is I 2
1 MR 2 .
2
The equation for the conservation of angular momentum
Lf = Li can be written I 2=
ω2 I1ω1=
⇒ ω2 ( I1/I 2 )ω1. Calculating I1 and I 2 ,
1
1

I1 =
M T R 2 + 2  M A L2A + M A d 2 
2
12

1
1

2
(40 kg)(0.10 m) + 2  (2.5 kg)(0.66 m) 2 + (2.5 kg)(0.33 m + 0.10 m) 2  =
=
1.306 kg m 2
2
12


I2
=
1
1
(1.306 kg m 2 )
2
2
MR=
(45 kg)(0.10 m)=
0.225 kg m 2 ⇒ ω=
(1.0 rev/s=
) 5.8 rev/s
2
2
2
(0.225 kg m 2 )
12.83. Model: The toy car is a particle located at the rim of the track. The track is a cylindrical hoop rotating about
its center, which is an axis of symmetry. No net torques are present on the track, so the angular momentum of the car
and track is conserved.
Visualize:
Solve: The toy car’s steady speed of 0.75 m/s relative to the track means that
vc − vt = 0.75 m/s ⇒ vc = vt + 0.75 m/s,
where vt is the velocity of a point on the track at the same radius as the car. Conservation of angular momentum
implies that
Li = Lf
0 = I cωc + I tωt = (mr 2 )ωc + ( Mr 2 )ωt = mωc + M ωt
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12-36
Chapter 12
The initial and final states refer to before and after the toy car was turned on. Table 12.2 was used for the track.
v
v
Since ωc = c , ωt = t , we have
r
r
=
0 mvc + Mvt
⇒ m(vt + 0.75 m/s) + Mvt =0
M
(0.200 kg)
(0.75 m/s) =
(0.75 m/s) =
⇒ vt =
−
−
−0.125 m/s
m+M
(0.200 kg + 1.0 kg)
The minus sign indicates that the track is moving in the opposite direction of the car. The angular velocity of the track is
vt (0.125 m/s)
=
=
wt =
0.417 rad/s clockwise.
r
0.30 m
In rpm,
 rev   60 s 
ωt= (0.417 rad/s) 


 2p rad   min 
= 4.0 rpm
Assess: The speed of the track is less than that of the car because it is more massive.
12.84. Model: Assume that the marble does not slip as it rolls down the track and around a loop-the-loop. The
mechanical energy of the marble is conserved.
Visualize:
Solve: The marble’s center of mass moves in a circle of radius R − r . The free-body diagram on the marble at its
highest position shows that Newton’s second law for the marble is
mv 2
mg + n = 1
R−r
The minimum height (h) that the track must have for the marble to make it around the loop-the-loop occurs when the
normal force of the track on the marble tends to zero. Then the weight will provide the centripetal acceleration
needed for the circular motion. For n → 0 N,
mg =
mv 2
⇒ v12 = g ( R − r )
(R − r)
Since rolling motion requires v12 = r 2ω12 , we have
ω12r 2= g ( R − r ) ⇒ ω12=
g (R − r)
r2
The conservation of energy equation is
1
1
( K f + U gf ) top of loop =( Ki + U gi )initial ⇒ mv12 + I ω12 + mgy1 =mgy0 =mgh
2
2
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Rotation of a Rigid Body
12-37
Using the above expressions and I = 52 mr 2 the energy equation simplifies to
1
1 2
 g (R − r) 
mg ( R − r ) +   mr 2 
 + mg 2( R − r ) =mgh ⇒ h =2.7( R − r )
2
2 5
 r2 
12.85. Model: The Swiss cheese wedge is of uniform density—or at least uniform enough that its center of mass is
at the same location as that of a solid piece. To find the angle at which the cheese starts sliding, the cheese will be
treated as a particle, and the model of static friction will be used.
Visualize:
Solve: The angle at which the cheese starts sliding, θS , will be compared to the critical angle θ c for stability. Use
Newton’s second law with the free body diagram.
( Fnet ) x= 0= fs − FG sin θs
( Fnet ) y= 0= n − FG cosθs
With FG = mg , the y-direction equation gives n = mg cosθ . The cheese starts sliding when µs is at its maximum
value. Combining that with the x-direction equation and fs = µs n,
0 ms (mg cosθs ) − mg sin θs
=
⇒ θs = tan −1 ( ms ) = tan −1 (0.90) = 42°
The cheese will start sliding at an angle of 42°.
The center of mass of the cheese wedge can be found using the result of Problem 12.52. There, the center of mass of
a triangle with the same proportions as the cheese wedge was found. So xcm is at the center of the cheese wedge
(by symmetry). The ycm can be found by proportional reasoning.
ycm
(30 cm − 20 cm)
=
⇒ ycm =
4.0 cm
12 cm
30 cm
Note that here we have measured ycm from the base of the wedge.
Stability considerations require that the center of mass be no farther than the left corner of the wedge. At the critical
angle geometry shown in the figure above, the right triangle formed by the wedge’s center of mass, lower-left corner,
and center point of the base is a 45°-45°-90° triangle. So θ=
c 45°.
The cheese will slide first as the incline reaches 42°. It would not topple until the angle reaches 45°. So Emily is
correct.
Assess: Both Emily’s and Fred’s suppositions are plausible. The calculation must be done to find out which is right.
12.86. Model: Define the system as the rod and cube. Energy and angular momentum are conserved in a perfectly
elastic collision in the absence of a net external torque. The rod is uniform.
Visualize: Please refer to Figure CP12.86.
Solve: Let the final speed of the cube be vf , and the final angular velocity of the rod be ω . Energy is conserved,
and angular momentum around the rod’s pivot point is conserved.
1 2 1 2 1
Ei =⇒
Ef
mv0 =mvf + I rodω 2
2
2
2
d 
d 
Li =
Lf ⇒ mv0   =
mvf   + I rodω
2
2
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12-38
Chapter 12
This is two equations in the two unknowns vf and ω . From Table 12.2,
=
I rod
1
1
1
=
Md 2
(2
=
m) d 2
md 2
12
12
6
From the angular momentum equation,
v0 = vf +
d
ω
3
⇒ω =
3
(v0 − vf )
d
Substituting into the energy equation,
1 2 1 2 11
 9 
mv0 =
mvf +  md 2  2  (v0 − vf ) 2
2
2
2 6
 d 
3
v02 =
vf2 + (v0 − vf ) 2
2
1
2 6
0 =−
vf
v0vf + v02
5
5
This is a quadratic equation in vf . The roots are
2
 v2 
6
6 
v0 ±  v0  − 4  0 
 5 
5
5 
2
  3
=
vf
= v0 ± v0
2
5
5
 1 v0
= 5
 v0
1
The answer vf = v0 means the ice cube missed the rod. So vf = v0 to the right.
5
12.87. Model: The clay ball is a particle. The rod is a uniform thin rod rotating about its center. Angular momentum
is conserved in the collision.
Visualize:
Solve: This is a two-part problem. Angular momentum is conserved in the collision, and energy is conserved as the
ball rises like a pendulum. The angular momentum conservation equation about the rod’s pivot point is
Li = Lf ⇒ mv0 r =
( I ball + rod )ω
Note r=
L
1
= 15 cm. The rod and ball are a composite object. From Table 12.2, I rod = ML2 , so
12
2
1
L2 1
L2 
M
ML2 =
m + ML2 =  m + 
12
4 12
4
3 
vf 2vf
= =
If vf is the final velocity of the clay ball, ω
since the ball sticks to the rod. Thus
r
L
I ball + rod =
I ball + I rod =
mr 2 +
mv0 L L2 
M  2vf 
=
 m + 

2
4
3  L 
mv
(0.010 kg)(2.5 m/s)
⇒ vf = 0 =
=0.714 m/s
M
(0.075 kg)
m+
(0.010 kg) +
3
3
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Rotation of a Rigid Body
12-39
Energy is conserved as the clay ball rises. Compare the energy of the ball-rod system just after the collision to when
the ball reaches the maximum height. Note that the center of mass of the rod does not change position.
1
Ei =
Ef ⇒ ( I rod + ball )ω 2 =
mgh
2
Thus
2
M  2vf 
M
1 L2 
2
 m + 
 = mgh ⇒ vf  m +
2 4
3  L 
3


 = mgL(1 − cosθ )

vf2 
M
cosθ
m +  =
mgL 
3 
Using the various values, cosθ = 0.393 ⇒ θ = 67°.
L
9.1 cm. This is about 2/3 of the height of the pivot point, and is
Assess: The clay ball rises h =(1 − cosθ ) =
2
reasonable.
⇒1−
12.88. Model: Because no external torque acts on the star during gravitational collapse, its angular momentum is
conserved. Model the star as a solid rotating sphere.
Solve: (a) The equation for the conservation of angular momentum is
2

2

Li =Lf ⇒ I iωi =I f ωf ⇒  mRi2  ωi = mRf2  ωf
5

5

⇒ Rf =
Ri
ωi
ωf
The angular velocity is inversely proportional to the period T. We can write
Rf = Ri
Tf
0.10 s
= (7.0 × 108 m)
= 1.3749 × 105 m = 137 km
Ti
2.592 × 106 s
(b) A point on the equator rotates with =
r Rf . Its speed is
v=
2π Rf 2π (137,490 m)
=
= 8.6 × 106 m/s
T
0.10 s
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.