Chapter 10 Rotation of a Rigid Object About a Fixed Axis

P10.3 (a) 212.0 rad s4.00 rad s

3.00 si

tω ω

α−

= = =

(b) ( )( )22 21 1 4.00 rad s 3.00 s 18.0 rad2 2i t tθ ω α= + = =

P10.5 100 rev 1 min 2 rad 10 rad s1.00 min 60.0 s 1.00 rev 3i

π πω ⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠, 0fω =

(a) ( )0 10 / 3

5.24 s2.00

f it sω ω π

α− −

= = =−

(b) 10 10rad s s 27.4 rad2 6 6

f if t t

ω ω π πθ ω+⎛ ⎞ ⎛ ⎞⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠⎝ ⎠

P10.9 5.00 rev s 10.0 rad sω π= = . We will break the motion into two stages: (1) a period

during which the tub speeds up and (2) a period during which it slows down.

While speeding up, ( )10 10.0 rad s

8.00 s 40.0 rad2

tπ

θ ω π+

= = =

While slowing down, ( )210.0 rad s 0

12.0 s 60.0 rad2

tπ

θ ω π+

= = =

So, total 1 2 100 rad 50.0 revθ θ θ π= + = =

P10.10 (a) v rω= ; 45.0 m s

0.180 rad s250 m

vr

ω = = =

(b) ( )22245.0 m s

8.10 m s toward the center of track250 mr

var

= = =

P10.14 (a) 25.0 m s

25.0 rad s1.00 m

vr

ω = = =

(b) ( )2 2 2f iω ω α θ= + Δ

( )

( )( ) ( )

22 2225.0 rad s 0

39.8 rad s2 2 1.25 rev 2 rad rev

f iω ωα

θ π

− −= = =

Δ ⎡ ⎤⎣ ⎦

(c) 2

25.0 rad s0.628 s

39.8 rad st ω

αΔ

Δ = = =

P10.21 (a) 2j j

j

I m=∑ r

In this case,

( ) ( )

[ ]

1 2 3 4

2 2

2

2

3.00 m 2.00 m 13.0 m

13.0 m 3.00 2.00 2.00 4.00 kg

143 kg m

r r r r

r

I

= = =

= + =

⎡ ⎤= + + +⎣ ⎦

= ⋅

(b) ( )( )22 21 1 143 kg m 6.00 rad s2 2RK Iω= = ⋅

32.57 10 J= ×

x (m)

y (m)

1

2

4

3

0 1 2 3

1

2

4

1 3

2.00 kg2.00 kg2.00 kg3.00 kg3.00 kg3.00 kg

2.00 kg2.00 kg2.00 kg 4.00 kg4.00 kg4.00 kg

FIG. P10.21 *P10.22 , 1 4.00 kgm = 1 1 3.00 mr y= =

, 2 2.00 kgm = 2 2 2.00 mr y= =

, 3 3.00 kgm = 3 3 4.00 mr y= =

2.00 rad sω = about the x-axis (a) 2 2

1 1 2 2 3 3xI m r m r m r= + + 2

( ) ( ) ( )

( )( )

2 2 2 2

22

4.00 3.00 2.00 2.00 3.00 4.00 92.0 kg m

1 1 92.0 2.00 184 J2 2

x

R x

I

K I ω

= + + =

= = =

⋅

FIG. P10.22

(b) ( )1 1 3.00 2.00 6.00 m sv r ω= = =

( )( )221 1 1

1 1 4.00 6.00 72.0 J2 2

K m v= = =

( )2 2 2.00 2.00 4.00 m sv r ω= = =

( )( )222 2 2

1 1 2.00 4.00 16.0 J2 2

K m v= = =

( )3 3 4.00 2.00 8.00 m sv r ω= = =

( )( )223 3 3

1 1 3.00 8.00 96.0 J2 2

K m v= = =

21 2 3

172.0 16.0 96.0 184 J2 xK K K K I ω= + + = + + = =

(c) The kinetic energies computed in parts (a) and (b) are the same. Rotational kinetic energy can be viewed as the total translational kinetic energy of the particles in the rotating object.

P10.23 ( )22I Mx m L x= + −

( )2 2dI Mx m L xdx

= − − = 0 (for an extremum)

mLxM m

∴ =+

2

2 2 2d I m Mdx

= + ; therefore I is at a minimum when the axis

of rotation passes through mLxM m

=+

which is also the

center of mass of the system. The moment of inertia about an axis passing through x is

2 2

2 2CM 1mL m MmI M m L L

M m M m M m2Lμ⎡ ⎤ ⎡ ⎤= + − =⎢ ⎥ ⎢ ⎥+ + +⎣ ⎦ ⎣ ⎦

x

MMM mmm

L

L−xx

L

L−xx

= FIG. P10.23

where

MmM m

μ =+

axis of rotation

z

x

y

P10.26 We assume the rods are thin, with radius much less than L. Call the junction of the rods the origin of coordinates, and the axis of rotation the z-axis.

For the rod along the y-axis, 213

I mL= from the table.

For the rod parallel to the z-axis, the parallel-axis theorem gives

2

2 21 12 2 4

LI mr m m⎛ ⎞= + ≅⎜ ⎟⎝ ⎠

L

FIG. P10.26

In the rod along the x-axis, the bit of material between x and x dx+ has mass m dxL

⎛ ⎞⎜ ⎟⎝ ⎠

and is at distance 2

2

2Lr x ⎛ ⎞= + ⎜ ⎟

⎝ ⎠ from the axis of rotation. The total rotational inertia

is:

22

2 2 2total

2

2 232

22

2 2 22

1 13 4 4

712 3 4

7 1112 12 4 12

L

L

L L

LL

L mI mL mL xL

m x mLmL xL

mL mL mLmL

−

−−

⎛ ⎞⎛ ⎞= + + + ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞= + +⎜ ⎟⎝ ⎠

= + + =

∫ dx

Note: The moment of inertia of the rod along the x axis can also be calculated from the

parallel-axis theorem as 2

2112 2

LmL m⎛ ⎞+ ⎜ ⎟⎝ ⎠

.

P10.27 Treat the tire as consisting of three parts. The two sidewalls are each treated as a hollow cylinder of inner radius 16.5 cm, outer radius 30.5 cm, and height 0.635 cm. The tread region is treated as a hollow cylinder of inner radius 30.5 cm, outer radius 33.0 cm, and height 20.0 cm.

Use ( 2 21 2

12

I m R R= + ) for the moment of inertia of a hollow cylinder.

Sidewall:

( ) ( ) ( )( )

( ) ( ) ( )

2 2 3 3 3

2 2 2 2side

0.305 m 0.165 m 6.35 10 m 1.10 10 kg m 1.44 kg

1 1.44 kg 0.165 m 0.305 m 8.68 10 kg m2

m

I

π −

−

⎡ ⎤= − × × =⎣ ⎦

⎡ ⎤= + = × ⋅⎣ ⎦

Tread:

( ) ( ) ( )( )

( ) ( ) ( )

2 2 3 3

2 2 2tread

0.330 m 0.305 m 0.200 m 1.10 10 kg m 11.0 kg

1 11.0 kg 0.330 m 0.305 m 1.11 kg m2

m

I

π ⎡ ⎤= − × =⎣ ⎦

⎡ ⎤= + = ⋅⎣ ⎦

Entire Tire: ( )2 2 2

total side tread2 2 8.68 10 kg m 1.11 kg m 1.28 kg mI I I −= + = × ⋅ + ⋅ = ⋅ 2

P10.30 We consider the cam as the superposition of the original solid disk and a disk of

negative mass cut from it. With half the radius, the cut-away part has one-quarter the face area and one-quarter the volume and one-quarter the mass of the original solid cylinder:

0M

0 014

M M− = M 043

M M=

By the parallel-axis theorem, the original cylinder had moment of inertia

2 2

2 2CM 0 0 0 0

1 32 2 4 4R RI M M R M M R⎛ ⎞+ = + =⎜ ⎟

⎝ ⎠

The negative-mass portion has 2 2

00

1 12 4 2 32

M RRI M⎛ ⎞⎛ ⎞= − = −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

. The whole cam has

2

2 2 200 0

3 23 23 44 32 32 32 3 24

M RI M R M R MR MR= − = = = 223 and

2 2 21 1 23 232 2 24 48

K I MR MR2 2ω ω ω= = =

P10.33 ( ) ( ) ( )0.100 m 12.0 N 0.250 m 9.00 N 0.250 m 10.0 N 3.55 N mτ = − − = −∑ ⋅ The thirty-degree angle is unnecessary information.

FIG. P10.33

P10.37 For , 1m

: y yF ma=∑ 1 0n m g+ − =

1 1 19.6 Nn m g= = 1 1 7.06 Nk kf nμ= =

: x xF ma=∑ ( )17.06 N 2.00 kgT− + = a (1)

For the pulley,

Iτ α=∑ : 21 2

12

aT R T R MRR

⎛ ⎞− + = ⎜ ⎟⎝ ⎠

( )1 21 10.0 kg2

T T a− + =

(2)(1 2 5.00 kgT T a− + = )

For , 2m 2 2 cos 0n m g θ+ − =

( )( )22 6.00 kg 9.80 m s cos30.0

50.9 Nn = °

=

FIG. P10.37

2 2k kf nμ= 18. : 3 N= 2 2 218.3 N sinT m m aθ− − + = ( )218.3 N 29.4 N 6.00 kgT a− − + = (3)

(a) Add equations (1), (2), and (3):

( )

2

7.06 N 18.3 N 29.4 N 13.0 kg4.01 N 0.309 m s13.0 kg

a

a

− − + =

= =

(b) ( )21 2.00 kg 0.309 m s 7.06 N 7.67 NT = + =

( )22 7.67 N 5.00 kg 0.309 m s 9.22 NT = + =

P10.38 ( )( )22 21 1 100 kg 0.500 m 12.5 kg m2 2

I mR= = = ⋅

2

50.0 rev min 5.24 rad s0 5.24 rad s

0.873 rad s6.00 s

i

f i

t

ωω ω

α

= =− −

= = = −

( )2 212.5 kg m 0.873 rad s 10.9 N mIτ α= = ⋅ − = − ⋅ FIG. P10.38

The magnitude of the torque is given by 10.9 N mfR = ⋅ , where f is the force of friction.

Therefore,

10.9 N m0.500 m

f ⋅= and kf nμ=

yields 21.8 N 0.31270.0 Nk

fn

μ = = =

P10.53 (a) ( )( )22trans

1 1 10.0 kg 10.0 m s 500 J2 2

K mv= = =

(b) ( )( )2

22 2rot 2

1 1 1 1 10.0 kg 10.0 m s 250 J2 2 2 4

vK I mrr

ω⎛ ⎞⎛ ⎞= = = =⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

(c) total trans rot 750 JK K K= + =

P10.56 2 22

1 1 12 2 2

IK mv I m vR

ω ⎡ ⎤= + = +⎢ ⎥⎣ ⎦2 where v

Rω = since no slipping

occurs. Also, , =iU mgh 0fU = , and 0iv =

Therefore, 22

12

Im v mR

⎡ ⎤+ =⎢ ⎥⎣ ⎦gh

Thus, ( )

=⎡ ⎤+⎣ ⎦

22

21

ghv

I mR

For a disk, 212

I mR=

So 212

21

ghv =

+ or disk

43gh

v =

For a ring, so 2I mR= 2 22gh

v = or ringv g= h

Since , disk ringv v> the disk reaches the bottom first.

*P10.62 (a) We consider the elevator-sheave-counterweight-Earth system, including n

passengers, as an isolated system and apply the conservation of mechanical energy. We take the initial configuration, at the moment the drive mechanism switches off, as representing zero gravitational potential energy of the system.

Therefore, the initial mechanical energy of the system is Ei = Ki + Ui = (1/2) mev

2 + (1/2) mcv2 + (1/2)Isω

2

= (1/2) mev2 + (1/2) mcv

2 + (1/2)[(1/2)msr2](v/r)2

= (1/2) [me + mcv2 + (1/2)ms] v

2

The final mechanical energy of the system is entirely gravitational because the system is momentarily at rest:

Ef = Kf + Uf = 0 + megd − mcgd

where we have recognized that the elevator car goes up by the same distance d that the counter- weight goes down. Setting the initial and final energies of the system equal to each other, we have

(1/2) [me + mc + (1/2)ms] v2 = (me − mc) gd

(1/2) [800 kg + n 80 kg + 950 kg + 140 kg](3 m/s)2 = (800 kg + n 80 kg − 950 kg)(9.8

m/s2) d d = [1890 + 80n](0.459 m)/(80n − 150) (b) d = [1890 + 80×2](0.459 m)/( 80×2 − 150) = 94.1 m (c) d = [1890 + 80×12](0.459 m)/( 80×12 − 150) = 1.62 m (d) d = [1890 + 80×0](0.459 m)/( 80×0 − 150) = −5.79 m

(e) The rising car will coast to a stop only for n ≥ 2. For n = 0 or n = 1, the car would accelerate upward if released. (f) The graph looks roughly like one branch of a hyperbola. It comes down steeply from 94.1 m for n = 2, flattens out, and very slowly approaches 0.459 m as n becomes large. (g) The radius of the sheave is not necessary. It divides out in the expression (1/2)Iω2 = (1/4)msheavev

2 .

(h) In this problem, as often in everyday life, energy conservation refers to minimizing use of electric energy or fuel. In physical theory, energy conservation refers to the constancy of the total energy of an isolated system, without regard to the different prices of energy in different forms. (i) The result of applying ∑F = ma and ∑τ = Iα to elevator car, counterweight, and sheave, and adding up the resulting equations is

(800 kg + n 80 kg − 950 kg)(9.8 m/s2) = [800 kg + n 80 kg + 950 kg + 140 kg]a a = (9.80 m/s2)(80n − 150)/(1 890 + 80n) downward

P10.69 fτ will oppose the torque due to the hanging object:

fI TRτ α τ= = −∑ : f TR Iτ α= − (1)

Now find T, I and α in given or known terms and substitute into equation (1).

: yF T mg ma= − = −∑ ( )T m g a= − (2)

also 2

2iaty v tΔ = + 2

2ya

t= (3)

FIG. P10.69

and

2

2yaR Rt

α = = (4)

2

212 2 8

RI M R MR⎡ ⎤⎛ ⎞= + =⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

25 (5)

Substituting (2), (3), (4), and (5) into (1), we find

( )2

2 2 2

22 25 58 4f

MR y2

y y Mym g R R m g

t Rt tτ

t⎡ ⎤⎛ ⎞ ⎛ ⎞= − − = − −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

P10.75 (a) Let represent the radius of the Earth. The base of the building moves east at ER

1v REω= where ω is one revolution per day. The top of the building moves east at ( )2 Ev Rω= + h . Its eastward speed relative to the ground is

2 1v v hω− = . The object’s time of fall is given by 2102

y gtΔ = + , 2htg

= .

During its fall the object’s eastward motion is unimpeded so its deflection

distance is ( )1 2

3 22 1

2 2hx v v t h hg g

ω ω⎛ ⎞

Δ = − = = ⎜ ⎟⎝ ⎠

(b) ( )1 22

3 22 rad 2 s50 m 1.16 cm86 400 s 9.8 mπ ⎛ ⎞

=⎜ ⎟⎝ ⎠

(c) The deflection is only 0.02% of the original height, so it is negligible in many

practical cases.

P10.77 21:2

aF T Mg Ma TR I MRR

τ α ⎛ ⎞= − = − = = = ⎜ ⎟⎝ ⎠∑ ∑

(a) Combining the above two equations we find ( )T M g a= −

and 2TaM

= FIG. P10.77

thus 3

MgT =

(b) 2 2 23 3

MgTa gM M

⎛ ⎞= = =⎜ ⎟⎝ ⎠

(c) ( )2 2 2f i f iv v a x x= + − ( )2 20 2 03fv g h⎛ ⎞= + −⎜ ⎟

⎝ ⎠

43fgh

v =

For comparison, from conservation of energy for the system of the disk and the

Earth we have

+ + = + +rot trans rot transgi i i gf fU K K U K K f :

22 21 1 10 0 0

2 2 2f

f

vMgh MR Mv

R⎛ ⎞⎛ ⎞+ + = + +⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

43fgh

v =

rm

h RP

P10.79 (a) rot trans 0K K UΔ + Δ + Δ = Note that initially the center of mass of the sphere is a

distance above the bottom of the loop; and as the mass reaches the top of the loop, this distance above the reference level is 2

h r+

R r− . The conservation of energy requirement gives

( ) ( ) 2 21 1+ = 2 + +2 2

mg h r mg R r mv Iω− FIG. P10.79

For the sphere 225

I mr= and v rω= so that the expression becomes

272 210

gh gr gR v+ = + (1)

Note that when the speed of the sphere at the top of the loop satisfies

the condition minh h=

( )

2mvF mgR r

= =−∑ or ( )2v g R r= −

Substituting this into Equation (1) gives ( ) ( )min 2 0.700h R r R= − + − r or ( )min 2.70 2.70h R r= − = R

(b) When the sphere is initially at 3h R= and finally at point P, the conservation of

energy equation gives

( ) 2 21 13 + + +2 5

mg R r mgR mv mv= , or

( )2 10 27

v R r= + g

Turning clockwise as it rolls without slipping past point P, the sphere is

slowing down with counterclockwise angular acceleration caused by the

torque of an upward force f of static friction. We have yF ma= y∑ and

Iτ α=∑ becoming f mg m rα− = − and 225

fr mr α⎛ ⎞= ⎜ ⎟⎝ ⎠

.

Eliminating f by substitution yields 57gr

α = so that 57yF m= −∑ g

( )2 10/ 7 (2 ) 20

=7x

R r mgmvF n mgR r R r

+ −= − = − = −

− −∑ (since ) R r>>

P10.83 (a) CMxF F f Ma= + =∑

n f

Mg

F

FR fR Iτ α= − =∑

( ) CMCM

IaFR Ma F RR

− − = CM43

FaM

= FIG. P10.83

(b) CM4 13 3

Ff Ma F M F FM

⎛ ⎞= − = − =⎜ ⎟⎝ ⎠

(c) ( )2 2 2f i f iv v a x x= + −

83fFdvM

=

P10.84 Call the frictional force exerted by each roller

backward on the plank. Name as the rolling resistance exerted backward by the ground on each roller. Suppose the rollers are equally far from the ends of the plank.

tf

bf

M

m R m R

F

For the plank, FIG. P10.84

x xF ma=∑ ( )6.00 N 2 6.00 kgt pf a− =

The center of each roller moves forward only half as far as the plank. Each roller has

acceleration 2pa

and angular acceleration

( ) ( )

25.00 cm 0.100 m

p pa a=

Then for each,

x xF ma=∑ ( )2.00 kg2p

t b

af f+ − =

Iτ α=∑ ( ) ( ) ( ) ( )215.00 cm 5.00 cm 2.00 kg 5.00 cm2 10.0 cm

pt b

af f+ =

So 1 kg2t bf f a⎛ ⎞+ = ⎜ ⎟

⎝ ⎠ p

Add to eliminate : bf ( )2 1.50 kgt pf a=

(a) And ( ) ( )6.00 N 1.50 kg 6.00 kgp pa a− =

( )( )

26.00 N0.800 m s

7.50 kgpa = =

For each roller, 20.400 m s2pa

a= =

(b) Substituting back, ( ) 22 1.50 kg 0.800 m stf =

( )2

0.600 N

10.600 N kg 0.800 m s2

0.200 N

t

b

b

f

f

f

=

+ =

= −

The negative sign means that the horizontal force

of ground on each roller is 0.200 N forward

rather than backward as we assumed.

Mg

nt

6.00 N

ft ntft

ntft

ntft

fb nbfb nb

mg mg

FIG. P10.84(b)

P10.85 reads . If we take torques around the center of mass,xF ma=∑

fn

T

mg

x f T ma− + =

we can use Iτ α=∑ , which reads 2 1fR TR Iα+ − = . For rolling without

slipping, 2

aR

α = . By substitution,

( )

( ) ( )

2 12 2

22 1 2

22 1

1 222

Ia IfR TR T fR R m

fR m TR R m IT If

f I mR T I mR R

I mR Rf TI mR

2

− = = −

− = −

+ = +

⎛ ⎞+= ⎜ ⎟+⎝ ⎠

FIG. P10.85

Since the answer is positive, the friction force is confirmed to be to the left .