solution guide to webassign problems - stephen · pdf filesolution guide to webassign problems...

.1

CHAPTER 8 & 9: Rotational Motion & Static Equilibrium

Solution Guide to WebAssign Problems

8.1 [1] (a) 30º( ) 2! rad 360º( ) = ! 6 rad = 0.52 rad

(b) 57º( ) 2! rad 360º( ) = 19 ! 60 rad = 0.99 rad

(c) 90º( ) 2! rad 360º( ) = ! 2 rad = 1.57 rad

(d) 360º( ) 2! rad 360º( ) = 2! rad = 6.28 rad

(e) 420º( ) 2! rad 360º( ) = 7! 3 rad = 7.33 rad

8.2 [4] The initial angular velocity is !o= 6500

rev

min

"#$

%&'

2( rad

1 rev

"#$

%&'

1 min

60 sec

"#$

%&'= 681 rad s.

Use the definition of angular acceleration.

! ="#

"t=

0 $ 681 rad s

3.0 s= $227 rad s

2% $2.3&10

2 rad s

2

8.3 [8] The angular speed of the merry-go-round is 2! rad 4.0 s = 1.57 rad s

(a) v =!r = 1.57 rad sec( ) 1.2 m( ) = 1.9 m s

(b) The acceleration is radial. There is no tangential acceleration.

aR=!

2r = 1.57 rad sec( )

2

1.2 m( ) = 3.0 m s2

towards the center

8.4 [11] The centripetal acceleration is given by a =!2r. Solve for the angular velocity.

! =a

r=

100,000( ) 9.8 m s2( )

0.070 m= 3741

rad

s

1 rev

2 "rad

#$%

&'(

60 s

1 min

#$%

&'(= 3.6 )104 rpm

PIM WebAssign Chapter 8 & 9 Answer Guide Physics: Principles with Applications, 6th Edition

2

8.5 [16] (a) For constant angular acceleration:

! =" #"

o

t=

1200 rev min # 4500 rev min

2.5 s=#3300 rev min

2.5 s

2$ rad

1 rev

%&'

()*

1 min

60 s

%&'

()*

= #1.4 +102

rad s2

(b) For the angular displacement, given constant angular acceleration:

! = 1

2"

o+"( )t = 1

24500 rev min +1200 rev min( ) 2.5 s( )

1 min

60 s

#$%

&'(= 1.2 )10

2 rev

8.6 [19] (a) The angular acceleration can be found from ! 2=!

o

2+ 2"#.

! =" 2 #"

o

2

2$=

0 # 850 rev min( )2

2 1500 rev( )= #241

rev

min2

%&'

()*

2+ rad

1 rev

%&'

()*

1 min

60 s

%&'

()*

2

= #0.42rad

s2

(b) The time to come to a stop can be found from ! = 1

2"

o+"( )t.

t =2!

"o+"

=2 1500 rev( )

850 rev min

60 s

1 min

#$%

&'(= 210 s

8.7 [20] Since there is no slipping between the wheels, the tangential component of the linear acceleration of each

wheel must be the same.

(a) atansmall

= atanargel

! "smallrsmall = " largerlarge !

! large = !small

rsmall

rlarge

= 7.2 rad s2( )2.0 cm

25.0 cm

"#$

%&'= 0.576 rad s2 ( 0.58 rad s

2

(b) Assume the pottery wheel starts from rest. Convert the speed to an angular speed, and then use Eq. 8-9a.

! = 65rev

min

"#$

%&'

2( rad

1 rev

"#$

%&'

1 min

60 s

"#$

%&'= 6.81 rad s

! =!0+)t * t =

! +!0

)=

6.81 rad s

0.576 rad s2= 12 s


3

8.8 [21] (a) The angular acceleration can be found from ! 2=!

o

2+ 2"#, with the angular velocities being

found from ! = v r .

! =" 2 #"

o

2

2$=

v2 # v

o

2( )2r

2$=

45 km h( )2

# 95 km h( )2%

&'(

1 m s

3.6 km h

)*+

,-.

2

2 0.40 m( )2

65 rev( )2/ rad

rev

)*+

,-.

= #4.133 rad s2 0 #4.1 rad s

2

(b) The time to stop can be found from ! =!o+"t, with a final angular velocity of 0.

t =! "!

o

#=v " v

o

r#=

" 45 km h( )1 m s

3.6km h

$%&

'()

0.40 m( ) "4.133 rad s2( )

= 7.6 s

8.9 [23] The torque is calculated by ! = rF sin". See the diagram, from the top view.

(a) For the first case, ! = 90º .

! = rF sin" = 0.74 m( ) 55 N( ) sin90º = 41 m •N

(b) For the second case, ! = 45º .

! = rF sin" = 0.74 m( ) 55 N( ) sin 45º= 29 m •N

8.10 [24] Each force is oriented so that it is perpendicular to its lever arm. Call counterclockwise torques positive.

The torque due to the three applied forces is given by

! appliedforces

= 28 N( ) 0.24 m( ) " 18 N( ) 0.24 m( ) " 35 N( ) 0.12 m( ) = "1.8m •N.

Since this torque is clockwise, we assume the wheel is rotating clockwise, and so the frictional torque is

counterclockwise. Thus the net torque is

8.11 [25] There is a counterclockwise torque due to the force of gravity on the left block, and a clockwise torque

due to the force of gravity on the right block. Call clockwise the positive direction.

!" = mgL2# mgL

1= mg L

2# L

1( ) , clockwise

9.12 [2] The torque is the force times the lever arm.

! = Fr = 58 kg( ) 9.8 m s2( ) 3.0 m( ) = 1.7 "103 m •N , clockwise


4

9.13 [9] The pivot should be placed so that the net torque on the board is

zero. We calculate torques about the pivot point, with

counterclockwise torques as positive. The upward force PF

! at

the pivot point is shown, but it exerts no torque about the pivot point.

The mass of the board is ,Bm and the CG is at the middle of the board.

(a) Ignore the force .Bgm

( )

( )( )

( ) adult from m 2.3m 25.2m 0.9kg 75kg 25

kg 25

0

!=+

=+

=

"=##=$

LMm

mx

xLmgMgx%

(b) Include the force .Bgm

( ) ( )

( )( )

( )( )

( ) adult from m 2.5m 54.2m 0.9kg 15kg 25kg 75

kg 7.5kg 252

02

B

B

!=++

+=

++

+=

=""""=#

LmmM

mmx

xLgmxLmgMgx

B

$

9.14 [13] The table is symmetric, so the person can sit near either edge and

the same distance will result. We assume that the person (mass M) is on

the right side of the table, and that the table (mass m) is on the verge of

tipping, so that the left leg is on the verge of lifting off the floor. There

will then be no normal force between the left leg of the table and the floor. Calculate torques about the right

leg of the table, so that the normal force between the table and the floor causes no torque. Counterclockwise

torques are taken to be positive. The conditions of equilibrium for the table are used to find the person’s

location.

( ) ( ) ( ) m 182.0kg 66.0

kg 0.20m 60.0m 60.0 0m 60.0 ===!="=#

M

mxMgxmg$

Thus the distance from the edge of the table is m 0.32m 182.0m 50.0 =!


5

9.15 [15] The beam is in equilibrium, and so both the net torque and

net force on it must be zero. From the free-body diagram,

calculate the net torque about the center of the left support, with

counterclockwise torques as positive. Calculate the net force,

with upward as positive. Use those two equations to findAF and .

BF

!" = FB x1 + x2 + x3 + x4( ) # F1x1 # F2 x1 + x2( ) # F3 x1 + x2 + x3( ) # mgx5

FB =F1x1 + F2 x1 + x2( ) + F3 x1 + x2 + x3( ) + mgx5

x1 + x2 + x3 + x4( )

=4300 N( ) 2.0 m( ) + 3100 N( ) 6.0 m( ) + 2200 N( ) 9.0 m( ) + 250 kg( ) 9.8 m s

2( ) 5.0 m( )

10.0 m

= 5925 N $ 5.9 %103 N

F" = FA + FB # F1 # F2 # F3 # mg = 0

FA = F1 + F2 + F3 + mg # FB = 9600 N + 250 kg( ) 9.8 m s2( ) # 5925 N = 6125 N $ 6.1%103 N

9.16 [20] The beam is in equilibrium. Use the conditions of equilibrium to calculate the

tension in the wire and the forces at the hinge. Calculate torques about the hinge, and

take counterclockwise torques to be positive.

!" = FT

sin#( )l2 $ m1gl

12 $ m

2gl

1= 0 %

FT=

1

2m

1gl

1+ m

2gl

1

l2 sin#

=1

2155 N( ) 1.70 m( ) + 245 N( ) 1.70 m( )

1.35 m( ) sin 35.0º( )

= 708.0 N & 7.08 '102

N

Fx" = FH x $ FT

cos# = 0 % FH x = FT

cos# = 708 N( ) cos 35.0º = 579.99 N & 5.80 '102

N

Fy" = FH y + FT

sin # $ m1g $ m

2g = 0 %

FH y = m1

g + m2g $ F

T sin # = 155 N + 245 N $ 708 N( ) sin 35.0º = $6.092 N & $6 N down( )


6

8.17 [31] (a) To calculate the moment of inertia about the y-axis (vertical), use

I = MiRix

2! = m 0.50 m( )2+ M 0.50 m( )

2+ m 1.00 m( )

2+ M 1.00 m( )

2

= m + M( ) 0.50 m( )2+ 1.00 m( )

2"#

$% = 4.9 kg( ) 0.50 m( )

2+ 1.00 m( )

2"#

$% = 6.1 kg •m2

(b) To calculate the moment of inertia about the x-axis (horizontal), use

I = Mi! Riy2= 2m + 2M( ) 0.25 m( )

2= 0.61 kg •m2 .

(c) Because of the larger I value, it is harder to accelerate the array about the vertical axis .

8.18 [32] The oxygen molecule has a “dumbbell” geometry, rotating about the dashed line, as

shown in the diagram. If the total mass is M, then each atom has a mass of M/2. If the distance

between them is d, then the distance from the axis of rotation to each atom is d/2. Treat each atom as a particle

for calculating the moment of inertia.

I = M 2( ) d 2( )2+ M 2( ) d 2( )

2= 2 M 2( ) d 2( )

2= 1

4Md

2 !

d = 4I M = 4 1.9 "10#46 kg •m2( ) 5.3"10#26 kg( ) = 1.2 "10#10 m

8.19 [35] The torque can be calculated from ! = I". The rotational inertia of a rod about its end is given by

I =1

3ML

2.

! = I" = 1

3ML

2 #$

#t= 1

32.2 kg( ) 0.95 m( )

2 3.0 rev s( ) 2% rad rev( )0.20 s

= 62 m •N

8.20 [36] The torque needed is the moment of inertia of the system (merry-go-round and children) times the

angular acceleration of the system. Let the subscript “mgr” represent the merry-go-round.

! = I" = Imgr + Ichildren( )#$

#t= 1

2MmgrR

2+ 2mchildR

2( )$ %$0

t

= 1

2760 kg( ) + 2 25 kg( )&' () 2.5 m( )

2 15 rev min( ) 2* rad rev( ) 1 min 60 s( )10.0 s

= 422.15 m •N + 4.2 ,102 m •N

The force needed is calculated from the torque and the radius. Assume that the force is all directed

perpendicularly to the radius.

! = F"R sin# $ F

"= ! R = 4.2215 %10

2m •N 2.5 m = 1.7 %10

2N


7

8.21 [43] The energy required to bring the rotor up to speed from rest is equal to the final rotational KE of the rotor.

KErot =1

2I! 2

= 1

23.75 "10#2 kg •m2( ) 8250

rev

min

2$ rad

1 rev

%&'

()*

1 min

60 s

%&'

()*

+

,-

.

/0

2

= 1.40 "104 J

8.22 [45] The total kinetic energy is the sum of the translational and rotational kinetic energies. Since the ball is

rolling without slipping, the angular velocity is given by ! = v R . The rotational inertia of a sphere about an

axis through its center is I = 2

5mR

2.

KEtotal = KEtrans + KErot =1

2mv

2+ 1

2I!

2= 1

2mv

2+ 1

2

2

5mR

2 v2

R2= 7

10mv

2

= 0.7 7.3 kg( ) 3.3 m s( )2= 56 J

8.23 [55] The skater’s angular momentum is constant, since no external torques are applied to her.

Li = Lf ! Ii" i = I f" f ! I f = Ii" i

" f

= 4.6 kg •m2( )0.50 rev s

3.0 rev s= 0.77 kg •m2

She accomplishes this by starting with her arms extended (initial angular velocity) and then pulling her arms

in to the center of her body (final angular velocity).

8.24 [57] (a) L = I! = 1

2MR

2! = 1

255 kg( ) 0.15 m( )

23.5

rev

s

"#$

%&'

2( rad

1 rev

"#$

%&'= 14 kg •m2 s

(b) If the rotational inertia does not change, then the change in angular momentum is strictly due to a

change in angular velocity.

! ="L

"t=

0 #14 kg •m2 s

5.0 s= #2.7 m •N

The negative sign indicates that the torque is in the opposite direction as the initial angular momentum.


8

8.25 [61] Since the person is walking radially, no torques will be exerted on the person-platform system, and so

angular momentum will be conserved. The person will be treated as a point mass. Since the person is initially

at the center, they have no initial rotational inertia.

(a) Li = Lf ! Iplatform" i = Iplatform + Iperson( )" f

! f =Iplatform

Iplatform + mR2! i =

920 kg •m2

920 kg •m2+ 75 kg( ) 3.0 m( )

22.0 rad s( ) = 1.154 rad s " 1.2 rad s

(b) KEi= 1

2Iplatform! i

2= 1

2920 kg •m2( ) 2.0 rad s( )

2= 1.8 "103 J

KEf =1

2Iplatform + Iperson( )! f

2= 1

2Iplatform + mpersonrperson

2( )! f

2

= 1

2920 kg •m2

+ 75 kg( ) 3.0 m( )2"

#$% 1.154 rad s( )

2= 1062 J & 1.1'103J

8.26 [63] Since the lost mass carries away no angular momentum, the angular momentum of the remaining mass

will be the same as the initial angular momentum.

Li = Lf ! Ii" i = I f" f ! " f

" i

=Ii

I f=

25MiRi

2

25M f Rf

2=

MiRi2

0.5 Mi( ) 0.01Rf( )2= 2.0 #104

" f = 2.0 #104 " i = 2.0 #104 2$ rad

30 day

%&'

()*

1 d

86400 s

%&'

()*= 4.848 #10+2 rad s , 5 #10+2 rad s

The period would be a factor of 20000 smaller, which would make it about 130 seconds.

The ratio of angular kinetic energies of the spinning mass would be

KEf

KEi

=

1

2I f! f

2

1

2Ii! i

2=

1

2

2

50.5Mi( ) 0.01Ri( )

2"#

$%

2.0 &104 ! i( )

2

1

2

2

5MiRi

2( )! i

2= 2.0 &10

4 ' KEf = 2 &10

4 KEi

t =! "!

o

#=v " v

o

r#=

0 " 60.0 km h( )1 m s

3.6 km h

$%&

'()

0.45 m( ) "1.6053 rad s2( )

= 23 s

solution guide to webassign problems - stephen · pdf filesolution guide to webassign problems...

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