Chapter 8: Rotational Motion

1
Chapter 8 Rotational Motion
2
Radians

• 1) 1 radian angle subtended by an arc (l)
  whose length is equal to the radius (r)
• 2) q l
• r
• 3600 2p radians 1 rev
• Radians are dimensionless

l
r
q
3

• Convert the following
• 20o to radians
• 20o to revolutions
• 5000o to radians
• 0.75 rev to radians
• 0.40 radians to degrees

4

• A bird can only see objects that subtend an angle
  of 3 X 10-4 rad. How many degrees is that?
• 3 X 10-4 rad 360o 0.017o
• 2p rad

5

• How small an object can the bird distinguish
  flying at a height of 100 m?
• q l
• r
• l q r
• l (3 X 10-4 rad)(100m)
• l 0.03 m 3 cm

q
r
l (approx.)
6

• How at what height would the bird be able to just
  distinguish a rabbit that is 30 cm long (and
  tasty)?
• (ANS 1000 m)

q
r
l (approx.)
7

• A tiny laser beam is aimed from the earth to the
  moon (3.8 X 108 m). The beam needs to have a
  diameter of 2.50 m on the moon. What is the
  angle that the beam can have?
• (6.6 X 10-9 radians)

8

• Convert
• 0.0200 rev/s to radians/s (0.126 rad/s)
• 30.0o/s to radians/s (0.534 rad/s)
• 1.40 rad/s to rev/s (0.223 rev/s)
• 3000 rpm to radians/s (314 rad/s)

9

• The Mighty Thor swings his hammer at 400 rev/min.
  Express this in radians/s.
• 400 rev 1 min 2p rad
• 1 min 60 s 1 rev
• 13.3p rad/s or
• 41.9 rad/s

10
Angular Velocity

• v Dx w Dq
• Dt Dt
• Angular Velocity radians an object rotates per
  second
• All points on an object rotate at the EXACT same
  angular velocity
• All points on an object DO NOT rotate with the
  same linear speed. (the farther out the faster)

11

• Merry-Go-Round Example
• wa wb
• (both rotate through 3600 in the same time
  period)
• va lt vb
• (Point b travels a longer distance around the
  circle)

vb
va
a
b
12

• If a person at point b flies off the
  merry-go-round, will they travel in a curve or
  straight line?

b
13
Angular Acceleration

• a Dv a Dw
• Dt Dt
• The angular acceleration of all points on a
  circle is the same.
• All points on an object DO NOT experience the
  same linear acceleration. (the farther out the
  more acceleration)

14
Frequency and Period

• Frequency Revolutions per second
• Period Time for one complete revolution
• f w
• 2p
• T 1
• f

15
Converting between Angular and Linear Quantities
atan

• Linear Radius X Angular
• v rw
• atan ra
• Note the use of atan to differentiate from
  centripetal acceleration, ac or ar

ar
16

• One child rides a merry-go-round (2 revolutions
  per minute) on an inside lion 2.0 m from the
  center. A second child rides an outside horse,
  3.0 m from the center.
• Calculate the angular velocity in rad/s. (0.209
  rad/s)
• Calculate the frequency and the period. (0.0333
  Hz, 30 s)
• Calculate the linear velocity of each rider.
  (0.419 m/s, 0.628 m/s)
• Who has a higher angular velocity?

17

• A clock has a seconds hand that is 7.00 cm long.
• Calculate the angular velocity of the second hand
  in radians/s. (0.105 rad/s)
• Calculate the frequency and period. (0.0167 Hz,
  60 s)
• Calculate the angular velocity in degrees/s.
  (6o/s)
• Calculate the linear velocity at the end of the
  seconds hand. (7.35 X 10-3 m/s)

18

• A baseball bat is found to have a linear
  acceleration of 13.9 m/s2 at the sweet spot.
  The sweet spot is at 59 cm from the handle.
• Calculate the angular acceleration in rad/s2 (a
  23.6 rad/s2)
• Convert the angular acceleration to rev/s2 (3.75
  rev/s2)

19

• A golf club has an angular acceleration of 20.0
  rad/s2 and is 114 cm long.
• Calculate the tangential acceleration of the
  club. (22.8 m/s2)
• Convert the angular acceleration to rev/s2.
  (3.18 rev/s2)
• Calculate the force the club could give to a
  45.93 g golf ball. (1.05 N)
• Calculate the velocity of the ball if the club is
  in contact with the ball for 50 cm. (4.77 m/s)

20
Angular Kinematics

• vvo at wwo at
• x vot ½ at2 q wot ½ at2
• v2 vo2 2ax w2 wo2 2aq

21

• A bike wheel starts at 2.0 rad/s. The cyclist
  accelerates at 3.5 rad/s2 for the next 2.0 s.
• Calculate the wheels new angular speed (9.0
  rad/s)
• Calculate the number of revolutions. (1.75)

22

• A DVD rotates from rest to 31.4 rad/s in 0.892 s.
• Calculate the angular acceleration. (35.2 rad/s2
  )
• How many revolutions did it make? (2.23)
• If the radius of the disc is 4.45 cm, find the
  linear speed of a point on the outside edge of
  the disc. (1.40 m/s)

23

• A bicycle slows from vo 8.4 m/s to rest over a
  distance of 115 m. The diameter of each wheel is
  68.0 cm.
• Calculate the angular velocity of the wheels
  before braking starts. (24.7 rad/s)
• How many revolutions did each wheel undergo?
  (HINT calculate the circumference of the circle
  first) (53.8 rev)
• Calculate the angular acceleration. (-0.903
  rad/s2)
• Calculate the time it took the bike to stop (27.4
  s)

24

• A spinning bike tire of radius 33.0 cm has an
  angular velocity of 50.0 rad/s. Twenty seconds
  later, its angular speed is 150.0 rad/s.
• Calculate the angular acceleration. (5.00
  rad/s2)
• Calculate the angular displacement over the 20 s.
  (2000 radians)
• Calculate the revolutions travelled in the 20 s
  (318 rev)
• Calculate the linear distance the tire travelled
  in 20 s. (660 m)

25

• A pottery wheel turning with an angular speed of
  30.0 rev/s is brought to rest in 60.0
  revolutions.
• Calculate the radians that the wheel travelled
  (377 rad)
• Calculate the angular acceleration. (-47.1
  rad/s2)
• Calculate the time required to stop. (4.00 s)
• If the radius of the wheel is 12.0 cm, calculate
  the linear distance the outside of the wheel
  travelled. (45.2 m)

26

• A game show wheel with a 90 cm radius is
  initially turning at 3.0 rev/s. A point on the
  outside of the wheel travels 147 meters before
  stopping.
• Calculate how many revolutions it when through.
  (26 rev)
• Calculate the angular deceleration. (1.09
  rad/s2)
• Calculate how long it took to stop. (17.3 s)
• Calculate the initial linear speed and linear
  acceleration on the outside of the wheel (17.0
  m/s, 0.98 m/s2)
• Calculate the initial centripetal acceleration on
  a point at the edge of the wheel. (319 m/s2)

27
Friction and Rolling Wheels

• Rolling uses static friction
• A new part of the wheel/tire is coming in contact
  with the road every instant

B
A
28

• Braking uses kinetic friction

Point A gets drug across the surface
A
29
Torque

• Torque tendency of a force to rotate a body
  about some axis (the force is always
  perpendicular to the lever arm)
• t Fr

r
pivot
F
30
Torque Sign Conventions
Counter-clockwise Torque is positive
Clockwise Torque is negative
31
Torque Example 1

• A wrench is 20.0 cm long and a 200.0 N force is
  applied perpendicularly to the end. Calculate
  the torque.
• t Fr
• t (200.0 N)(0.20 m)
• t 40.0 m-N

20.0 cm
200.0 N
32
Torque Example 2

• Suppose that same 200.0 N force is now applied at
  a 60o angle as shown. Calculate the Torque. Is
  it greater or less?

20.0 cm
200.0 N
60o
33

• First we need to resolve the Force vector into x
  and y components
• Only Fy has any effect on the torque
  (perpendicular)
• Fy Fsinq (200.0 N)(sin 60o) 173.2 N
• t Fr (173.2 N)(0.20 m) 34.6 m-N

200.0 N
Fy
60o
Fx
34
Torque Example 3

• The biceps muscle exerts a 700 N vertical force.
  Calculate the torque about the elbow.
• Fr (700 N)(0.050 m)
• 35 m-N

35

• A force of 200 N acts tangentially on the rim of
  a wheel 25 cm in radius.
• Calculate the torque.
• Calculate the torque if the force makes an angle
  of 40o to a spoke of the wheel.
• If the wheel is mounted vertically, draw a free
  body diagram of the wheel if the force is the one
  in (a).

36
Torque Example 4

• Two wheels, of radii r1 30 cm and r2 50 cm
  are connected as shown. Calculate the net
  torque on this compound wheel when two 50 N force
  act as shown.

50 N
30o
r2
Note that Fx will pull the wheel
r1
50 N
37

• First find the horizontal component of the top
  force
• Fx (50 N)(cos 30o) 43 N
• The top force is pulling clockwise (-) and the
  bottom force pulls counterclockwise ()
• St F1r1 F2r2
• St (50N)(0.30m) (43 N)(0.50 m) -6.5 m-N

38

• Two children push on a merry-go-round as shown.
  Calculate the net torque on the merry-go-round if
  the radius is 2.0 m.
• (1800 m N)

39

• A 60.0 cm diameter wheel is pulled by a 500.0 N
  force.  The force acts at an angle of 65.0o with
  respect to the spoke.  Assume that a frictional
  force of 300.0 N opposes this force at a radius
  of 2.00 cm.  Calculate the net torque on the
  wheel. (130.0 Nm)

40
Moment of Inertia (I)

• Measure of Rotational Inertia
• An objects resistance to a change in angular
  velocity
• Would it be harder to push a child on a
  playground merry-go-round or a carousel?

41
Deriving I

• Consider pushing a mass around in a circle (like
  the child on a merry-go-round)
• F ma t Fr
• a ra t mrar
• F mra t mr2a

F
r
m
42

• I moment of inertia
• I mr2
• More properly
• I Smr2 m1r12 m2r22 .
• St Ia
• Would it be harder (require more torque) to twirl
  a barbell in the middle (pt. M) or the end (Pt. E)

E
M
43
Moment of Inertia Example 1

• Calculate the moment of inertia (I) for the
  barbell when rotated about point M. We will
  assume the barbell is 1.0 m long, and that each
  weight is a point mass of 45.4 kg.
• I Smr2 (45.4 kg)(0.50 m)2 (45.4 kg)(0.50
  m)2
• I 22.7 kg-m2

M
44
Moment of Inertia Example 2

• Now calculate I assuming Mr. Fredericks uses his
  massive musculature to twirl the barbells from
  point E.
• I Smr2 (45.4 kg)(0 m)2 (45.4 kg)(1 m)2
• I 45.4 kg-m2

E
45
Moment of Inertia Example 3

• Calculate I for Bouncing Boy (75 kg, radius 1.2
  m). Use the formulas from the book.
• I 2/5 MR2
• I (2)(75 kg)(1.2 m)2
• 5
• I 43.2 kg-m2

46

• Calculate the moment of inertia of an 8.00 kg
  solid wheel with a radius of 25.0 cm.

47

• A wheel has a moment of inertia of 0.50 kg m2.
• Calculate the torque (St Ia) is required to give
  it an acceleration of 3 rad/s2. (1.5 mN)
• Calculate its angular speed (from rest) after
  5.00 s. (15 rad/s)
• Calculate the number of revolutions it goes
  through in 5.00 s. (37.5 rad, 5.97 rev)

48

• A 15.0 N force is applied to a cord wrapped
  around a pulley of radius 33.0 cm. The pulley
  reaches an angular speed (w) of 30.0 rad/s in
  3.00 s.
• Calculate the angular acceleration. (10.0 rad/s2)
• Calculate the torque (4.95 m-N)
• Calculate the moment of inertia of the pulley.
  (0.495 kg-m2)

33.0 cm
15.0 N
49

• A 25.0 kg wheel has a radius of 40.0 cm. A 1.20
  kg mass is hung on the end of the wheel by a
  string and falls freely.
• Calculate the moment of inertia of the wheel. (2
  kg m2)
• Calculate the torque on the pulley. (4.70 mN)
• Calculate the angular acceleration of the wheel.
  (2.35 rad/s2)
• Calculate the tangential acceleration of the
  wheel (and bucket) (0.94 m/s2)
• Calculate the tension in the string. (10.6 N)

50

• A 15.0 N force is applied to a cord wrapped
  around a pulley of radius 33.0 cm. The pulley
  reaches an angular speed (w) of 30.0 rad/s in
  3.00 s. Since this is a real pulley, there is a
  frictional torque (tfr 1.10 m-N) opposing
  rotation.
• Calculate net torque on the pulley. (3.85 m-N)
• Calculate the moment of inertia of the pulley.
  (0.385 kg-m2)

33.0 cm
15.0 N
51

• Using the same pulley as the previous problem
  (I0.385 kg m2), hang a 15.0 N bucket (1.53 kg)
  from the cord. There is a frictional torque of
  1.10 m-N opposing rotation. Calculate the
  angular and linear acceleration of the bucket.
  Also calculate FT.

33.0 cm
FT
52

• We will break this problem into two parts
  pulley and bucket. Lets first look at the
  pulley
• St FTr tfr
• St Ia
• Ia FTr tfr
• (0.385)a FT(0.33) 1.10

tfr
33.0 cm
FT
53

• Now we will deal with the bucket
• SF mg FT
• ma mg FT
• 1.53a 15 FT
• a ra
• 1.53ra 15 FT
• (1.53)(0.33)a 15 FT
• 0.505a 15 FT
• 0.385a 0.33FT 1.10
• a 7.07 rad/s2, a 2.33 m/s2, FT 11.5 N

FT
mg
54

• Now calculate the rotational speed of the pully
  (w) and the linear speed of the bucket after 3.00
  s.
• wwo at
• w 0 at (6.98 rad/s2)(3.00 s) 20.9 rad/s
• vrw
• v (0.330 m)(20.9 rad/s) 6.90 m/s

55

• A 2.00 kg bucket is attached to a 1.00 kg, 4.00
  cm radius cylindrical pulley. The bucket is
  suspended 2.00 m above the floor.
• Calculate the moment of inertia of the cylinder
  (1/2 MR2) (8X10-4 kgm2)
• Calculate the acceleration of the bucket. (7.84
  m/s2)
• Calculate the angular acceleration of the pulley.
  (196 rad/s2)
• Calculate how long it takes to reach the floor.
  (0.714 s)

4.00 cm
FT
56

• a) I 1/2 MR2 (1/2)(1.00 kg)(0.04 m)2
• I 8 X 10-4 kgm2
• Bucket Pulley
• ma mg FT t FTR
• 2a (2)(9.8) FT Ia FTR
• 2a 19.6 FT a aR
• Ia/R FTR
• 8 X 10-4 a FT(0.04)
• 0.04
• a -7.84 m/s2

57

• a aR
• a a/R 7.84/0.04 196 rad/s2
• y yo vt ½at2
• y ½ at2
• t v(2y/a) 0.714 s

58

• A 1000 kg elevator is suspended by a 100 kg
  cylinder pulley of radius 50.0 cm.
• Calculate the moment of inertia of the pulley.
  (12.5 kgm2)
• Calculate the linear acceleration if the elevator
  drops. (9.33 m/s2)
• Calculate the angular acceleration of the
  elevator. (18.7 rad/s2)
• Calculate the tension in the cord. (470 N)
• Are you glad you are not in the elevator?

59
Translational and Rotational Energy

• Does a rotating helicopter blade have kinetic
  energy before the helicopter takes off?
• How about afterwards?
• Does all of the energy of the fuel go into moving
  the helicopter?

60

• Translational Speed (v)
• speed of the center of a wheel with respect to
  the ground
• Can also be called linear speed
• Use regular KE ½ mv2
• Rotational speed (w)
• angular speed of the wheel
• Use KE ½ Iw2

61
Deriving the Rotational KE

• KE S½ mv2
• v rw
• KE S ½ m(rw)2
• KE S ½ mr2w2
• I S mr2
• KE ½ Iw2

62
Calculating Work

• A pottery wheel of I 0.200 kgm2 is accelerated
  from rest to 500 rpm.
• Calculate the final angular velocity in rad/s.
  (52.4 rad/s)
• Calculate the change in kinetic energy (274 J)
• How much work is done in the process (274 J)
• If the wheel takes 30 seconds to get up to speed,
  calculate the angular acceleration (1.75 rad/s2)

63

• Jupiter has a mass of 1.9 X 1027 kg and a radius
  of 7.0 X 107 m. Its day is only 9.92 hours long.
• Calculate Jupiters moment of inertia, treating
  it as a sphere. (3.72 X 1042 kgm2)
• Calculate Jupiters rotational speed in rad/s.
  (1.76 X 10-4 rad/s)
• Calculate Jupiters kinetic energy of rotation.
  (5.76 X 1034 J)
• Jupiter is 7.78 X 1011 m from the sun. Calculate
  Jupiters moment of inertia, assuming it is a
  point mass from the sun. (1.15 X 1051 kgm2)
• Calculate Jupiters orbital speed around the sun
  if it takes11.9 year to orbit the sun. (1.67 X
  10-8 rad/s)
• Calculate the kinetic energy about the sun. (1.60
  X 1035 J)

64
Law of Conservation of Mechanical Energy

• (KEt KEr PEt)i (KEt KEr PEt)f
• When using this equation, we will ignore
  friction.
• Also the work done in rotating object is
• W tDq

65
Rotational KE Example 1

• What will be the translational speed of a log
  (100 kg, radius 0.25 m, I ½ mr2) as it rolls
  down a 4 m ramp from rest?

4 m
66

• Calculate I
• I ½ mr2 (1/2)(100 kg)(0.25m)2
• I 3.125 kgm2
• Law of Conservation of Energy
• (KEt KEr PEt)i (KEt KEr PEt)f
• (0 0 mgy)i ( ½ mv2 ½ Iw2 0)f
• mgy ½ mv2 ½ Iw2
• 3920 50v2 1.56w2
• v wr w v/r w2 v2/r2
• 3920 50v2 1.56v2/(0.25m)2
• 3920 75v2 v 7.2 m/s

67

• (KEt KEr PEt)i (KEt KEr PEt)f
• (0 0 mgy)i ( ½ mv2 ½ Iw2 0)f
• mgy ½ mv2 ½ Iw2
• 2mgy mv2 Iw2 (multiplied both sides by 2)
• v rw so w v/r
• 2mgy mv2 Iv2
• r2
• I 1/2 mr2
• 2mgy mv2 1mr2v2
• 2r2

68

• 2mgy mv2 1mr2v2
• 2r2
• 2gy v2 1v2
• 2
• 2gy 2v2 1v2
• 2 2
• 2gy 3v2
• 2
• v 4gy 4(9.8m/s2)(4.00 m) 7.23 m/s
• 3 3

69

• Now we can calculate the angular speed
• v rw so w v/r
• 7.23 m/s 29 rad/s
• 0.25 m
• 29 rad 1 rev 4.6 rev/s
• s 2p rad

70
Rotational KE Example 2

• What will be the translational speed of Bouncing
  Boy (75 kg, radius 1.2 m) as he rolls down a 3
  m ramp from rest? (Treat him as a sphere)

3 m
71

• (KEt KEr PEt)i (KEt KEr PEt)f
• (0 0 mgy)i ( ½ mv2 ½ Iw2 0)f
• mgy ½ mv2 ½ Iw2
• 2mgy mv2 Iw2 (multiplied both sides by 2)
• v rw so w v/r
• 2mgy mv2 Iv2
• r2
• I 2/5 mr2
• 2mgy mv2 2mr2v2
• 5r2

72

• 2mgy mv2 2mr2v2
• 5r2
• 2gy v2 2v2
• 5
• 2gy 5v2 2v2
• 5 5
• 2gy 7v2
• 5
• v 10gy 10(9.8m/s2)(3.00 m) 6.50
  m/s
• 7 7

73

• Now we can calculate the angular speed
• v rw so w v/r
• 6.50 m/s 5.4 rad/s
• 1.2 m
• 5.4 rad 1 rev 0.86 rev/s
• s 2p rad

74

• What would his speed be if he just slid down the
  ramp?
• (KEt KEr PEt)i (KEt KEr PEt)f
• (0 0 mgy)i ( ½ mv2 0 0)f
• mgy ½ mv2
• gy ½ v2
• v2 2gy
• v \/2gy (2 X 9.8 m/s X 3.00 m)1/2
• v 7.7 m/s
• Why is this larger than if he rolls?

75

• A wooden dowel (mass 0.50 kg, r 1.00 cm)
  rolls down a 1.25 m tall ramp.
• Calculate the moment of inertia (2.5 X 10-5
  kgm2)
• Calculate the linear speed at the bottom. (4.04
  m/s)
• Calculate the angular speed at the bottom. (404
  rad/s, 64.3 rev/s)

76

• A bike tire can be modeled as a thin hoop
  (Imr2). The wheel has a diameter of 66.0 cm and
  a mass of 2.00 kg. The wheel is allowed to roll
  down a ramp and has a linear (translational)
  speed of 2.00 m/s at the bottom of the ramp.
• Calculate the moment of inertia of the wheel.
  (0.218 kgm2)
• Calculate the height of the ramp (40.8 cm)

77
Conservation of Angular Momentum

• The total angular momentum of a rotating body
  remains constant if the net torque acting on it
  is zero
• Angular Momentum (L) is conserved
• p mv (linear momentum)
• L Iw (angular momentum)
• Iiwi Ifwf

78
(No Transcript)
79
The skater in more detail

• For now, lets just consider the skaters arms
• Arms out (initial) Arms in (final)

80
Iiwi Ifwf I mr2 Her mass will not change
when she moves her arms mri2wi
mrf2wf ri2wi rf2wf Since rf will be smaller
(arms are in) wf must increase to compensate
81
I wanna go fast!

• A child sits on a merry-go-round at point A. If
  the child wants to go faster, should he walk
  towards the center or the outside?

A
82
Cons. of Angular Momentum Ex. 1

• A skater holds her arms at a length of 56 cm.
  She spins at 9.43 rad/s. What will be her new
  speed if she pulls her arms tight to her body, 20
  cm?
• Iiwi Ifwf
• I mr2
• mri2wi mrf2wf
• ri2wi rf2wf
• wf ri2wi /rf2
• wf (0.56 m)2(9.43 rad/s)/(0.20 m)2 74 rad/s

83
Cons. of Angular Momentum Example 2

• A mass m is attached to the end of a string which
  passes through a hole in a table. Initially the
  r1 is 0.80 m and the block moves at 2.4 m/s.
  What will be the new speed as the radius of the
  string is reduced to 0.48 m?

84

• Iiwi Ifwf
• I mr2
• mri2wi mrf2wf
• ri2wi rf2wf
• v rw so w v/r
• ri2vi rf2vf
• ri rf
• rivi rfvf
• vf rivi (0.80m)(2.4 m/s) 4.0 m/s
• rf (0.48 m)

85

• Old record players rotate at 33.3 rev/min.
  Assume a record has a radius of 15.24 cm and a
  mass of 100.0 grams.
• Calculate the angular speed in rad/s (3.49
  rad/s)
• Calculate the moment of inertia. (1/2 mr2) (1.16
  X 10-3 kgm2)
• Calculate angular momentum of the record. (4.05 X
  10-3 kg m2/s)
• A second record drops on top of the first.
  Calculate the new moment of inertia and new
  angular speed. (2.32 X 10-3 kg m2, 1.74 rad/s)

86
Vectors and Angular Momentum

• Direction of the vector w is defined by the
  right-hand rule.
• L Iw
• L is defined as going in the same direction.

87
Merry-Go-Round

• What will happen if you start to walk on a still
  merry-go-round? Can you walk around it without
  it moving?

88

• Assume this merry-go-round is initially at rest
  (Ltotal 0)
• As the man walks counter-clockwise, the platform
  spins counterclockwise
• Total Angular Momentum is still zero.

89

• What will happen if the student flips the bicycle
  wheel while standing on the merry-go-round?

90

• A merry-go-round (500 kg) has a radius of 1.5 m.
  A 40.0 kg student is sitting on the
  merry-go-round as it spins at 30.0 rev/min.
• Calculate the angular speed (3.14 rad/s)
• Calculate the moment of inertia of the
  merry-go-round and student, assuming she sits at
  the edge. (652.5 kg m2)
• Calculate the angular momentum. (2049 kgm2/s)
• A second student (80.0 kg) now jumps on the
  merry-go-round. Calculate the new moment of
  inertia and angular speed. (832.5 kgm2, 2.46
  rad/s)

91

• 2. RSun 6.5 X 105 km
• 4. 2900 m
• 6. 188 rad/s
• 8. 1.9 rad/s2
• 10. -260 rad/s2
• 12. 3300 revolutions
• 14. a) 1.99 X 10-7 rad/s b) 7.27 X 10-5 rad/s
• 16. 3.6 X 104 rpm
• 0.56 rad/s2
• 22. a) -84 rad/s2 b) 150 rev
• 24. 38 m

92

• 22. a) -84 rad/s2 b) 150 rev
• 24. 38 m
• 26. a) 0.58 rad/s2 b) 12 s
• 30. a) 38 m N b) 33 m N
• 32. 650 m N
• 34. 1.89 kg m2
• 36. a) 6.1 kg m2 b) 0.61 kg m2
• 38. a) 0.851 kg m2 b) 0.0720 m N
• 1.99 X 10-3 kg m2 b) 0.0682 m N
• 590 m N, 240 N
• 46. 2250 kg m2, 8800 mN

93

• 42. 590 m N, 240 N
• 44. a) 7.5 m N b) 300 N
• 46. 2250 kg m2, 8800 mN
• 50. 94 J
• 52. a) 2.6 X 1029 J b) 2.7 X 1033 J
• a) 8.37 m/s, 41.8 rad/s b) 2.50 (mass
  independent)

94

• 58. a) angular momentum conserved b) 1.6
• 60. 0.77 kg m2 (pull arms close to body)
• 62. a) 14 kg m2/s b) -2.7 m N
• 66.a) 1.2 rad/s b) 2000 J, 1200 J (loss of
  800J, 40 loss)