Chapter 8: Torque and Angular
Momentum
• Concept. Questions: 2, 4.
• Problems: 5,13,18, 27, 39, 43, 55, 69, 73.
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Rotational Kinetic Energy
Torque & Angular Acceleration
Torque & Angular Momentum
(Vector Nature of)
Rotational Dynamics:
Newton’s 2nd Law for Rotation
Net Torque
Angular Accel. 
Rotational Inertia
Clockwise (CW)
Counter-clockwise (CCW)
2
Rotational Inertia & Energy
KE   mv   m(r  )
1
2

1
2
2
1
2
 mr 
2
I   mr
2
KE  I
1
2
2
2
2
2
Central Axis
I  ma  ma  ma  ma
2
2
I  4ma
2
2
2
4
Axis on End
I  m0  m0  m(2a)  m(2a)
2
2
2
I  8ma
2
2
5
Calculated Rot. Inertias
• rotational inertias of solid objects can be
calculated
• The calculated values are listed in your
textbook on p.263
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6
Ex: Rotational Inertia: A 0.3kg meter stick is held
horizontally from one end. Its rotational inertia about one
end is:
I  ML
1
3
2
 (0.3kg)(1.0m)
1
3
 0.10kg  m
2
2
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Net Torque
Angular Accel. 
Rotational Inertia
Torque = lever-arm x force
meter-newton
ft-lb
torque
• lever-arm is the shortest distance from
axis to line of the force
• Torque (giam7-11)
  rF sin 
 (r sin  ) F  F
 r ( F sin  )  rF
9
Ex: Zero and Non-Zero Torque
• Zero Torque
• Large Torque
Ex: Torque due to Gravity: A 0.3kg meter stick is held
horizontally from one end. The torque due to gravity about
the end is:
 grav  Mgxcm
 (0.3kg)(9.8N / kg)(0.5m)
 1.47 m  N
12
Newton’s 2nd Law (Rotation)
Net Torque
Angular Accel. 
Rotational Inertia
 Net  I
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Ex: Angular Acceleration: A 0.3kg meter stick is held
horizontally from one end. Its angular acceleration when
released is:

 net
I

 grav
1
2
2
ML
1.47 m  N

2
0.1kg  m
 14.7rad / s / s
14
Ex: A merry-go-round has a rotational inertia of 100kgm^2
and a radius of 1.0 meter. A force of 250 N is applied
tangentially at its edge. The angular acceleration is:
 net
F


I
I
(1.0m)( 250 N )

2
100kg  m
 0.25rad / s / s
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Equilibrium Problems
• Equilibrium is state when: Net force = 0 &
Net torque = 0
• You can choose the axis anywhere, so we
choose it where an unknown force acts.
• 1st Step: Storque-ccw = Storque-cw
• 2nd Step: Sforce-up = Sforce-down
• /
16
Ex: The drawing shows a
person whose weight is
584N. Calculate the net
force with which the floor
pushes on each end of his
body.
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Rotational Kinetic Energy
• Rotational K = ½(I)2.
• Example: Constant Power Source has
100kg, 20cm radius, solid disk rotating at
7000 rad/s.
• I = ½MR2 = ½(100kg)(0.2m)2 = 2kgm2.
• Rot K = ½ (2kgm2)(7000/s)2 = 49 MJ
18
Rotational Work-Energy Theorem
• (Work)rot = Dq.
• Example: torque of 50 mN is applied for
one revolution.
• rotational work = (50Nm)(2prad) = 314 J
• (Rotational Work)net = DKrot.
• /
19
Angular Momentum (L)
• analog of translational momentum
• L = I [kgm2/s]
• Example: Disk R = 1m, M = 1kg,  = 10/s
• I = ½MR2 = ½(1)(1)2 = 0.5 kgm2
• L = I = (0.5kgm2)(10/s) = 5kgm2/s
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Conservation of Angular
Momentum
• For an isolated system
• (I)before = (I)after
• Example: Stationary disk M,R is dropped
on rotating disk M, R, i.
• (I)before = (I)after
• (½MR2)(i) = (½MR2 + ½MR2)(f)
• (f) = ½ (i)
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8 Summary
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Mass  Rotational Inertia
Force  Torque
Rotational KE
Angular Acceleration
Work and Energy
Angular Momentum
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Concept Review
• Torque: rotational action
• Rotational Inertia: resistance to change in
rotational motion.
• Torque = force x lever-arm
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Mass-Distribution
Larger
radius

Larger
Speed

Rotational
Inertia
Larger
Effort
~
R2
24
Torque
[m·N]
()
F
F


  F
 = lever-arm
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Rotational Inertia ( I )
I  m r  m r 
2
1 1
2
2 2
kg(m)2
Example
I  (4kg)(3m) 2  (5kg)( 2m) 2
 36  20  56kg  m2
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27
Problem 33
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Pivot at left joint, Fj = ?, but torque = 0.
ccw (Fm)sin15(18) = mg(26) = cw
ccw (Fm)sin15(18) = (3)g(26) = cw
(Fm) = (3)g(26)/sin15(18) = 160N
Note: any point of arm can be considered
the pivot (since arm is at rest)
28
If ball rolls w/o slipping at 4.0m/s, how large is the height h in
the drawing?
rolling w/o
slipping
29
#39
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Left force = mg = 30g, Right = 25g
mg = 30g + 25g m = 55kg
ccw mg(xcg) = cw 30g(1.6)
(55)g(xcg) = 30g(1.6)
(55)(xcg) = 30(1.6)
Xcg = (30/55)(1.6)
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#60, z-axis
• Each mass has r2 = 1.52 + 2.52.
• I = sum mr2 = (2+3+1+4)(1.52 + 2.52)
31
#65
• First with no frictional torque, then with
frictional torque as specified in problem.
• M = 0.2kg, R = 0.15m, m1 = 0.4, m2 = 0.8
32
#83
• Pulley M, R. what torque causes it to
reach ang. Speed. 25/s in 3rev?
• Alpha: use v-squared analog eqn.
• Torque = I = (½MR2)(
33
#89, uniform sphere part
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Rolling at v = 5m/s, M = 2kg, R = 0.1m
K-total = ½mv2 + ½I2.
= ½(2)(5x5) + ½[(2/5)(2)(0.1x0.1)](5/0.1)2.
= 25 + 10 = 35J
Roll w/o slipping, no heat created, mech
energy is conserved, goes all to Mgh.
• 35 = Mgh h = 35/Mg = 35(19.6) = 1.79m
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#111
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Ice skater, approximate isolated system
Therefore:
(I)before = (I)after
(100)(i) = (92.5)(f)
(f) = (100/92.5)(i)
K-rot increases by this factor squared
times new rot. Inertia x ½.
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Example: Thin rod formulas.
36
Angular Momentum
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Symbol: L
Unit: kg·m2/s
L = mvr = m(r)r = mr2 = I.
v is perpendicular to axis
r is perpendicular distance from axis to line
containing v.
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Angular Momentum
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Symbol: L
Unit: kg·m2/s
L = mvr = m(r)r = mr2 = I.
v is perpendicular to axis
r is perpendicular distance from axis to line
containing v.
38
13) Consider a bus designed to obtain its motive power from a large
rotating flywheel (1400. kg of diameter 1.5 m) that is periodically
brought up to its maximum speed of 3600. rpm by an electric motor
at the terminal. If the bus requires an average power of 12. kilowatts,
how long will it operate between recharges?
Answer: 39. minutes
Diff: 2 Var: 1 Page Ref: Sec. 8.4
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6) A 82.0 kg painter stands on a long horizontal board 1.55 m from one
end. The 15.5 kg board is 5.50 m long. The board is supported at each
end.
(a) What is the total force provided by both supports?
(b) With what force does the support, closest to the painter, push
upward?
40
28) A 4.0 kg mass is hung from a string which is wrapped around a
cylindrical pulley (a cylindrical shell). If the mass accelerates
downward at 4.90 m/s2, what is the mass of the pulley?
A) 10.0 kg
B) 4.0 kg
C) 8.0 kg
D) 2.0 kg
E) 6.0 kg
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19) A solid disk with diameter 2.00 meters and mass 4.0 kg freely rotates about a
vertical axis at 36. rpm. A 0.50 kg hunk of bubblegum is dropped onto the disk and
sticks to the disk at a distance d = 80. cm from the axis of rotation.
(a) What was the moment of inertia before the gum fell?
(b) What was the moment of inertia after the gum stuck?
(c) What is the angular velocity after the gum fell onto the disk?
(a) 2.0 kg-m2
(b) 2.3 kg-m2
(c) 31. rpm
42
1. A pair of forces with equal magnitudes and
opposite directions is acts as shown. Calculate
the torque on the wrench.
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3. The drawing shows the top view of
two doors. The doors are uniform and
identical. The mass of each door is M
and width as shown below is L. How do
their rotational accelerations compare?
44
A Ring, a Solid-Disk, and a Solid-Sphere are released from
rest from the top of an incline. Each has the same mass and
radius. Which will reach the bottom first?
45
5. The device shown below is spinning with rotational rate i when the
movable rods are out. Each moveable rod has length L and mass M. The central
rod is length 2L and mass 2M.
Calculate the factor by which the angular velocity
is increased by pulling up the arms as shown.
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Rotational Review
s  rq
(angles in radians)
 avg
Dq

Dt
 avg
D

Dt
vt  r
at  r
ac  r
2
  
a  ac  at
+ 4 kinematic equations
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Angular Momentum Calculation
L = I
Example: Solid Disk M = 2kg R = 25cm
Spins about its center-of-mass at 35 rev/s
L  I 

1
2

(2kg)(.25m) 2 (35rev / s)( 2prad / rev) 
 13.7kgm2 / s
48
4. A one-meter-stick has a mass of 480grams.
a) Calculate its rotational inertia about an axis perpendicular to the
stick and through one of its ends.
b) Calculate its rotational inertia about an axis perpendicular to the
stick and through its center-of-mass.
c) Calculate its angular momentum if spinning on axis (b) at a rate of
57rad/s.
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Conservation of Angular
Momentum
• Example: 50 grams of putty shot at 3m/s
at end of 200 gram thin 80cm long rod free
to rotate about its center.
• Li = mvr = (0.050kg)(3m/s)(0.4m)
• Lf = I = {(1/12)(0.200kg)(0.8m)2 +
(0.050kg)(0.4m)2}()
• final rotational speed of rod&putty =
50
 avg
Dq

Dt
 avg
D

Dt
51
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