PHY 113 C General Physics I
11 AM - 12:15 PM MWF Olin 101
Plan for Lecture 11:
Chapter 10 – rotational motion
1. Angular variables
2. Rotational energy
3. Moment of inertia
4. Torque
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PHY 113 C Fall 2013-- Lecture 11
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PHY 113 C Fall 2013-- Lecture 11
2
Comments about Exam 1
• Solutions posted online
• Please review the problems while the ideas are
“fresh” in your mind
iclicker question (serious)
In future exams, would you like to have an additional
“take” home component? Of course it would be
assumed that all portions of the exam would be
subject to strict honor code guidelines
A. Yes
B. No
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PHY 113 C Fall 2013-- Lecture 11
3
Review:
iclicker question
In which of the following cases does gravity do
positive work?
yf
yi
a.
yi
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b.
yf
PHY 113 C Fall 2013-- Lecture 11
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Review:
iclicker question
In which of the following cases does the work of
gravity have a great magnitude?
rf
rf
a.
ri
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b.
ri
PHY 113 C Fall 2013-- Lecture 11
5
Review of center of mass:
General relation for center of mass :
Define :
rCM 
 m r 
i i
M   mi 
i
M
i
d 2rCM
i Fi  Ftotal  M dt 2
Center of mass measured relative to
1 kg mass :
Example:
1m
i i
rCM
1 kg
 m r 

 m 
3 kg
i
i
rCM
3 1

m  0.75 m
4
i
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PHY 113 C Fall 2013-- Lecture 11
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Center of mass example from Webassign
Finding the center of mass
For a solid object composed of constant density
material, the center of mass is located at the center
of the object.
 m r  m r 


 m  m 1
i i
rCM
i
i
i
i
i
i

3  5  2 15  1  25ˆi  3  5  1 15  2  25ˆj

cm
6
 11.6667ˆi  13.3333ˆj cm

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PHY 113 C Fall 2013-- Lecture 11

7
Another Webassign question from Assignment #9


p  mv  2.91  3.05ˆi  3.99ˆj kg  m/s
 8.8755ˆi  11.6109ˆj kg  m/s

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PHY 113 C Fall 2013-- Lecture 11

8
Another Webassign question from Assignment #9
p
i initial
i

p
i final
i
mC v Ci  mT v Ti  mC v Cf  mT v Tf
v Tf  ?
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vTf 
PHY 113 C Fall 2013-- Lecture 11
mC v Ci  v Cf 
mT
 vTi
9
Examples of two-dimensional collision; balls moving on a
frictionless surface
Suppose : m1  m2  0.06kg , v1i  2m / s,
v2 f  1m / s,
  20o
m1v1i  m1v1 f cos   m2 v2 f cos 
0  m1v1 f sin   m2 v2 f sin 
v1 f sin   v2 f sin 
 1m / s sin 20o  0.342m / s
v1 f cos   v1i  v2 f cos 


 2m / s   1m / s  cos 20o  1.060m / s
0.342
tan  
   17.88o
1.060
0.342m / s 1.060m / s
v1 f 

 1.11m / s
o
o
sin 17.88
cos17.88
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PHY 113 C Fall 2013-- Lecture 11
10
Angular motion
angular “displacement”  (t)
dθ
ω
(t)

angular “velocity” 
dt dω
angular “acceleration”  α(t) 
dt
s
“natural” unit == 1 radian
 radians  3.14159 radians  180o
Relation to linear
variables: s = r (f-i)
v = r w
a = r a
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PHY 113 C Fall 2013-- Lecture 11
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Rotation at constant angular velocity w
w
r
v
v = r w
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PHY 113 C Fall 2013-- Lecture 11
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v1=r1w
r1
w
r2
v2=r2w
Special case of constant angular acceleration: a = a0:
wt = wi + a0 t
t = i + wi t + ½ a0 t2
( wt2  wi2 + 2 a0 t - i 
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PHY 113 C Fall 2013-- Lecture 11
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A wheel is initially rotating at a
rate of f=30 rev/sec.
R
What is the angular velocity?
ω  2πf  2 (30) rad/s
 188.495 rad/s
What is the speed of a dot on the rim
of the wheel at a radius R  0.5m?
v  wR  (188.495 rad/s)(0.5m)  94.247 m / s
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PHY 113 C Fall 2013-- Lecture 11
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A wheel is initially rotating at a
rate of f=30 rev/sec. Because of a
constant angular deceleration, the
wheel comes to rest in 3 seconds.
R
What is the angular deceleration?
0  2πf  2 (30) rad/s
a

3s
3s
 62.83 rad/s 2
What is the deceleration of a dot on the rim
of the wheel at a radius R  0.5m?
a  aR  (62.83 rad/s )(0.5m)  31.42m / s
2
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PHY 113 C Fall 2013-- Lecture 11
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2
Example: Compact disc motion
w1
w2
In a compact disk, each spot on the disk passes the laser-lens
system at a constant linear speed of v = 1.3 m/s.
w1=v/r1=56.5 rad/s
w2=v/r2=22.4 rad/s
What is the average angular acceleration of the CD over the time
interval Dt=4473 s as the spot moves from the inner to outer radii?
a = (w2-w1)/Dt =-0.0076 rad/s2
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PHY 113 C Fall 2013-- Lecture 11
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Object rotating with constant angular velocity (a = 0)
w
R
v=Rw
v=0
Kinetic energy associated with rotation:
K   12mi vi2   12mi ri2ω2 
i
i
1
2
I
ω
;
2
where : I   mi ri2 “moment of inertia”
i
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PHY 113 C Fall 2013-- Lecture 11
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Moment of inertia:
I   mi ri
2
i
iclicker exercise:
A. a
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Which case has the larger I?
B. b
PHY 113 C Fall 2013-- Lecture 11
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Moment of inertia:
I   mi ri
2
i
I  2Ma
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2
I  2Ma  2mb
2
PHY 113 C Fall 2013-- Lecture 11
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19
Note that the moment of inertia depends on
both
a) The position of the rotational axis
b) The direction of rotation
m
m
d
d
I=2md2
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m
m
d
d
I=m(2d)2=4md2
PHY 113 C Fall 2013-- Lecture 11
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iclicker question:
Suppose each of the following objects each has the same total
mass M and outer radius R and each is rotating counterclockwise at an constant angular velocity of w=3 rad/s. Which
object has the greater kinetic energy?
(a) (Solid disk)
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(b) (circular ring)
PHY 113 C Fall 2013-- Lecture 11
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Various moments of inertia:
R
R
R
solid cylinder:
I=1/2 MR2
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solid sphere:
solid rod:
I=2/5 MR2
I=1/3 MR2
PHY 113 C Fall 2013-- Lecture 11
22
Calculation of moment of inertia:
Example -- moment of inertia of solid rod through an axis
perpendicular rod and passing through center:
R
M
M R 2
1


2
I   mi ri    dr r     r dr  MR 2
3
i
 2 R  R
R  2R 
2
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R
PHY 113 C Fall 2013-- Lecture 11
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iclicker exercise:
Three round balls, each having a mass M and
radius R, start from rest at the top of the incline.
After they are released, they roll without slipping
down the incline. Which ball will reach the bottom
first?
B
A
C
I A  MR 2
1
I B  MR 2  0.5MR 2
2
2
I C  MR 2  0.4 MR 2
5
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PHY 113 C Fall 2013-- Lecture 11
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Total kinetic energy of
rolling object :
K total  K rolling  K CM
1 2 1
2
 Iw  MvCM
2
2
Note that :
d
w
dt
ds
d
R
 Rw  vCM
dt
dt
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K total  K rolling  K CM
1 I
1
2
2
Rw   MvCM

2
2R
2
1 I
 2
  2  M vCM
2 R

PHY 113 C Fall 2013-- Lecture 11
25
iclicker exercise:
Three round balls, each having a mass M and
radius R, start from rest at the top of the incline.
After they are released, they roll without slipping
down the incline. Which ball will reach the bottom
first?
B
A
C
I A  MR 2
1
I B  MR 2  0.5MR 2
2
2
I C  MR 2  0.4 MR 2
5
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PHY 113 C Fall 2013-- Lecture 11
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Review of rotational energy associated with a rigid body
Rotational energy :
1
1
2
2
K rot   mi vi   mi wri 
2 i
2 i
1
1 2
2 2
  mi ri w  Iw
2 i
2
where I   mi ri
2
i
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PHY 113 C Fall 2013-- Lecture 11
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Note that for a given center of rotation, any solid
object has 3 moments of inertia; some times two or
more can be equal
m d
j
d
m
i
k
iclicker exercise:
Which moment of inertia is the smallest?
(A) i
(B) j
(C) k
IA=0
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IB=2md2
IC=2md2
PHY 113 C Fall 2013-- Lecture 11
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From Webassign:


I yy   mi xi2  32   2 2   2 2   4 2  kg  m 2
2
2
2
2
i
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PHY 113 C Fall 2013-- Lecture 11
29
Total kinetic energy of
rolling object :
K total  K rot  K CM
CM
CM
1 2 1
2
 Iw  MvCM
2
2
Note that :
d
w
dt
ds
d
R
 Rw  vCM
dt
dt
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K total  K rot  K CM
1 I
1
2
2
Rw   MvCM

2
2R
2
1 I
 2
  2  M vCM
2 R

PHY 113 C Fall 2013-- Lecture 11
30
iclicker exercise:
Three round balls, each having a mass M and
radius R, start from rest at the top of the incline.
After they are released, they roll without slipping
down the incline. Which ball will reach the bottom
first?
B
A
C
I A  MR 2
1
I B  MR 2  0.5MR 2
2
2
I C  MR 2  0.4 MR 2
5
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PHY 113 C Fall 2013-- Lecture 11
31
How to make objects rotate.

r
r sin 
Define torque:
t=rxF

t = rF sin 
F
F  ma
r  F  τ  r  ma  Iα
Note: We will define and use the “vector cross product”
next time. For now, we focus on the fact that the direction
of the torque determines the direction of rotation.
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PHY 113 C Fall 2013-- Lecture 11
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Another example of torque:
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PHY 113 C Fall 2013-- Lecture 11
33
Torque from T1 :
t 1   R1T1
(clockwise)
Torque from T2 :
t 2  R2T2
Example :
(counter clockwise)
R1  1m, T1  5N
R2  0.5m, T2  15N
Total torque :
t  R2T2  R1T1
 (0.5)(15)  (1)(5)Nm
 2.5 Nm (counter clockwise)
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PHY 113 C Fall 2013-- Lecture 11
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Newton’s second law applied to center-of-mass motion
dv i
dv CM
 Ftotal  M
 Fi   mi
dt
dt
i
i
Newton’s second law applied to rotational motion
Fi  mi
dv i
dv
 ri  Fi  ri  mi i
dt
dt
I   mi d i2
τ i  ri  Fi
i
v i  ω  ri
ri
d ω  ri 
 τ i  mi ri 
dt
i
dm
Fi
i
dω
 τ total  I
 Iα (for rotating about principal axis)
dt
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PHY 113 C Fall 2013-- Lecture 11
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An example:
A horizontal 800 N merry-go-round is a solid disc of radius
1.50 m and is started from rest by a constant horizontal
force of 50 N applied tangentially to the cylinder. Find the
kinetic energy of solid cylinder after 3 s.
R
K = ½ I w2
t Ia
In this case I = ½ m R2
FR  Ia
w  at 
FR
t
I
F
and
I
w  wi  at = at
t = FR
1 mg 2
R
2 g
1 2 1  FR 
1 F 2t 2
F 2t 2
2 50 N 
2
K  Iw  I 
t 

g

9
.
8
m/s
(
3
s
)
 275.625 J
2
2
2  I 
2 I/R
mg
800 N
2
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2
PHY 113 C Fall 2013-- Lecture 11
36
Re-examination of “Atwood’s” machine
R
T1
T1-m1g = m1a
I
T2
T2-m2g = -m2a
t =T2R – T1R = I a = I a/R
T1
T2
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

m2  m1

a  g
2
 m2  m1  I / R 

Ig 
m2  m1

τ  
2
R  m2  m1  I / R 
PHY 113 C Fall 2013-- Lecture 11
37
Another example: Two masses connect by a frictionless pulley having
moment of inertia I and radius R, are initially separated by h=3m.
What is the velocity v=v2= -v1 when the masses are at the same
height? m1=2kg; m2=1kg; I=1kg m2 ; R=0.2m .
Conservation of energy :
Ki  U i  K f  U f
0  m1 gh  12 m1v 2  12 m2 v 2  12
 m1 g  12 h   m2 g  12 h 
I
R2
v2
m1  m2
v
m1  m2  I / R 2
h
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v1
m1
2  1
m2

 0.19m / s
2
v
2
h/2
2  1  1 / 0.2 
PHY 113 C Fall 2013-- Lecture 11
38
Rolling motion reconsidered:
Kinetic energy associated with rotation:
I   mi ri 2
K rot  12 Iw 2
i
Distance to axis
of rotation
K tot  K com  K rot
Rolling:
If there is no slipping :
vcom  Rw
I  2

 K tot  M 1 
v
2  com
 MR 
1
2
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PHY 113 C Fall 2013-- Lecture 11
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Rolling motion reconsidered:
Kinetic energy associated with rotation:
I   mi ri 2
K rot  12 Iw 2
i
Distance to axis
of rotation
K tot  K com  K rot
Rolling:
If there is no slipping :
vcom  Rw
I  2

 K tot  M 1 
v
2  com
 MR 
1
2
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PHY 113 C Fall 2013-- Lecture 11
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Note that rolling motion is caused by the torque of friction:
Newton’s law for torque:
τ total  I
dω
 Iα
dt
F  f s  MaCM
F
fs
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f s R  Ia  IaCM / R
 aCM


1


fs  F
2
 1  MR / I 
For a solid cylinder, I  12 MR 2

fs R2

I

PHY 113 C Fall 2013-- Lecture 11
 f s  13 F
41
Bicycle or automobile wheel:
f s  MaCM
τ - Rf s  Iα  IaCM / R
t
τ/R
fs 
1  I/MR 2

fs
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For I  MR
1
f s  τ/R
2
PHY 113 C Fall 2013-- Lecture 11

2
42
iclicker exercise:
What happens when the bicycle skids?
A. Too much torque is applied
B. Too little torque is applied
C. The coefficient of kinetic friction is too small
D. The coefficient of static friction is too small
E. More than one of these
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