PHY 113 A General Physics I
9-9:50 AM MWF Olin 101
Plan for Lecture 18:
Chapter 10 – rotational motion
1. Torque
2. Conservation of energy including
both translational and rotational
motion
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PHY 113 A Fall 2012 -- Lecture 18
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PHY 113 A Fall 2012 -- Lecture 18
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Review of rotational energy associated with a rigid body
Rotational energy :
1
1
2
2
K rot   mi vi   mi ri 
2 i
2 i
1
1 2
2 2
  mi ri   I
2 i
2
where I   mi ri
2
i
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PHY 113 A Fall 2012 -- Lecture 18
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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 A Fall 2012 -- Lecture 18
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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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Total kinetic energy of
rolling object :
K total  K rot  K CM
CM
CM
1 2 1
2
 I  MvCM
2
2
Note that :
d

dt
ds
d
R
 R  vCM
dt
dt
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K total  K rot  K CM
1 I
1
2
2
R   MvCM

2
2R
2
1 I
 2
  2  M vCM
2 R

PHY 113 A Fall 2012 -- Lecture 18
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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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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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Another example of torque:
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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 A Fall 2012 -- Lecture 18
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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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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 2
t Ia
In this case I = ½ m R2
FR  Ia
  at 
FR
t
I
F
and
I
  i  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  I  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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PHY 113 A Fall 2012 -- Lecture 18
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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 A Fall 2012 -- Lecture 18
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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 
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Rolling motion reconsidered:
Kinetic energy associated with rotation:
I   mi ri 2
K rot  12 I 2
i
Distance to axis
of rotation
K tot  K com  K rot
Rolling:
If there is no slipping :
vcom  R
I  2

 K tot  M 1 
v
2  com
 MR 
1
2
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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 A Fall 2012 -- Lecture 18
 f s  13 F
16
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 A Fall 2012 -- Lecture 18

2
17
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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