10/8/2013
PHY 113 C General Physics I
11 AM - 12:15 PM MWF Olin 101
Plan for Lecture 12:
Chapters 10 & 11 – Rotational motion,
torque, and angular momentum
1. Torque
2. Angular momentum
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PHY 113 C Fall 2013-- Lecture 12
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PHY 113 C Fall 2013-- Lecture 12
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Angular motion
angular “displacement”  q(t)
dq
angular “velocity”  w(t) 
dt dw
angular “acceleration”  a(t) 
dt
s
“natural” unit == 1 radian
 r di ns  3.14159 r di ns  180o
Relation to linear
variables: sq = r (qf-qi)
vq = r w
aq = r a
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PHY 113 C Fall 2013-- Lecture 12
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Rotational motion
q(t)
w(t)
angular “displacement”  q(t)
dq
angular “velocity”  w(t) 
dt dw
angular “acceleration”  a(t) 
dt
Special case of constant angular acceleration: a = a0:
w(t) = wi + a0 t
q(t) = qi + wi t + ½ a0 t2
( w(t))2  wi2 + 2 a0 (q(t) - qi )
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PHY 113 C Fall 2013-- Lecture 12
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Webassign Assignment #10:
A centrifuge in a medical laboratory rotates at an angular
speed of 3800 rev/min. When switched off, it rotates 48.0
times before coming to rest. Find the constant angular
acceleration of the centrifuge.
Recall:
Special case of constant angular acceleration: a = a0:
w(t) = wi + a0 t
q(t) = qi + wi t + ½ a0 t2
( w(t))2  wi2 + 2 a0 (q(t) - qi )
For w(t)  0
a
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(3800 rev/min )2   1 min  2  2 r d


2(q (t )  q i )
2  (48 rev)
rev
 60 s 
wi2

PHY 113 C Fall 2013-- Lecture 12
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Review of rotational energy associated with a rigid body
Rot tion l energy :
K rot 
1
1
2
mi vi2   mi (wri )

2 i
2 i
1
1
  mi ri 2w 2  Iw 2
2 i
2
here I   mi ri 2
i
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PHY 113 C Fall 2013-- Lecture 12
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2
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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
j
m d
d
m
i
k
iclicker exercise:
Which moment of inertia is the smallest?
(A) i
(B) j
(C) k
IA=0
IB=2md2
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IC=2md2
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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 12
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Digression – use of rotational energy in energy storage
http://www.beaconpower.com/products/about-flywheels.asp
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PHY 113 C Fall 2013-- Lecture 12
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Beacon Power company
(went bankrupt in 2011)
Continuing efforts –
http://www.scientificamerican.com/article.cfm?id=energy-storage-role-in-electric-grid
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PHY 113 C Fall 2013-- Lecture 12
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Physics of rolling --
Tot l kinetic energy of
rolling object :
CM
CM
K total  K rot  K CM
1 2 1
2
Iw  MvCM
2
2
Note th t :

w
K total  K rot  K CM
dq
dt
1 I
(Rw )2  1 MvCM 2
2 R2
2
1 I
 2
  2  M vCM
2R


ds
dq
R
 Rw  vCM
dt
dt
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PHY 113 C Fall 2013-- Lecture 12
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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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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?
2
2
B
A
I A  MR  1.0 MR
1
I B  MR 2  0.5MR 2
2
2
I C  MR 2  0.4 MR 2
5
C
Ki  Ui  K f  U f
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0  Mgh 
1 
I  2
M 1 
vCM  0
2  MR 2 
 vCM 
2 gh
1  I / MR 2
PHY 113 C Fall 2013-- Lecture 12
How can you make objects rotate?
)
13
q
r sin q
Define torque:
(
r
t=rxF
q
t = rF sin q
F
Fm
r  F  t  r  m  Ia
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Another example of torque:
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Tor ue from T1 :
t 1   R1T1
(clock ise)
Tor ue from T2 :
t 2  R2T2
Ex mple :
(counter clock ise)
R1  1m, T1  5N
R2  0.5m, T2  15N
Tot l tor ue :
t  R2T2  R1T1
 ((0.5)(15)  (1)(5))Nm
 2.5 Nm (counter clock ise)
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PHY 113 C Fall 2013-- Lecture 12
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Example form Webassign #11
t3
X
iclicker exercise
When the pivot point
is O, which torque is
zero?
A. t1?
B. t2?
C. t3?
t2
t1
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PHY 113 C Fall 2013-- Lecture 12
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Newton’s second law applied to center-of-mass motion
 Fi   mi
i
i
dv i
dv
 Ftotal  M CM
dt
dt
Newton’s second law applied to rotational motion
Fi  mi
dv i
dv
 ri  Fi  ri  mi i
dt
dt
I   mi d i2
t i  ri  Fi
i
v i  w  ri
ri
d (w  ri )
dt
i
dm
Fi
i
dw
I
 Ia (for rot ting bout princip l xis)
dt
 t i  mi ri 
 t total
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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
F
t Ia
K = ½ I w2
In this case I = ½ m R2
and
FR  Ia
w  at 
FR
t
I
I
w  wi  at = at
t = FR
1 mg 2
R
2 g
(50 N ) (3s) 2  275.625J
1 2 1  FR 
1 F 2t 2
F 2t 2
Iw  I 
t 
g
 9.8m/s 2
2
2  I 
2 I / R2
mg
800 N
2
K
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2
PHY 113 C Fall 2013-- Lecture 12
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Re-examination of “Atwood’s” machine
R
T1
T1-m1g = m1a
I
T2-m2g = -m2a
T2
t =T2R – T1R = I a = I a/R
T1


m2  m1

 g 
2 
 m2  m1  I/R 
T2

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
Ig 
m2  m1


R  m2  m1  I/R 2 
PHY 113 C Fall 2013-- Lecture 12
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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 .
Conserv tion 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 )
v
h
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v1
m1
I
R2
v2
m1  m2
m1  m2  I / R 2
(2)  (1)
m2
 0.19m / s

h/2 v2
(2)  (1)  (1 / (0.2)2 )
PHY 113 C Fall 2013-- Lecture 12
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Example from Webassign #10
Tu r  Tl r  Ia
I
Tl
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1
MR 2
2
PHY 113 C Fall 2013-- Lecture 12
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Note that rolling motion is caused by the torque of friction:
Newton’s law for torque:
t total  I
dw
 Ia
dt
F  f s  MaCM
F
f s R  Ia  IaCM / R
 aCM 


1
f s  F 

2
 1  MR / I 
For solid cylinder, I  12 MR 2
(
fs
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f s R2
I
)
 f s  13 F
PHY 113 C Fall 2013-- Lecture 12
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Bicycle or automobile wheel:
f s  MaCM
- Rf s  Ia  IaCM / R
t
fs 
fs
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/R
(1  I/MR )
2
For I  MR 2
1
f s  /R
2
PHY 113 C Fall 2013-- Lecture 12
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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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Vector cross product; right hand rule
C  AB
C  A B sin q
iˆ  ˆi  ˆj  ˆj  kˆ  kˆ  0
iˆ  ˆj   ˆj  ˆi  kˆ
ˆj  kˆ  kˆ  ˆj  ˆi
kˆ  iˆ  ˆi  kˆ  ˆj
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For unit vectors:
ˆi  ˆi  ˆj  ˆj  kˆ  kˆ  0
ˆi  ˆj  ˆj  ˆi  kˆ
k
ˆj  kˆ  kˆ  ˆj  ˆi
kˆ  ˆi  ˆi  kˆ  ˆj
j
i
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More details of vector cross products:
ˆi
 
A  B  Ax
Ay
ˆj
Az 
kˆ
Bx
By
Bz
Ay
Az
By
Bz
ˆi  Ax
Bx
Az ˆ Ax
j
Bx
Bz
Ay
By
kˆ
 
A  B  ( Ay Bz  Az By ) ˆi  ( Ax Bz  Az Bx ) ˆj  ( Ax By  Ay Bx ) kˆ
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iclicker exercise:
What is the point of vector products
A. To terrify physics students
B. To exercise your right hand
C. To define an axial vector
D. To keep track of the direction of rotation
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From Newton’s second law:
Fm
rF  t  rm  rm
iclicker exercise:
Consider
r
d (mv ) d
 (r  p )
dt
dt
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dv
d (mv ) d
 r
 (r  p )
dt
dt
dt
Is this
A. Wrong?
B. Approximately right?
C. Exactly right?
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From Newton’s second law – continued –
conservation of angular momentum:
d
(r  p )
dt
Define :
L  rp
dL
If t  0
0
dt
 L  (const nt)
rF  t 
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Torque and angular momentum
Define angular momentum: L  r  p
For composite object: L = Iw
Newton’s law for torque:
t total  I
dw dL

dt
dt
If t total  0 then
dL
0
dt
 L  const nt
In the absence of a net torque on a system,
angular momentum is conserved.
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iclicker exercise:
A student sits on a rotatable stool
holding a spinning bicycle wheel with
angular momentum Li. What
happens when the wheel is inverted?
( ) The student ill rem in t rest.
(b) The student ill rot te
counterclock ise.
(c) The student ill rot te clock ise.
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11
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More details:
Lbf  Lwheelf  Lbi  Lwheeli
Lbf  Lwheel  0  Lwheel
Lbf  2 Lwheel
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PHY 113 C Fall 2013-- Lecture 12
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Other examples of conservation of angular momentum
w1
m
w2
m
m
d1
d2
d1
m
d2
I2=2md22
I1=2md12
I1w1=I2w2 w2=w1 I1/I2
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What about centripetal acceleration?
w
ar = v2/R = w2 R
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Webassign problem:
A disk with moment of inertia I1 is initially rotating at angular
velocity wi. A second disk having angular momentum I2, initially is
not rotating, but suddenly drops and sticks to the second disk.
Assuming angular moment to be conserved, what would be the
final angular velocity wf?
I1wi  (I1  I 2 )w f
w f  wi
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I1
I1  I 2
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Summary – conservation laws we have studied so far
Conserved quantity
Linear momentum p
Angular momentum L
Mechanical energy E
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Necessary condition
Fnet = 0
tnet = 0
No dissipative forces
PHY 113 C Fall 2013-- Lecture 12
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