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Catching up from Yesterday
In the two cases shown below identical ladders are leaning against
frictionless walls. In which case is the force of friction between the
ladder and the ground bigger?
A) Case 1
B) Case 2
Case 1
C) Same
Case 2
Mechanics Lecture 18, Slide 1
CheckPoint
In the two cases shown below identical ladders are leaning against
frictionless walls. In which case is the force of friction between the
ladder and the ground bigger?
A) Case 1
B) Case 2
C) Same
Case 1
A) Because the bottom of the ladder is
further away from the wall.
B) The angle is steeper, which means there
is more normal force and thus more friction.
C) Both have same mass.
Mechanics Lecture 18, Slide 2
Case 2
CheckPoint
Suppose you hang one end of a beam from the ceiling by a rope
and the bottom of the beam rests on a frictionless sheet of ice.
The center of mass of the beam is marked with an black spot.
Which of the following configurations best represents the
equilibrium condition of this setup?
A)
B)
C)
Mechanics Lecture 18, Slide 3
CheckPoint
Which of the following configurations best represents the
equilibrium condition of this setup?
A)
B)
C)
Demo
If the tension has any horizontal component, the
beam will accelerate in the horizontal direction.
Objects tend to attempt to minimize their potential
energy as much as possible. In Case C, the center of
mass is lowest.
Mechanics Lecture 18, Slide 4
Stability & Potential Energy
CM Demos
footprint
footprint
Mechanics Lecture 18, Slide 5
Physics 211
Lecture 19
Today’s Concepts:
Angular Momentum
Mechanics Lecture 19, Slide 6
Linear and Rotation
Mechanics Lecture 19, Slide 7
New Cast of Characters with Old Theme

Remember
Mechanics Lecture 19, Slide 8
Angular Momentum
We have shown that for a system of particles
FEXT
Momentum is conserved if
dp

dt
FEXT  0
What is the rotational version of this?
The rotational analogue of force F is torque
  r F
Define the rotational analogue of momentum p to be
Angular Momentum:
Lrp
For a symmetric solid object
L  I
Mechanics Lecture 19, Slide 9
Torque & Angular Momentum
 EXT
dL

dt
where
Lrp
L  I
In the absence of external torques
and
 EXT
 ext  r  FEXT
dL

0
dt
Total angular momentum is conserved
Mechanics Lecture 19, Slide 10
Example – Disk dropped on Disk
top,initial  o
Demo
top, final  bottom, final   f
bottom,initial  0
Friction is an “internal force” because it is “internal” to the
“system”. Friction is between two parts of the “system”.
Friction may change the Kinetic Energy.
But friction, being an internal force, does not change the
Angular Momentum (nor the linear momentum in linear
situations).
Mechanics Lecture 19, Slide 11
Example – Disk dropped on Disk
top,initial  o
top, final  bottom, final   f
bottom,initial  0
1
Linitial  MR 2o  0
2
1
MR 2o  MR 2 f
2

1
1
2
L final  MR  f  MR 2 f
2
2
1
 f  0
2
Mechanics Lecture 19, Slide 12
What about Kinetic Energy?
top,initial  o
top, final  bottom, final   f
bottom,initial  0
2
1 2 L
K  I 
2
2I
K before
L2

2 I before
same
K after
L2

2 I after
x2
Mechanics Lecture 19, Slide 13
CheckPoint
In both cases shown below a solid disk of mass M, radius R,
and initial angular velocity o is dropped onto an initially
stationary second disk having the same radius. In Case 2 the mass of
the bottom disk is twice as big as in Case 1. If there are no external
torques acting on either system, in which case is the final kinetic
energy of the system bigger?
A) Case 1
B) Case 2
C) Same
top,initial  o
top,initial  o
M
M
M
Same initial L
bottom,initial  0
bottom,initial  0
Case 1
Case 2
2M
Mechanics Lecture 19, Slide 14
CheckPoint
In which case is the final kinetic energy of the system bigger?
A) Case 1
B) Case 2
C) Same
M
2M
M
M
Case 1
A) L is equal but bigger mass in case 2
Case 2
L2
K
2I
Demo: wheel rim drop
B) Case 2 has greater mass
C) Conservation of angular momentum
Mechanics Lecture 19, Slide 15
Your Comments
Sort of . . .
Work is done by friction so KE is not conserved.
Okay; now what does that mean?
Find the final KE in both cases.
True but what about the final KE?
That’s for omega but what about the final KE?
True but what about the final KE?
True but what about the
????
final KE?
????
This doesn’t mean anything.
Mechanics Lecture 19, Slide 16
Point Particle Moving in a Straight Line
L  R p
axis

R1
1

p
  
L1  R1  p
 R1 p sin( 1 )
 pD

R2

R3
2

p

 
L2  R2  p
 R2 p sin( 2 )
 pD
D
3

p

 
L3  R3  p
 R3 p sin( 3 )
 pD
Mechanics Lecture 19, Slide 17
L  mvD  pD
L  R p
axis
D

p
Demo: train on bike wheel
Direction given by right hand rule
(out of the page in this case)
Mechanics Lecture 19, Slide 18
Playground Example
Before
After
M

Top
View
R
m
v
LBefore  mvR
LAfter
1
 I  ( MR 2  mR2 )
2
Disk
LBefore  LAfter
Kid
v
1

R 1  M 2m 
Mechanics Lecture 19, Slide 19
Algebra Details
LBefore  mvR
w=
LAfter
mvR
1
 I  ( MR 2  mR2 )
2
1
( MR 2 + mR 2 )
2
w=
mv
1
( MR + mR )
2
w=
LBefore  LAfter
mv
1
( M + m)R
2
Disk
Kid
v
1

R 1  M 2m 
Mechanics Lecture 19, Slide 20
CheckPoint
The magnitude of the angular momentum of a freely rotating
disk around its center is L. You toss a heavy block onto the disk
along the direction shown. Friction acts between the disk and the
block so that eventually the block is at rest on the disk and rotates
with it. What is the magnitude of the final angular momentum of
the disk-block system:
Instead of a block, imagine it’s a kid hopping onto a merry go round
A) > L
B)  L
C) < L
Top View
The block provides no angular momentum since its perpendicular distance from
the axis of rotation is 0. No external forces act on this systems, so angular
momentum is conserved.
Mechanics Lecture 19, Slide 21
CheckPoint
What is the magnitude of the final angular momentum of the diskblock system:
A) > L
B)  L
C) < L
Top View
The block provides no angular momentum since its perpendicular distance from
the axis of rotation is 0. No external forces act on this systems, so angular
momentum is conserved.
Mechanics Lecture 19, Slide 22
Your Comments
Will it be conserved in the next CheckPoint?
Good! Altho’ I wouldn’t call it an external force.
Good! The block’s perpendicular distance from the axis of rotation is zero. .
But the block is NOT at rest.
What is the final angular momentum?
What does that mean for the final angular momentum?
What about the
Zero!>!>! How can that be?
next
CheckPoint?
okay
Tell me more.
What about the block?
Tell me more.
!!!!!
Mechanics Lecture 19, Slide 23
ACT
The magnitude of the angular momentum of a freely rotating disk
around its center is L. You toss a heavy block onto the disk along the
direction shown. Friction acts between the disk and the block so that
eventually the block is at rest on the disk and rotates with it. Is the
total angular momentum of the disk-block system conserved during
this?
A) Yes
B) No
Top View
Friction decreases the angular momentum of the disk
But the TOTAL angular momentum of the block+disk is CONSTANT!
Mechanics Lecture 19, Slide 24
CheckPoint
The magnitude of the angular momentum of a freely rotating disk
around its center is L. You toss a heavy block onto the disk along the
direction shown. Friction acts between the disk and the block so that
eventually the block is at rest on the disk and rotates with it. What is
the magnitude of the final angular momentum of the disk-block
system:
A) > L
B)  L
C) < L
Top View
Instead of a block, imagine it’s a kid hopping onto a merry go round
Mechanics Lecture 19, Slide 25
CheckPoint
What is the magnitude of the final angular momentum of the
disk-block system:
A) > L
B)  L
C) < L
Top View
A) Since the block has angular momentum to begin with,
that is added into L.
B) L is conserved so Linitial=Lfinal
C) The block adds momentum going in the opposite
direction, so the total momentum is smaller than the
momentum of the disk.
Mechanics Lecture 19, Slide 26
Your Comments
Bravo!
Bravo!
What about the block?
What about the block?
Why bigger? Angular momentum has a direction connected to it.
What about the angular momentum of the block?
Any better by now?
No, not this time.
Good.
Okay but tell me more.
Smaller than what? In the previous CheckPoint, the lever arm was zero.
Tell me more.
Mechanics Lecture 19, Slide 27
ACT
A student holding a heavy ball sits on the outer edge a merry go
round which is initially rotating counterclockwise. Which way
should she throw the ball so that she stops the rotation?
A) To her left
B) To her right
C) Radially outward
C
A
B

top view: initial
final
Mechanics Lecture 19, Slide 28
ACT
A student is riding on the outside edge of a merry-go-round
rotating about a frictionless pivot. She holds a heavy ball at
rest in her hand. If she releases the ball, the angular
velocity of the merry-go-round will:
A) Increase
B) Decrease C) Stay the same
1
Mechanics Lecture 19, Slide 29
Mechanics Lecture 19, Slide 30
m2
I disk
1
2
 m1 R
2
m1
0
Linitial  I disko
Mechanics Lecture 19, Slide 31
m2
I disk
1
 m1 R 2
2
m1
o
1
K initial  I disko2
2
Mechanics Lecture 19, Slide 32
m2
m1
0
Linitial  I disk 0
f
L final  ( I disk  I rod ) f
L final  Linitial
I rod 
1
m2 L2
12
Solve for f
Mechanics Lecture 19, Slide 33
Lfinal  Linitial
Mechanics Lecture 19, Slide 34
M
f
K final
1
 ( I disk  I rod ) 2f
2
Mechanics Lecture 19, Slide 35
Just like
Faverage
P

t
for linear momentum
M
f
Lrod  I rod ( f  0 )
 average
L

t
for angular momentum
 average 
I rod ( f  0 )
t
Mechanics Lecture 19, Slide 36
Conservation Ideas
Internal forces do not change the momentum.
Internal torques do not change the angular
momentum.
However, internal forces or internal torques
can still do work and, thus,
change the Kinetic Energy.
Mechanics Lecture 19, Slide 37
Heads Up! for next preLecture
Mechanics Lecture 19, Slide 38