Transcript Physics 131: Lecture 14 Notes

Physics 151: Lecture 24
Today’s Agenda

Topics
Angular Momentum
More fun demos !!!
Gyroscopes Ch. 11.6
Ch. 11.3-5
Physics 151: Lecture 24, Pg 1

See text: 11.3
p=mv
Angular Momentum:
Definitions & Derivations

We have shown that for a system of particles
FEXT

dp

dt
Momentum is conserved if
FEXT  0
Now we also know,
t EXT
•
dL

dt
Angular momentum is conserved if
With angular momentum L = r x p
tEXT = 0
Animation
Physics 151: Lecture 24, Pg 2
See text: 11.5
What does it mean?


t EXT 
dL
dt
where
L r p
and
In the absence of external torques t EXT 
t EXT  r  FEXT
dL
0
dt
Total angular momentum is conserved
Also, remember that
L of a rigid body about
a fixed axis is:
L I

Analogue of p = mv !!
Physics 151: Lecture 24, Pg 3
See text: 11.4
Angular momentum of a rigid body
about a fixed axis:

In general, for an object rotating about a fixed (z) axis we
can write LZ = I 

The direction of LZ is given by the
right hand rule (same as ).

z
We will omit the ”Z” subscript for simplicity,
and write L = I 
LZ  I 

Physics 151: Lecture 24, Pg 4
Demonstration of
Conservation of Angular Momentum

Figure Skating :
z
z
Arm
A
Arm
B
LA =
IA
<
LB
IB
A > B
Physics 151: Lecture 24, Pg 5
Lecture 23, ACT 2
Angular momentum

Two different spinning disks have the same angular momentum,
but disk 1 has more kinetic energy than disk 2.
Which one has the biggest moment of inertia ?
(a) disk 1
(b) disk 2
K1
L  I1 1
(c) not enough info
>
K2
1
2
I1
disk 1
L  I2  2
<
I2
disk 2
Physics 151: Lecture 24, Pg 6
Example: Two Disks

A disk of mass M and radius R rotates around the z axis
with angular velocity 0. A second identical disk, initially
not rotating, is dropped on top of the first. There is friction
between the disks, and eventually they rotate together
with angular velocity F. What is F ?
z
z
0
F
Physics 151: Lecture 24, Pg 7
Example: Two Disks

First realize that there are no external torques acting on the twodisk system.
Angular momentum will be conserved !

Initially, the total angular momentum
is due only to the disk on the bottom:
z
2
LINI
1
 I1 1  MR 2  0
2
1
0
Physics 151: Lecture 24, Pg 8
Example: Two Disks

First realize that there are no external torques acting on the twodisk system.
Angular momentum will be conserved !

Finally, the total angular momentum is due
to both disks spinning:
z
LFIN  I1 1  I2  2  MR 2  F
2
1
F
Physics 151: Lecture 24, Pg 9
Example: Two Disks

1
MR 2  0  MR 2  F
2
Since LINI = LFIN
F 
z
1
0
2
z
LINI
0
LFIN
F
Physics 151: Lecture 24, Pg 10
Example: Two Disks

Now let’s use conservation of energy principle:
EINI = EFIN
1/2 I 02 = 1/2 (I + I) F2
F2 = 1/2 02
F = 0 / 21/2
z
z
EINI
0
EFIN
F
Physics 151: Lecture 24, Pg 11
Example: Two Disks

So we got a different answers !
Conservation of energy !
Conservation of momentum !
F’ = 0 / 21/2
F = 0 / 2
Which one is correct ?
Because this is an inelastic collision,
since E is not conserved (friction) !
Physics 151: Lecture 24, Pg 12
Lecture 25, Act 2

A mass m=0.1kg is attached to a cord passing through a
small hole in a frictionless, horizontal surface as in the
Figure. The mass is initially orbiting with speed i = 5rad/s
in a circle of radius ri = 0.2m. The cord is then slowly pulled
from below, and the radius decreases to r = 0.1m. How
much work is done moving the mas from ri to r ?
(A) 0.15 J
(B) 0 J
(C) - 0.15 J
ri
i
animation
Physics 151: Lecture 24, Pg 13
Lecture 25, Act 3

A particle whose mass is 2 kg moves in the xy
plane with a constant speed of 3 m/s along the
direction r = i + j. What is its angular momentum
(in kg · m2/s) relative to the origin?
a. 0 k
b. 6 (2)1/2 k
c. –6 (2)1/2 k
d. 6 k
e. –6 k
Physics 151: Lecture 24, Pg 14
See text: Ex. 11.11
Example: Bullet hitting stick

A uniform stick of mass M and length D is pivoted
at the center. A bullet of mass m is shot through
the stick at a point halfway between the pivot and
the end. The initial speed of the bullet is v1, and
the final speed is v2.
What is the angular speed F of the stick after
the collision? (Ignore gravity)
M
m
D
F
D/4
v1
v2
before
after
Physics 151: Lecture 24, Pg 15
See text: 11.5
Angular Momentum Conservation



A freely moving particle has a definite angular
momentum about any given axis.
If no torques are acting on the particle, its angular
momentum will be conserved.
In the example below, the direction of L is along the
z axis, and its magnitude is given by LZ = pd = mvd.
y
x
d
m v
Physics 151: Lecture 24, Pg 16
See text: Ex. 11.11
Example: Bullet hitting stick...


Conserve angular momentum around the pivot (z) axis!
The total angular momentum before the collision is due
only to the bullet (since the stick is not rotating yet).
LBEFORE  p x ( distance of closest approach )  mv1
D
4
L r  p
M
m
D
D/4

v1
before
Physics 151: Lecture 24, Pg 17
See text: Ex. 11.11
Example: Bullet hitting stick...


Conserve angular momentum around the pivot (z) axis!
The total angular momentum after the collision has
contributions from both the bullet and the stick.
LAFTER  mv 2
D
 I  F where I is the moment of inertia
4
of the stick about the pivot.
F
D/4
v2
after
Physics 151: Lecture 24, Pg 18
See text: Ex. 11.11
Example: Bullet hitting stick...

Set LBEFORE = LAFTER using I 
mv1
D
D 1
 mv 2  MD 2  F
4
4 12
1
MD 2
12
F 
3m
v1  v 2 
MD
M
m
D
F
D/4
v1
v2
before
after
Physics 151: Lecture 24, Pg 19
See text: Ex. 11.11
Example: Throwing ball from stool

A student sits on a stool which is free to rotate. The
moment of inertia of the student plus the stool is I. She
throws a heavy ball of mass M with speed v such that its
velocity vector passes a distance d from the axis of
rotation.
What is the angular speed F of the student-stool
system after she throws the ball ?
M
v
F
I
top view: before
See example 13-9
d
I
after
Physics 151: Lecture 24, Pg 20
See text: Ex. 11.11
Example: Throwing ball from stool...

Conserve angular momentum (since there are no external
torques acting on the student-stool system):
LBEFORE = 0
Mvd
F 
LAFTER = 0 = IF - Mvd
I
M
v
F
I
top view: before
See example 13-9
d
I
after
Physics 151: Lecture 24, Pg 21
See text: 11.6
Gyroscopic Motion:


Suppose you have a spinning gyroscope in the
configuration shown below:
If the left support is removed, what will happen ??
support

pivot
g
Physics 151: Lecture 24, Pg 22
See text: 11.6
Gyroscopic Motion...


Suppose you have a spinning gyroscope in the
configuration shown below:
If the left support is removed, what will happen ?
The gyroscope does not fall down !

pivot
g
Physics 151: Lecture 24, Pg 23
See text: 11.6
Gyroscopic Motion...

... instead it precesses around its pivot axis !

This rather odd phenomenon can be easily understood
using the simple relation between torque and angular
momentum we derived in a previous lecture.

pivot
Physics 151: Lecture 24, Pg 24
See text: 11.6
Gyroscopic Motion...


The magnitude of the torque about the pivot is t = mgd.
The direction of this torque at the instant shown is out of the page
(using the right hand rule).
The change in angular momentum at the instant shown must
also be out of the page!
d
dL
t
dt
L

pivot
mg
Physics 151: Lecture 24, Pg 25
See text: 11.6
Gyroscopic Motion...

Consider a view looking down on the gyroscope.
The magnitude of the change in angular momentum in a time
dt is dL = Ld.
So
dL
d
L
 L
dt
dt
where  is the “precession frequency”
L(t)
dL
top view
d
pivot
L(t+dt)
Physics 151: Lecture 24, Pg 26
See text: 11.6
Gyroscopic Motion...
dL
t
 L
dt

t
L

So

In this example t = mgd and L = I:

The direction of precession is given by applying the right
hand rule to find the direction of t and hence of dL/dt.
d

L


mgd
I
pivot
mg
Physics 151: Lecture 24, Pg 27
Lecture 24, Act 1
Statics
Suppose you have a gyroscope that is supported on
a gymbals such that it is free to move in all angles,
but the rotation axes all go through the center of
mass. As pictured, looking down from the top, which
way will the gyroscope precess?
(a) clockwise (b) counterclockwise (c) it won’t precess


Physics 151: Lecture 24, Pg 28
Lecture 24, Act 1
Statics
Remember that t/L.
So what is t?
t=rxF
r in this case is zero. Why?
Thus  is zero.
It will not precess. At All. Even if you move the base.
This is how you make a direction finder for an airplane.

Answer (c) it won’t precess

Physics 151: Lecture 24, Pg 29
Recap of today’s lecture

Chapter 11,
Conservation of Angular Momentum
Spinning Demos
Physics 151: Lecture 24, Pg 30