Physics 207: Lecture 28
Announcements
•Final hwk assigned this week, final quiz next week
• Review session on Thursday May 19, 2:30 – 4:00pm, Here
Today’s Agenda
z
z
z
z
z
Recap Angular Momentum
Rotation about a fixed axis
ÍL = Iω
ÍExample: Two disks
ÍStudent on rotating stool
Angular momentum of a freely moving particle
ÍBullet hitting stick
ÍStudent throwing ball
Comment about τ = Iα (not true if I is changing!!)
Vector considerations of angular momentum
ÍBike wheel and rotating stool
1
Angular momentum, L
z
z
τ EXT =
dL
dt
where
L=r ×p
and
In the absence of external torques τ EXT =
τ EXT = r × FEXT
dL
=0
dt
Total angular momentum is conserved
2
Page 1
Angular momentum of a rigid body
about a fixed axis:
z
Consider a rigid distribution of point particles rotating in the x-y
plane around the z axis, as shown below. The total angular
momentum around the origin is the sum of the angular
momenta of each particle:
L = ∑ ri × p i = ∑ m i ri × v i = ∑ m i ri v i k̂
i
i
i
We see that L is in the z direction.
m2
Using vi = ω ri , we get
v2
L = ∑ mi ri ω kˆ
2
i
L = Iω
(since ri and vi are
perpendicular)
v1
j
r2
ω
i r1 m1
m3
r3
v3
Analogue of p = mv!!
3
Angular momentum of a rigid body
about a fixed axis:
z
In general, for an object rotating about a fixed (z) axis we
can write LZ = I ω
z
The direction of LZ is given by the
right hand rule (same as ω).
z
z
We will omit the Z subscript for simplicity,
and write L = I ω
LZ = I ω
ω
4
Page 2
Example: Two Disks
z
A disk of mass M and radius R rotates around the z axis
with angular velocity ωi. 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 ωf.
z
z
ωi
ωf
5
Example: Two Disks
z
First realize that there are no external torques acting on the
two-disk system.
ÍAngular momentum will be conserved!
z
Initially, the total angular momentum
is due only to the disk on the bottom:
z
2
Li = I1 ω1 =
1
MR 2ω f
2
1
ω0
6
Page 3
Example: Two Disks
z
First realize that there are no external torques acting on the
two-disk system.
ÍAngular momentum will be conserved!
z
Finally, the total angular momentum is due
to both disks spinning:
z
2
1
Lf = I1 ω1 + I2 ω 2 = MR 2ω f
ωf
7
Example: Two Disks
z
1
MR 2 ω i = MR 2 ω f
2
Since Li = Lf
ωf =
1
ω
2 i
z
An inelastic collision,
since E is not
conserved (friction)!
z
Li
Lf
ωf
ωf
8
Page 4
Example: Rotating Table
z
A student sits on a rotating stool with his arms extended
and a weight in each hand. The total moment of inertia is
Ii, and he is rotating with angular speed ωi. He then pulls
his hands in toward his body so that the moment of inertia
reduces to If. What is his final angular speed ωf?
ωf
ωi
If
Ii
9
Example: Rotating Table...
z
Again, there are no external torques acting on the studentstool system, so angular momentum will be conserved.
ÍInitially: Li = Iiωi
ω f Ii
=
ÍFinally: Lf = If ωf
ω
I
i
ωf
ωi
If
Ii
Li
f
Lf
10
Page 5
Lecture 28, Act 1
Angular Momentum
z
A student sits on a freely turning stool and rotates with
constant angular velocity ω1. She pulls her arms in, and due
to angular momentum conservation her angular velocity
increases to ω2. In doing this her kinetic energy:
(a) increases
(b) decreases
(c) stays the same
ω2
ω1
I2
I1
L
L
11
Lecture 28, Act 1
Solution
1 2 L2
Iω =
2
2I
(using L = Iω)
z
K=
z
L is conserved:
K2 > K1
I2 < I1
K increases!
ω2
ω1
I2
I1
L
L
12
Page 6
Lecture 28, Act 1
Solution
z
z
Since the student has to force her arms to move toward her
body, she must be doing positive work!
The work/kinetic energy theorem states that this will increase
the kinetic energy of the system!
ω2
ω1
I2
I1
L
L
13
Angular Momentum of a
Freely Moving Particle
z
We have defined the angular momentum of a particle about
the origin as L = r × p
z
This does not demand that the particle is moving in a circle!
ÍWe will show that this particle has a constant angular
momentum!
y
x
v
14
Page 7
Angular Momentum of a
Freely Moving Particle...
z
Consider a particle of mass m moving with speed v along
the line y = -d. What is its angular momentum as
measured from the origin (0,0)?
y
x
d
m
v
15
Angular Momentum of a
Freely Moving Particle...
z
z
We need to figure out L = r × p
The magnitude of the angular momentum is:
L = r × p = rp sin θ = p[r sin θ] = pd =
p x (dis tan ce of closest approach )
z
Since r and p are both in the x-y plane, L will be in the z
direction (right hand rule): LZ = pd
y
d
θ
r
x
p=mv
θ
16
Page 8
Angular Momentum of a
Freely Moving Particle...
z
z
So we see that the direction of L is along the z axis, and its
magnitude is given by LZ = pd = mvd.
L is clearly conserved since d is constant (the distance of
closest approach of the particle to the origin) and p is
constant (momentum conservation).
y
x
d
p
17
Example: Bullet hitting stick
z
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.
ÍWhat is the angular speed ωF of the stick after the
collision? (Ignore gravity)
M
D
m
ωF
D/4
v1
v2
initial
final
18
Page 9
Example: Bullet hitting stick...
z
z
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).
Li = p x ( dis tan ce of closest approach ) = mv 1
D
4
M
D
m
D/4
v1
initial
19
Example: Bullet hitting stick...
z
z
Conserve angular momentum around the pivot (z) axis!
The total angular momentum after the collision has
contributions from both the bullet and the stick.
Í
Lf = mv 2
D
where I is the moment of inertia
+ Iω F
of the stick about the pivot.
4
ωF
D/4
v2
final
20
Page 10
Example: Bullet hitting stick...
z
I=
Set Li = Lf using
mv1
1
MD 2
12
D
D 1
= mv 2 + MD 2 ω F
4
4 12
ωF =
3m
(v − v 2 )
MD 1
M
D
m
ωF
D/4
v1
v2
initial
final
21
Example: Throwing ball from stool
z
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.
ÍWhat is the angular speed ωF of the student-stool
system after she throws the ball?
M
v
ωF
I
d
I
top view: initial
final
Page 11
22
Example: Throwing ball from stool...
z
Conserve angular momentum (since there are no external
torques acting on the student-stool system):
ÍLi = 0
Mvd
ωF =
ÍLf = 0 = IωF - Mvd
I
M
v
ωF
I
d
I
top view: initial
final
23
Review...
z
z
z
A freely moving particle has a definite angular momentum
about any 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
24
Page 12
Lecture 28, Act 2
Angular momentum
z
Two different spinning disks have the same angular momentum,
but disk 1 has more kinetic energy than disk 2.
ÍWhich one has the biggest moment of inertia?
(a) disk 1
(b) disk 2
(c) not enough info
25
Lecture 28, Act 2
Solution
1
1
1
K = I ω2 = I 2 ω2 = L2
2
2I
2I
(using L = Iω )
If they have the same L, the one with the biggest I
will have the smallest kinetic energy.
L = I1 ω1
L = I2 ω2
ω1
ω2
I1
<
I2
disk 1
disk 2
26
Page 13
When does τ = Iα not work ?
dL
dt
z
Last time we showed that τ EXT =
z
This is the fundamental equation for understanding rotation.
z
If we write L = Iω, then
dL d ( I ω )
dI
dω
dI
= Iα +ω
=
=I
+ω
dt
dt
dt
dt
dt
τ EXT = I α + ω
dI
dt
We can’t assume τ = Iα when the moment of inertia is changing!
27
When does τ = Iα not work?
τ EXT = I α + ω
dI
dt
Now suppose τEXT = 0:
Iα +ω
dI
=0
dt
α=−
ωdI
I dt
So in this case we can have an α without an external torque!
28
Page 14
Example...
z
A puck in uniform circular motion will experience rotational
acceleration if its moment of inertia is changed.
z
Changing the radius changes the moment of inertia, but
produces no torque since the force of the string is along the
radial direction. (since r X F = 0)
I1 > I2
ω1
ω2
ω2 > ω1
The puck accelerates without external torque!!
29
Lecture 28, Act 3
Rotations
z
A puck slides in a circular path on a horizontal frictionless table.
It is held at a constant radius by a string threaded through a
frictionless hole at the center of the table. If you pull on the
string such that the radius decreases by a factor of 2, by what
factor does the angular velocity of the puck increase?
(a) 2
(b) 4
(c) 8
ω
30
Page 15
Lecture 28, Act 3
Solution
z
Since the string is pulled through a hole at the center of rotation,
there is no torque: Angular momentum is conserved.
2
L1 = I1ω1 = mR2ω1
mR2ω1 = m
ω1 =
m
1 2
R ω2
4
1
ω
4 2
ω2 = 4ω1
ω1
R
R
L2 = I2ω2 = m ⎛⎜ ⎞⎟ ω2
⎝2⎠
=
m
ω2
R/2
31
Review: Angular Momentum
z
z
τ EXT =
dL
dt
where
L=r ×p
In the absence of external torques
and
τ EXT =
τ EXT = r × FEXT
dL
=0
dt
Total angular momentum is conserved
z
z
This is a vector equation.
Valid for individual components.
32
Page 16
Review...
z
In general, for an object rotating about a fixed (z) axis we
can write LZ = I ω
z
The direction of LZ is given by the
right hand rule (same as ω).
z
LZ = I ω
ω
33
Angular momentum is a vector!
Demo: Turning the bike wheel.
z
A student sits on the rotatable stool holding a bicycle wheel
that is spinning in the horizontal plane. She flips the
rotation axis of the wheel 180o, and finds that she herself
starts to rotate.
ÍWhat’s going on?
34
Page 17
Turning the bike wheel...
Since there are no external torques acting on the studentstool system, angular momentum is conserved.
ÍInitially: LINI = LW,I
ÍFinally: LFIN = LW,F + LS
z
LS
LW,I
LW,I = LW,F + LS
LW,F
35
Lecture 28, Act 4
Rotations
z
A student is initially at rest on a rotatable chair, holding a
wheel spinning as shown in (1). He turns it over and starts
to rotate (2). If he keeps twisting, turning the wheel over
again (3), his rotation will:
(a) stop
(1)
(b) double
(2)
(c) stay the same
??
(3)
36
Page 18
Lecture 28, Act 4
Solution
LNET
LW
[1]
LNET
LW
LS
[2]
LNET
LW
not
turning
[3]
37
Page 19