Physics 211: Lecture 23 Todays Agenda

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Physics 211 Lecture 23Todays Agenda

• Combining moments of inertia
• Comment about ? I? (not true if I is
  changing!!)
• General expression for the angular momentum of a
  system
• Sliding beam example
• Vector considerations of angular momentum
• Bike wheel and rotating stool
• Gyroscopic Motion
• Comments about moving rotation axis

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Just making sure that this is clear

• If you have some object made up of several parts,
  the total moment of inertia about some axis is
  the sum of the moments of inertia of the parts
  about the same axis

axis
L
M
m
Itotal Istick Imass 1/3ML2 mL2
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Lecture 23, Act 1Angular 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 (c) not
enough info
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Lecture 23, Act 1 Solution
(using L I? )
If they have the same L, the one with the biggest
I will have the smallest kinetic energy.
w1
w2
I1 lt I2
disk 2
disk 1
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When does ? I??not work ?

• Last time we showed that

• This is the fundamental equation for
  understanding rotation.

• If we write L I?, then

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When does ? I??not work?
Now suppose ?EXT 0
So in this case we can have an ? without an
external torque!
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Example...

• A puck in uniform circular motion will experience
  rotational acceleration if its moment of inertia
  is changed.
• 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 gt I2
w2
w2 gt w1
The puck accelerates without external torque!!
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Review Angular Momentum

• where
  and
• In the absence of external torques

Total angular momentum is conserved

• This is a vector equation.
• Valid for individual components.

9
Review...

• In general, for an object rotating about a fixed
  (z) axis we can write LZ I ?
• The direction of LZ is given by theright hand
  rule (same as ?).

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Review...

• 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
v
m
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Lecture 23, Act 2Rotations
Puck on ice

• 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

w
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Lecture 23, Act 2 Solution

• Since the string is pulled through a hole at the
  center of rotation, there is no torque Angular
  momentum is conserved.

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A general expression for L of a system

• For a system of particles we can write
• Express position and velocity in terms of the
  center of mass

ri Rcm ri
where ri and vi are the position andvelocity
measured in the CM frame.
vi Vcm vi
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A general expression for L of a system...

• So we can write

Expanding this
Which becomes
MVcm 0
MRcm 0
Lcm
L
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A general expression for L of a system...
L Lcm L

• So finally we get the simple expression
• Where is the angular
  momentum of the CM
• and L is the angular momentum about the CM.
• The total angular momentum of a system about a
  given axis is the sum of the angular momentum of
  the center of mass about this axis and the
  angular momentum about an axis through the center
  of mass.

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A general expression for L of a system...

• We have just showed that L Lcm L
• Picture it this way

y
origin (axis)
x
d
m,I
CM
?
v
due to movement of CM
due to rotation about CM
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Example 1 (Homework Problem)

• A rod of length d and mass m1 is sliding on a
  frictionless surface with speed vo as shown
  (without rotating). An initially stationary
  block having mass m2, sticks to the end of the
  rod as it goes by.
• What is the final angular velocity ?F of the
  block-rod system?

d
vo
cm
?F
m1
m2
top view initial top view
final
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Example 1...

• Choose the origin to be at the location of the
  block before the collision. We can determine the
  y-position of the center of mass before the
  collision.

y
vo
d/2
m1
x
top view initial
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Example 1...

• It is best to take z component of angular
  momentum about the point (0,ycm). The angular
  momentum before the collision is due entirely to
  the center of mass motion of the rod since the
  rod is not rotating.

y
vo
d/2
ycm
m1
x
top view initial
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Example 1...

• The z component of angular momentum about the
  point (0,ycm) after the collision is due to
  rotation about the center of mass of the
  rodblock

0
y
Icm
(0,ycm)
vF
?F
x
top view final
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Example 1...

• We need to know the moment of inertia Icm about
  the center of mass of the system.

Irod (using axis thm.)
Iblock
m1
cm of rod
d/2 - ycm
cm of block-rod system
d/2
ycm
m2
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Example 1...

• Using conservation of angular momentum

And plugging in for Icm and ycm
Icm
y
vF
ycm
?F
x
top view final
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Example 1...

• Suppose m1 2m2 2m

initial
final
Icm
d
vo
cm
?F
2m
m
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Angular momentum is a vector!Demo Turning the
bike wheel.

• 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.
• Whats going on?

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Turning the bike wheel...

• Since there are no external torques acting on the
  student-stool system, angular momentum is
  conserved.
• Initially LINI LW,I
• Finally LFIN LW,F LS

LS
LW,I
LW,I LW,F LS
LW,F
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Lecture 23, Act 3Rotations

• 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 (b) double (c)
stay the same
??
(1)
(2)
(3)
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Lecture 23, Act 3 Solution
LNET
LNET
LNET
not turning
LW
LW
LS
LW
1
2
3
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Gyroscopic Motion

• Suppose you have a spinning gyroscope in the
  configuration shown below
• If the left support is removed, what will happen??

pivot
support
?
g
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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
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Gyroscopic Motion...
Bicycle wheel

• ... 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 Lecture
  22.

pivot
?
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Gyroscopic Motion...

• The magnitude of the torque about the pivot is ?
  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
L
pivot
?
mg
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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
• where?? is the precession frequency

L(t)
dL
d?????????
pivot
L(tdt)
top view
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Gyroscopic Motion...
Toy Gyroscope

• So
• In this example, ? mgd and L I?
• The direction of precession is given by applying
  the right hand rule to find the direction of ?
  and hence of dL/dt.

d
?
L
pivot
?
mg

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Aside (optional) Why we can write ? I? when
the CM is accelerating

• We have shown that for any system
• Express position and velocity in terms of the
  center of mass
• Write

where ri is the position measured in the CM
frame.
ri Rcm ri
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Aside...

• So becomes

But
which leaves
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Aside...

• The sum on the left side is the total torque
  about the CM.
• For a rigid, symmetric, solid object,so

Is always true, regardless ofthe motion of the
CM!
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Recap of todays lecture

• General expression for the angular momentum of a
  system
• Sliding beam example
• Vector considerations of angular momentum
• Bike wheel and rotating stool
• Gyroscopic Motion
• Comments about moving rotation axis