Quick Quiz 44
More Angular Momentum
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A block is sliding along a surface, following a straight
line, and not rotating about its centre of mass. The
angular momentum vector of the block:
General motion of a rigid body
Collisions involving rotation
a) points out of the surface
b) lies in the surface, but perpendicular to the motion
c) is zero – the block is sliding in a straight line
d) not enough information
Text Section 11.6
Physics 1D03 - Lecture 26
Angular momentum of a particle:
of a rotating rigid body:
1
L = r × p = r × ( mv )
Physics 1D03 - Lecture 26
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Collisions: Collisions can conserve angular momentum
as well as linear momentum.
L = I ω.
Total linear momentum is conserved if there is no external
force during the collision.
In general, for a moving, rotating rigid body,
L = r × ( mv CM ) + I CM ω
Total angular momentum is conserved if there is no external
torque during the collision.
The first term is called the “orbital” angular momentum
and the second term is the “spin” angular momentum.
Physics 1D03 - Lecture 26
Angular momentum may be calculated about any axis. Usually
it is convenient to use an axis through the centre of mass,
unless one of the colliding objects actually rotates about
some other fixed axis.
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Physics 1D03 - Lecture 26
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Quick Quiz 45
The metre stick is resting on a frictionless
surface (not attached to anything ) before the
ball hits the end at right angles. Assuming the ball
still stops, what should we write for the stick?
A student sits on a tree branch overhanging a merrygo-round platform that rotates freely.
I.
First, he jumps straight down and lands on the platform.
The angular velocity of the platform:
Now there is no external force. So,
Linear momentum is conserved:
a) increases
b) decreases
c) stays the same
M vCM = m v0
CM
Angular momentum is conserved:
ICM ω = mv0 L/2
II. He then jumps straight up and hangs onto the tree branch.
The angular velocity of the platform:
If the collision is elastic (it may not be),
kinetic energy is conserved:
a) increases
b) decreases
c) stays the same
v0
½ mv02 = ½ IP ω2 + ½ M v2CM
Physics 1D03 - Lecture 26
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Rotating
Rotatingrods
rodsdropped
droppedon
onbearing
bearingdemo
demo
Quick Quiz 46
A metre stick (mass M, length L= 1m) is suspended
from one end by a frictionless pivot at P. A ball of
mass m, velocity v0, strikes the other end of the
(stationary) stick at right angles, and stops (final
velocity of the ball is zero).
Physics 1D03 - Lecture 26
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Quick Quiz 47
A stick (uniform thin rod) is lying on the ice. A
hockey puck hits the stick, at right angles, and
the stick starts to slide. Point P is on the end
farthest from where the puck hits.
P
Which of the following describe the motion of the stick
after the collision? (Answer True, False, or Maybe for
each one.)
P
Immediately after the collision, the end P will
start to move:
A)
B)
C)
D)
A) ICM ω = mv0L/2
B) IP ω = mv0L
C) MvCM = mv0
D) ½ mv02 = ½ IP ω 2
in a direction parallel to v0
in a direction opposite to v0
at an angle (not 0o or 180o) to v0
It depends where the puck hits
CM
v0
v0
Physics 1D03 - Lecture 26
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Physics 1D03 - Lecture 26
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2
Example 25
Where (on the bat) should a baseball player hit the ball so that
the bat doesn’t hurt his hands?
An equivalent, simpler problem: Suppose the bat
is a stick, and we hold it at the end (point P).
Where should we hit the stick with a ball so that
it will (momentarily) rotate about point P after
the collision, without any external force applied
at P ?
P
r
CM
The ball applies a brief impulse F∆t = −∆pball
to the stick.
F∆t
PhysicsReal
1D03 - Lecture
26
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bat suspended
Real bat suspended
Summary
In general, for a rigid body,
L = r × ( mv CM ) + I CM ω
In collisions, angular momentum will be conserved it
there is no external torque.
Physics 1D03 - Lecture 26
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