Transcript Angular Momentum - UW-Madison Department of Physics

Chapt. 10: Angular Momentum
Angular momentum conservation
And applications
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Angular Momentum Conservation
 EXT  r  FEXT
where

and
FEXT
dp

dt
In the absence of external torques
 EXT
dL

0
dt
Ii ωi = If ωf
Total angular momentum is conserved.
(demos)
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Spinning skater:
Because the torque exerted by the ice is
small, the angular momentum of the skater
is approximately constant.
When she reduces her moment of inertia
by drawing in her arms, her angular speed
increases.
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A student sitting on a stool that rests on
a turntable with frictionless bearings
is holding a rapidly spinning bicycle
Wheel (a). The rotation axis of the wheel
is initially horizontal, and the magnitude
of the spin-angular-momentum vector
of the spinning wheel is
What will happen if the student
suddenly tips the axle of the wheel (b)
so that after the rotation the spin axis
of the wheel is vertical and the wheel
is spinning counterclockwise
(when viewed from above)?
Answer: The turntable, stool, and student will be rotating clockwise with an
angular momentum about the vertical axis of the turntable of magnitude
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Example: Two Disks
• A disk of mass M and radius R rotates around the z axis with
angular velocity  i. 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 the relation ship between  i and  ?
f
•
analogous to an inelastic collision: p1 + p2(=0)  p3
z
z
f
i
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Example: Two Disks
• First realize that there are no external torques acting on
the two-disk system of combined mass M.
– Angular momentum will be conserved!
• Initially, the total angular momentum
is due only to the disk on the bottom:
z
z
2
f
1
i
1
Li  I1  1  MR 2 i
2
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L f  I1 1  I2 2  MR2 f
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Example: Two Disks
1
MR 2 i  MR 2 f
2
• Since Li = Lf
1
 f  i
2
z
z
Li
Lf
i
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An inelastic collision,
since E is not
conserved (friction)!
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f
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Example: bullet hitting a 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. Knowing that the initial speed of the bullet is v1, and the final speed is v2.
•
What is the angular speed ωf of the stick immediately after the collision?
(Ignore gravity)
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Example: bullet hitting a stick
Initial angular momentum:
D
Li  px  mv1
4
Final angular momentum:
D
D 1
L f  mv2  I f  mv2  MD 2 f
4
4 12
where
1
I
MD 2
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Conservation of angular momentum around pivot axis:
Li  L f
1
mv1  mv2  MD f
3
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m
f  3
v1  v 2 

MD
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Example: throw ball from stool
Li  0  L f
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L f  I f  Mvd
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Mvd
f 
I
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Angular momentum
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Angular momentum
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A 25-kg child in a playground runs with an initial
speed of 2.5 m/s along a path tangent to the rim
of a merry-go-round, whose radius is 2.0 m.
The merry-go-round, which is initially at rest,
has a moment of inertia of 500 kg · m2
The child then jumps on to it.
Find the final angular velocity of the child and
the merry-go-round together.
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Summary of Dynamics
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•
Dynamics (cause – effect)
– Force – Linear acceleration
• Change in linear momentum, Fnet = dp/dt
– Torque – angular acceleration
• Change in angular momentum, τ = dL/dt
•
Conservation of momentum
– When net external force = zero
•
Conservation of angular momentum
– When net external torque = zero
•
Conservation of Energy
– Mechanical energy = kinetic + potential
– Non-conservative forces e.g. friction
• Energy lost to the environment, e.g., heat
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Dynamical Reprise: Linear
• The change of motion of an object is described by Newton
as F=ma where F is the force, m is the mass, and a is the
acceleration.
• For a set of discrete point particles, all forces act on the
center of mass
• The center of mass is a calculable property of the object.
m r
c.m. 
m
r
Fnet  Ma
i i
i
i
i
If F=0, then there is no change of motion -if at rest, the object will remain at rest.
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Dynamical Reprise: Rotation
• About a fixed rotation axis, you can always write   I 
where  is the torque, I is the moment of inertia, and 
is the angular acceleration.
2
• For a set of discrete point particles, I   m i ri
• The parallel axis theorem lets you calculate the moment of
inertia about an axis parallel to an axis through the CM if

you know ICM :
D
M
CM
x
IPARALLEL = ICM + MD2
L
IPARALLEL ICM
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