Transcript Angular Momentum

PHYSICS 111
Rotational Momentum and Conservation of Energy
Equations:
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KE rotation = 1/ 2 Iω
I = mr 2
Ke rotating and translation = 1/2mv^2 ( 1+ I /mr 2 )
W (work) = t ( torque) (Delta ɵ)
W = τΔθ = ΔKE rotation
ᾠ = V final / r
Total E rotation and translation = .5 mv^2 + .5I ᾠ ^2 + mgh
L ( angular momentum) = Iω = p (translational momentum) * r(
perpendicular)
P = mv
If angular momentum is conserved = Ii * ωi = If *ωf
Concept Checker:
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A cylinder is rolling without slipping down an inclined
plane. The friction at the contact point P is
1. Static and points up the inclined plane.
 2. Static and points down the inclined plane.
 3. Kinetic and points up the inclined plane.
 4. Kinetic and points down the inclined plane.
 5. Zero because it is rolling without slipping.
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Answer:
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Answer 1. The friction at the contact point P is static
and points up the inclined plane. This friction
produces a torque about the center of mass that
points into the plane of the figure. This torque
produces an angular acceleration into the plane,
increasing the angular speed as the cylinder rolls
down.
Last Concept
A long narrow uniform stick lies motionless on ice (assume the
ice provides a frictionless surface). The center of mass of the
stick is the same as the geometric center (at the midpoint of
the stick). A puck (with putty on one side) slides without
spinning on the ice toward the stick, hits one end of the stick,
and attaches to it.
Which quantities are constant?
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1. Angular momentum of puck about center of mass of stick.
 2. Momentum of stick and ball.
 3. Angular momentum of stick and ball about any point.
 4. 1 and 3
 5. 2 and 3
 6. 1 and 2
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Solution:
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(2) and (3) are correct. There are no external forces
acting on this system so the momentum of the center of
mass is constant (1).
There are no external torques acting on the system so the
angular momentum of the system about any point is
constant (3) .
However there is a collision force acting on the puck, so the
torque about the center of the mass of the stick on the
puck is non-zero, hence the angular momentum of puck
about center of mass of stick is not constant.
The mechanical energy is not constant because the
collision between the puck and stick is inelastic.
Concept Checker
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A certain star, of mass m and radius r, is rotating with a
rotational velocity ω . After the star collapses, it has the same
mass but with a much smaller radius. Which statement below is
true?
A) The star's moment of inertia I has decreased, and its
angular momentum L has increased.
B) The star's moment of inertia I has decreased, and its angular
velocity ω has decreased.
C) The star's moment of inertia I remains constant, and its
angular momentum L has increased.
D) The star's angular momentum L remains constant, and its
rotational kinetic energy has decreased.
E) The star's angular momentum L remains constant, and its
rotational kinetic energy has increased.
Answer:
The correct answer is e.
 According to conservation of angular momentum,
the angular momentum L of the star remains
constant, so when its moment of inertia I decreases
(due to the decreased radius), its angular velocity
ω goes up proportionally
 L initial = L final
 Iᾠ final = Iᾠ initial
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Practice Problem:
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A skater spins with an angular speed of
12.0 rad/s with her arms
outstretched. She lowers her arms,
decreasing her moment of inertia
from to .
Calculate her initial and final rotational
kinetic energy.
How do you account for the change in
kinetic energy?
Solution:
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Her initial energy is:
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Her angular momentum is:
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L = Iᾠ = 41 (12) = 492 kg* m^2/s
You know that angular momentum is conserved. Meaning that
Lfinal and L inital = 492 kg* m^2/s
Then you need to solve for wfinal
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KE = .5 I ᾠ^2 = .5 (41)(12) = 2952 Joules
L = I ᾠ  492 = 36 (ᾠ f )
ᾠ = 13.67 rad/s
So her final rotational kinetic energy is:
Ke= .5I ᾠ^2 = .5 (36)(13.67) =
3364 Joules
Why? Because it took her energy to pull her arms in
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Practice Problem:
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A 65.0 kg woman stands at the rim of a horizontal
turntable that has a moment of inertia of 400 kg ∙ m²
and a radius of 3.00 m. The turntable is initially at rest
and is free to rotate about a frictionless, vertical axle
through its center. The woman then starts walking around
the rim clockwise (as viewed from above the system) at a
constant speed of 2.0 m/s relative to the Earth.
A) In what direction and with what angular speed does
the turntable rotate?
Solution:
M= 65 kg
 I = 400 kg*m^2
R=3m
 V = 2 m/s
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Recognize that this problem focuses on the law of conservation
of angular momentum. So…We know that initially
L=0 because we are told that the turntable is initially at rest. Letting
the subscript w stand for the woman and t for the table, we can
write: 0 = Iw * ωw + It *ωt
 0 =mr^2(v/r) + Itwt
 0 =65 *3^2( -2/3)+400*wt
 Wt = .98 rad/ s and since it is positive, means ccw
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