Asked by Tarren
A skater spinning with angular speed of 1.5 rad/s draws in her outstretched arms thereby reducing her moment of inertia by a factor a 3.
a) Find her angular speed.
b)What is the ratio of her final angular momentum to her initial angular momentum?
c)Did her mechanical energy change? (Yes but why?)
a) Find her angular speed.
b)What is the ratio of her final angular momentum to her initial angular momentum?
c)Did her mechanical energy change? (Yes but why?)
Answers
Answered by
Tarren
I forgot to add one more part.
d) Determine the ratio of her final kinetic energy to her initial kinetic energy.
d) Determine the ratio of her final kinetic energy to her initial kinetic energy.
Answered by
drwls
I*w stays the same due to the requirement for angular momentum conservation.
If I is cut in half, w must double.
KE is (1/2)I*w^2 = (1/2)(I*w)^2/I
Since I drops by half while I*w is constant, the kinetic energy must also double.
The extra KE comes from work done by the skater pulling in her arms. There is a lot of centrifugal resistance.
If I is cut in half, w must double.
KE is (1/2)I*w^2 = (1/2)(I*w)^2/I
Since I drops by half while I*w is constant, the kinetic energy must also double.
The extra KE comes from work done by the skater pulling in her arms. There is a lot of centrifugal resistance.
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