Posted by **Tarren** on Thursday, December 2, 2010 at 1:17pm.

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?)

- Physics -
**Tarren**, Thursday, December 2, 2010 at 1:19pm
I forgot to add one more part.

d) Determine the ratio of her final kinetic energy to her initial kinetic energy.

- Physics -
**drwls**, Thursday, December 2, 2010 at 2:26pm
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.

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