Posted by Tarren on .
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?)
I forgot to add one more part.
d) Determine the ratio of her final kinetic energy to her initial kinetic energy.
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.