- #1
klopez
- 22
- 0
An ice skater starts a spin with her arms stretched out to the sides. She balances on the tip of one skate to turn without friction. She then pulls her arms in so that her moment of inertia decreases by a factor of two. In the process of her doing so, what happens to her kinetic energy?
-It undergoes a change by an amount that obviously depends on how fast the skater pulls her arms in.
-It decreases by a factor of two.
-It is zero because her center of mass is stationary.
-It increases by a factor of four.
-It decreases by a factor of four.
-It remains constant.
-It increases by a factor of two.
I know that Ki = (1/2)Iω2 , so isn't her Kf = (1/2)(I/2)(2ω2) = (1/2)Iω2 = Ki ?
Doesn't the angular speed increase by a factor of two as well?
My guess would be "It remains constant" (And I can only choose one)
Can anyone give me advice on this problem? Thank you
Kevin
-It undergoes a change by an amount that obviously depends on how fast the skater pulls her arms in.
-It decreases by a factor of two.
-It is zero because her center of mass is stationary.
-It increases by a factor of four.
-It decreases by a factor of four.
-It remains constant.
-It increases by a factor of two.
I know that Ki = (1/2)Iω2 , so isn't her Kf = (1/2)(I/2)(2ω2) = (1/2)Iω2 = Ki ?
Doesn't the angular speed increase by a factor of two as well?
My guess would be "It remains constant" (And I can only choose one)
Can anyone give me advice on this problem? Thank you
Kevin