A figure skater stands on one spot on the ice(assumed frictionless) and spins around with her arms extended. When she pulls in her arms, her moment of inertia decreases to 3/4 of the initial value. Assume that her spin angular momentum is conserved. What is the ratio of her final rotational kinetic energy to her initial rotational kinetic energy?
3 answers
4/3
I omega, angular momentum, does not change so:
I1 omega1 = I2 omega2
I1 omega1 = (3/4) I1 omega2
so
omega2 = omega 1 * 4/3
then
original ke = (1/2) I1 omega1&2
final Ke = (1/2) I2 omega2^2
so
final Ke=(1/2)(3/4)I1 *(4/3)^2 omega1^2
= 4/3 of original
(she had to do work to pull her arms in)
I1 omega1 = I2 omega2
I1 omega1 = (3/4) I1 omega2
so
omega2 = omega 1 * 4/3
then
original ke = (1/2) I1 omega1&2
final Ke = (1/2) I2 omega2^2
so
final Ke=(1/2)(3/4)I1 *(4/3)^2 omega1^2
= 4/3 of original
(she had to do work to pull her arms in)
Thank you Damon
2 answers