Question

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?

Answer 1

I₁ = initial moment of inertia before pulling in the arms

I₂ = final moment of inertia after pulling in the arms = I₁ /2

w₁ = initial angular velocity before pulling in the arms

w₂ = final angular velocity after pulling in the arms

using conservation of angular momentum

I₁ w₁ = I₂ w₂

I₁ w₁ = (I₁/2 ) w₂

w₂ = 2 w₁

KE₁ = initial rotational kinetic energy before pulling in the arms = (0.5) I₁ w²₁

KE₂ = final rotational kinetic energy after pulling in the arms = (0.5) I₂ w²₂

Ratio of final rotational kinetic energy to initial rotational kinetic energy is given as

KE₂ /KE₁ = (0.5) I₂ w²₂/((0.5) I₁ w²₁ )

KE₂ /KE₁ = ((I₁/2 ) (2 w₁)²)/(I₁ w²₁)

KE₂ /KE₁ = 2

KE₂ = 2 KE₁

hence the kinetic energy becomes twice