Question

A figure skater rotating on one spot with both arms and one leg extended has a moment of inertia . She then pulls in her arms and the extended leg, reducing her moment of inertia to 0.75 . What is the ratio of her final to initial kinetic energy?

Answer 1

Given that the initial moment of inertia is

`I_i`

The final moment of inertia is

`I_f=0.75I_i`

Let the initial angular velocity be

`\omega_i`

Let the final angular velocity be

`\omega_f`

The ratio of final and initial kinetic energy is

`\begin{gathered} (K_f)/(K_i)=((1)/(2)I_i(\omega_i)^2)/((1)/(2)I_f(\omega_f)^2) \\ =(I_i(\omega_i)^2)/(0.75I_i(\omega_f)^2) \\ =\frac{(\omega_i)^2_{}}{0.75(\omega_f)^2} \\ =1.3((\omega_i)^2)/((\omega_f)^2) \end{gathered}`