Answer
The ratio of the new rotational kinetic energy to the initial rotational kinetic energy is $~~3$
Work Step by Step
Let the initial rotational inertia be $~~I_i$
Then the new rotational inertia is $~~I_f = \frac{I_i}{3}$
Let the initial angular speed be $\omega_i$
We can use conservation of angular momentum to find an expression for the new angular speed:
$L_f = L_i$
$I_f~\omega_f = I_i~\omega_i$
$\frac{I_i}{3}~\omega_f = I_i~\omega_i$
$\omega_f = 3~\omega_i$
We can write an expression for the initial rotational kinetic energy:
$K_i = \frac{1}{2}I_i\omega_i^2$
We can write an expression for the new rotational kinetic energy:
$K_f = \frac{1}{2}I_f\omega_f^2$
$K_f = \frac{1}{2}(\frac{I_i}{3})(3~\omega_i)^2$
$K_f = 3\times \frac{1}{2}I_i\omega_i^2$
$K_f = 3~K_i$
Therefore:
$\frac{K_f}{K_i} = 3$
The ratio of the new rotational kinetic energy to the initial rotational kinetic energy is $~~3$
