The rotational inertia of a collapsing spinning star drops to $\frac{\mathbf{1}}{\mathbf{3}}$ its initial value. What is the ratio of the new rotational kinetic energy to the initial rotational kinetic energy?

The relation between new and initial moment of inertia of the spinning star is,

${\text{I}}_{\text{f}}=\frac{1}{3}{\text{I}}_{\text{i}}$

Using the law of conservation of angular momentum, find the relation between the initial and the new angular speed. Then by using this relation and formula for rotational K.E energy, find the ratio of the new rotational K.E to the initial rotational K.E of the spinning star.

Formula:

Initial angular momentum of the system = Final angular momentum of the system

$\mathbf{K}\mathbf{.}\mathbf{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{I\omega }}^{\mathbf{2}}$

According to the law conservation of angular momentum

Initial angular momentum of the system = Final angular momentum of the system

${\mathrm{I}}_{\mathrm{i}}{\mathrm{\omega }}_{\mathrm{i}}={\mathrm{I}}_{\mathrm{f}}{\mathrm{\omega }}_{\mathrm{f}}$

But,${\text{I}}_{\text{f}}=\frac{1}{3}{\text{I}}_{\text{i}}$

$\begin{array}{c}\Rightarrow {\text{I}}_{\text{i}}{\text{ω}}_{\text{i}}=\frac{1}{3}{\text{I}}_{\text{f}}{\text{ω}}_{\text{f}}\\ \Rightarrow \frac{\text{K}.{\text{E}}_{\text{f}}}{\text{K}.{\text{E}}_{\text{i}}}=\frac{\frac{1}{2}{\text{I}}_{\text{f}}{\text{ω}}_{\text{f}}^2}{\frac{1}{2}{\text{I}}_{\text{i}}{\text{ω}}_{\text{i}}^2}\\ \Rightarrow \frac{\text{K}.{\text{E}}_{\text{f}}}{\text{K}.{\text{E}}_{\text{i}}}=\frac{\frac{1}{2}\times \frac{1}{3}{\text{I}}_{\text{i}}{\text{ω}}_{\text{f}}^2}{\frac{1}{2}{\text{I}}_{\text{i}}{\left(\frac{1}{3}{\text{ω}}_{\text{f}}\right)}^2}\\ \Rightarrow \frac{\text{K}.{\text{E}}_{\text{f}}}{\text{K}.{\text{E}}_{\text{i}}}=3\end{array}$

Therefore, the ratio of new rotational $\mathrm{K}.\mathrm{E}.$ to the initial rotational $\mathrm{K}.\mathrm{E}.$ of the spinning star is 3.