Question

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?

Short answer

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

Step-by-step solution

Given

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

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

To understand the concept

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:

i. The law conservation of angular momentum is

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

ii. The rotational K.E is,

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

Calculate the ratio of the new rotational kinetic energy to the initial rotational kinetic energy

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.