A uniform metal rod of length L and mass M is rotating with angular speed $\omega$ about an axis passing through one of the ends and perpendicular to the rod. If the temperature increases by $t°C$, then the change in its angular speed is proportional to 

At $t°\mathrm{C}$, the length of the rod becomes $L^{\text{'}}=L(1+\alpha t)$, where $\alpha$ is the coefficient of linear expansion. From the law of conservation of angular momentum, we have
$I\omega =I^{\text{'}}{\omega }^{\text{'}}$

or $\frac{1}{3}M{L}^2\omega =\frac{1}{3},M{L}^{\text{'}2}{\omega }^{\text{'}}$

or $\frac{{\omega }^{\text{'}}}{\omega }={\left(\frac{L}{L^{\text{'}}}\right)}^2=\frac{1}{(1+\alpha t{)}^2}$

Now, for a given value of $t,(1+\alpha t{)}^{-2}$is constant, say k.

$$\begin{gathered} \therefore \frac{{\omega }^{\text{'}}}{\omega }=k \\ or \frac{{\omega }^{\text{'}}-\omega }{\omega }=k-1 \\ or {\omega }^{\text{'}}-\omega =(k-1)\omega \end{gathered}$$

i.e.$\left({\omega }^{\text{'}}-\omega \right)\propto \omega .$