Question

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

• A. $\sqrt{\omega }$
• B. $\omega$
• C. ${\omega }^2$
• D. $\frac{1}{\omega }$

Answer 1

The correct option is B

$\omega$

At $t^{\circ }C$, the length of the rod becomes $L^{\prime }=L(1+\alpha t)$ where $\alpha$ is the coefficeing of linear expansion.
From the law of conservation of angular momentum we have, $I\omega =I^{\prime }{\omega }^{\prime }\Rightarrow \frac{1}{3}M{L}^2\omega =\frac{1}{3}M{L}^{\prime 2}{\omega }^{\prime }\Rightarrow \frac{{\omega }^{\prime }}{\omega }={\left(\frac{L}{L^{\prime }}\right)}^2=\frac{1}{(1+\alpha t{)}^2}$
Now, for a given value of t, $(1+at{)}^{-2}$ is a constant, say k.
$\therefore \frac{{\omega }^{\prime }}{\omega }=k\Rightarrow \frac{{\omega }^{\prime }-\omega }{\omega }=k-1\Rightarrow {\omega }^{\prime }-\omega =(k-1)\omega \Rightarrow ({\omega }^{\prime }-\omega )\propto \omega$
Hence, the correct choice is (b).