Question

A uniform cylindrical rod of mass $m$ and length $L$ is rotating with an angular velocity $\omega$. The axis of rotation is perpendicular to its axis of symmetry and passes through one of its edge faces. If the room temperature increases by $t$ and the coefficient of linear expansion is $\alpha$, the change in its angular velocity is :

• A. $2\omega \alpha t$
• B. $\omega \alpha t$
• C. $\frac{3}{2}\omega \alpha t$
• D. $\frac{\omega \alpha t}{2}$

Answer 1

The correct option is A $2\omega \alpha t$

If the temp is increased by t, $l^{\prime }=l(1+\alpha t)$

Initial MI $I=\frac{m{l}^2}{3}$

Final MI $I^{\prime }=\frac{m{l}^{\prime 2}}{3}$

From the conservation of angular momentum,

$I\omega =I^{\prime }{\omega }^{\prime }$

$\Rightarrow \frac{m{l}^2}{3}\omega =\frac{m{l}^{\prime 2}}{3}{\omega }^{\prime }=\frac{m{l}^2(1+\alpha t{)}^2}{3}{\omega }^{\prime }$

$\Rightarrow \frac{{\omega }^{\prime }}{\omega }=(1+\alpha t{)}^{-2}$

$\Rightarrow {\omega }^{\prime }=\omega (1-2\alpha t)$

Neglecting 2nd order term as it is negligible

$\Delta \omega =-2\omega \alpha t$

Angular velocity decreases since I increases.