Question

A uniform metal rod of length 'L' and mass 'M' is rotating about an axis passing through one of the ends perpendiculars to the rod with angular speed '$\omega$'. If the temperature increases by $t^{0}C$ then the change in its angular velocity is proportional to which of the following? (Coefficient of linear expansion of rod $=\alpha$)

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

Answer 1

The correct option is B $\omega$

The moment of inertia of a rod about an axis through one of its ends and perpendicular to the axis is $I=\frac{1}{3}M{L}^2$

As the temperature increases, the length of the rod expands.

Thus new length of the rod $L_f=L(1+\alpha t)$

New moment of inertia of the rod $I_f=\frac{M{L}_f^3}{3}=\frac{1}{3}M{L}^2(1+\alpha t{)}^2$

Let the angular momentum decreases by $\Delta \omega$.

By the principle of conservation of angular momentum, we get $L_f=L$

$\Rightarrow I\omega =I_f(\omega -\Delta \omega )$

$\Rightarrow \Delta \omega =\frac{I_f-I}{I_f}\omega$

Thus, $\Delta \omega \propto \omega$