Question

A uniform cylinder of steel of mass $M$, radius $R$ is placed on frictionless bearings and set to rotate about its vertical axis with angular velocity $\omega_{0}$. After the cylinder has reached the specified state of rotation. It is heated without any mechanical contact from temperature $T_{0}$ to $T_{0}+\Delta T$. If $\frac{\Delta I}{I}$ is the fractional change in moment of inertia of the cylinder and $\frac{\Delta \omega}{\omega_{0}}$ be the fractional change in the angular velocity of the cylinder and $\alpha$ be the coefficient of linear expansion, then
(1) $\frac{\Delta}{I}=\frac{2 \Delta R}{R}$
(2) $\frac{\Delta I}{I}=-\frac{\Delta \omega}{\omega_{0}}$
(3) $\frac{\Delta \omega}{\omega_{0}}=-2 \alpha \Delta T$
(4) $\frac{\Delta I}{I}=-\frac{2 \Delta R}{R}$

• A. 1, 2 and 3 are correct
• B. 1 and 2 are correct
• C. 2 and 4 are correct
• D. 1 and 3 are correct

Answer 1

Answer: (A)

Since no mechanical contact is there, so angular momentum is conserved.
$I \omega=$ constant
$\Delta(I \omega)=0$
$I \Delta \omega+\omega \Delta I=0$
$\frac{\Delta \omega}{\omega}=-\frac{\Delta I}{I}$
$\frac{\Delta \omega}{\omega}=-\frac{\Delta I}{I}=-2 \,\alpha \,\Delta\, T $
$\left\{\because \frac{\Delta I}{I}=\frac{2 \Delta R}{R}=2 \,\alpha \,\Delta \,T\right\}$