A uniform metal rod of length L and mass M is | vedclass

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(B) The angular momentum $J$ of the rod remains constant because no external torque acts on the system.$J = I \omega = 	ext{constant}$.Initially, the

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$\omega$

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(B) The angular momentum $J$ of the rod remains constant because no external torque acts on the system.$J = I \omega = 	ext{constant}$.Initially, the moment of inertia is $I_1 = \frac{1}{3} M L^2$ and angular velocity is $\omega_1 = \omega$.When the temperature increases by $t^{\circ} C$, the length of the rod becomes $L' = L(1 + \alpha t)$.The new moment of inertia is $I_2 = \frac{1}{3} M (L')^2 = \frac{1}{3} M L^2 (1 + \alpha t)^2 = I_1 (1 + \alpha t)^2$.From conservation of angular momentum, $I_1 \omega_1 = I_2 \omega_2$.$\omega_2 = \omega_1 \left( \frac{I_1}{I_2} ight) = \omega \left( \frac{I_1}{I_1(1 + \alpha t)^2} ight) = \omega (1 + \alpha t)^{-2}$.Using binomial approximation for small $\alpha t$, $\omega_2 \approx \omega (1 - 2 \alpha t)$.The change in angular velocity is $\Delta \omega = \omega_2 - \omega_1 = \omega (1 - 2 \alpha t) - \omega = -2 \omega \alpha t$.Thus, the magnitude of the change in angular velocity $|\Delta \omega|$ is proportional to $\omega$.

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