A uniform rod of length L pivoted at one end | vedclass

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(B) Since the rod is rotating freely, the angular momentum $L$ is conserved.$L = I \omega = 	ext{constant}$.Initial moment of inertia $I = \frac{mL^2

Vedclass · Accepted Answer

$\frac{1}{2 \alpha}$

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(B) Since the rod is rotating freely, the angular momentum $L$ is conserved.$L = I \omega = 	ext{constant}$.Initial moment of inertia $I = \frac{mL^2}{3}$.After the temperature increase, the new length is $L' = L(1 + \alpha \Delta T)$.The new moment of inertia is $I' = \frac{m(L')^2}{3} = \frac{m L^2 (1 + \alpha \Delta T)^2}{3} = I(1 + \alpha \Delta T)^2$.Using conservation of angular momentum: $I \omega = I' \omega'$.Given $\omega' = \frac{\omega}{2}$, we have $I \omega = I(1 + \alpha \Delta T)^2 \frac{\omega}{2}$.$1 = \frac{(1 + \alpha \Delta T)^2}{2} \implies (1 + \alpha \Delta T)^2 = 2$.Taking the square root: $1 + \alpha \Delta T = \sqrt{2}$.Since $\alpha \ll 1$, we use the binomial approximation $(1 + x)^n \approx 1 + nx$.$1 + 2 \alpha \Delta T \approx 2$.$2 \alpha \Delta T = 1$.$\Delta T = \frac{1}{2 \alpha}$.

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