A thin uniform rod of mass ' m ' and length ' L ' is pivoted at one… | MHT CET 2025 (19 Apr, Shift 1)

Q19MHT CET 2025 · 19 APR · SHIFT 1Medium

Rotational Kinetic Energy, Work & Power

A thin uniform rod of mass ' $m$ ' and length ' $L$ ' is pivoted at one end so that it can rotate in a vertical plane. The free end is held vertically above pivot and then released. The angular acceleration of the rod when it makes an angle ' $\theta$ ' with the vertical is [consider negligible friction at the pivot] ( $g =$ acceleration due to gravity)

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Answer: (a)

Torque of gravity about the pivot: weight acts at $L/2$ , so $\tau = mg\frac{L}{2}\sin\theta$ .
$I = \frac{mL^2}{3}$ about the end.
$\alpha = \frac{\tau}{I} = \frac{mg(L/2)\sin\theta}{mL^2/3} = \frac{3g\sin\theta}{2L}$
So the answer is (a) $\frac{3g\sin\theta}{2L}$ .

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