Two masses m and m/2 are connected at the two ends of a massless… | JEE Main 2019 (9 Jan, Shift 1)

Q61JEE Main 2019 · 9 JAN · SHIFT 1Difficult

Rotational Kinematics & Dynamics

Two masses m and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length l. The rod is suspended by a thin wire of torsional constant k, at the centre of mass of the rod-mass system(see figure). Because of torsional constant k, the restoring torque is $\tau$ = k $\theta$ for angular displacement $\theta$ . If the rod is rotated by $\theta$ 0 and released, the tension in it when it passes through its mean position will be:

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Answer: (c)

CM is at $l/3$ from $m$ and $2l/3$ from $m/2$ ; $I = m\left(\tfrac{l}{3}\right)^2 + \tfrac{m}{2}\left(\tfrac{2l}{3}\right)^2 = \dfrac{ml^2}{3}$ .
Energy: $\tfrac{1}{2}k\theta_0^2 = \tfrac{1}{2}I\omega^2 \Rightarrow \omega^2 = \dfrac{3k\theta_0^2}{ml^2}$
Rod tension = centripetal force on $m$ : $T = m\omega^2\dfrac{l}{3} = \dfrac{k\theta_0^2}{l}$
So the answer is (c).

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Rotational Kinematics & Dynamics question from JEE Main 2019 (9 Jan, Shift 1)

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