Two masses m and m 2 are connected at the two ends of a massless rigid rod of length l . The rod is…

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:

1. $k{\theta }_0^2$
2. $\frac{3k{\theta }_0^{ 2}}{l}$
3. $\frac{2k{\theta }_0^{ 2}}{l}$
4. $\frac{k{\theta }_0^{ 2}}{l}$

Solution

When rod is rotated by angle $\alpha ,$ the twist in the wire is $\varphi .$

angular frequency of wire, $\mathrm{\Omega }=\sqrt{\frac{k}{I}}$

angular frequency of rod, $\omega ={\theta }_0.\mathrm{\Omega }$

Max. Tension in the rod $=m{\omega }^2.\frac{l}{3}$

$=m {\theta }_0^{ 2}.\frac{k}{I}.\frac{l}{3} ,$ here $I=m\frac{l^2}{3}$

$={\theta }_0^{ 2}\frac{k}{l}$

Asked in: JEE Main 2019 (09 Jan Shift 1)

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