Question 128 of 280

Two masses $m$ and $\dfrac{m}{2}$ 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:

A$k{\theta }_{0}^{2}$B$\dfrac{3k{\theta }_{0}^{2}}{l}$C$\dfrac{2k{\theta }_{0}^{2}}{l}$D$\dfrac{k{\theta }_{0}^{2}}{l}$

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Answer: (D) $\dfrac{k{\theta }_{0}^{2}}{l}$

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

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

angular frequency of rod, $\omega ={\theta }_{0}.\Omega$

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

$=m{\theta }_{0}^{2}.\dfrac{k}{I}.\dfrac{l}{3},$ here $I=m\dfrac{{l}^{2}}{3}$

$={\theta }_{0}^{2}\dfrac{k}{l}$