# Q. 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 0. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:

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Solution:

## $\omega = \sqrt{\frac{k}{I}}$ $\omega = \sqrt{\frac{3k}{m\ell^{2}}}$ $\Omega =\omega\theta_{0} =$ average velocity $T=m\Omega^{2}r_{1}$ $T = m\Omega^{2} \frac{\ell}{3}$ $= m\omega^{2} \theta^{2}_{0} \frac{\ell}{3}$ $= \frac{k\theta^{2}_{0}}{\ell}$ $I = \mu\ell^{2} = \frac{\frac{m^{2}}{2}}{\frac{3m}{2}} \ell^{2}$ $= \frac{m\ell^{2}}{3}$ $\frac{r_{1}}{r_{2}} = \frac{1}{2} \Rightarrow r_{1} = \frac{\ell}{3}$

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