Question

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 rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:

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

Answer 1

The Correct Option is D

$\omega = \sqrt{\frac{k}{I}}$
$ \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} $