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 the 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. $\frac{k{\theta }_0^2}{l}$
• B. $\frac{k{\theta }_0^2}{2l}$
• C. $\frac{2k{\theta }_0^2}{l}$
• D. $\frac{3k{\theta }_0^2}{l}$

Answer 1

The correct option is A $\frac{k{\theta }_0^2}{l}$


The angular frequency of the system is given by,

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

$\text{Here},I=\frac{m}{2}{\left(\frac{2l}{3}\right)}^2+m{\left(\frac{l}{3}\right)}^2$

$\Rightarrow I=\frac{m{l}^2}{3}$

$\Rightarrow \omega =\sqrt{\frac{3k}{m{l}^2}}$

Let ${\omega }_m$ be the angular velocity when the system crosses the mean position.

The angular velocity is maximum at mean position and is given by,

${\omega }_m=\omega {\theta }_0$

Considering the mass $m$ and tension $T$ in the rod when it passes through the mean position,

$T=m{r}_1{\omega }_m^2$

$\Rightarrow T=m{\omega }^2{\theta }_0^2\frac{l}{3}$

$\Rightarrow T=m\frac{3k}{m{l}^2}{\theta }_0^2\frac{l}{3}$

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

Hence, option $\left(C\right)$ is correct.