Question

Two masses $m$ and $m/2$ 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 as shown in the figure.
Because of torsional constant $K$, the restoring torque is $\tau =K\theta$ for angular displacement $\theta .$
If the rod is rotated by ${\theta }_{\circ }$ and released so that rod oscillates. The tension in the rod, when it passes through its mean position will be

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

Answer 1

The correct option is A $\frac{3K{\theta }_{\circ }^2}{l}$

Finding position of COM


$r_1=\frac{m}{m+\frac{m}{2}}l=\frac{2l}{3}$
$r_2=l-r_1=\frac{l}{3}$
When rod is tilted by ${\theta }_0$,


Centripetal force on $m$ is
$F_c=\frac{m{v}^2}{l/3}$
$\Rightarrow v=\left(\frac{l}{3}{\theta }_{\circ }\right)\omega$
$\omega$ is frequency of torsional pendulum.


$\Rightarrow \omega =\sqrt{\frac{K}{I}}$

Tension in the rod is
$T=\frac{m{v}^2}{l/3}=\frac{3m{v}^2}{l}=\frac{3m}{l}{\left(\frac{l}{3}{\theta }_{\circ }\omega \right)}^2$
$T=m{\theta }_0^2\times \frac{l}{3}\times \frac{K}{I}$
Moment of inertia about COM is
$I=\frac{m}{2}{\left(\frac{2l}{3}\right)}^2+m{\left(\frac{l}{3}\right)}^2=\frac{m}{2}\frac{4^2{l}^2}{9}+\frac{m{l}^2}{9}=\frac{m{l}^2}{3}$
Hence, $T=m{\theta }_0^2\times \frac{l}{3}\times \frac{K\left(3\right)}{m{l}^2}=\frac{K{\theta }_0^2}{l}$