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 $0$. If the rod is rota ted by ${\theta }_0$ and released, the tension in it when it passes through its mean position will be:

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

Answer 1

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

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

$\omega =\sqrt{\frac{3k}{m{\ell }^2}}$ (Ref. image 1)

$\Omega =\omega {\theta }_0=$ average velocity

$T=m{\Omega }^2{r}_1$

$T=m{\Omega }^2\frac{\ell }{3}$

$=m{\omega }^2{\theta }_0^2\frac{\ell }{3}$

$=m\frac{3k}{m{\ell }^2}{\theta }_0^2\frac{\ell }{3}$

$=\frac{k{\theta }_0^2}{\ell }$

$I=\mu {\ell }^2=\frac{\frac{m^2}{2}}{\frac{3m}{2}}{\ell }^2$

$=\frac{m{\ell }^2}{3}$ (Ref. image 2)

$\frac{r_1}{r_2}=\frac{1}{2}\Rightarrow r_1=\frac{\ell }{3}$