Question

Two small balls, each of mass m are connected by a light rigid rod of length L. The system is suspended from its centre by a thin wire of torsional constant k. The rod is rotated about the wire through an angle ${\theta }_0$ and released. Find the force exerted by the rod on one of the balls as the system passes through the mean position.

Answer 1

The M.I. of the two ball system,

$I=2m{\left(\frac{L}{2}\right)}^2=\frac{m{L}^2}{2}$

At any position $\theta$ during the oscillation

Torque = $k\theta$

So, work done during the displacement 0 to ${\theta }_0.$

$\omega ={\theta }_{{\int }_0^0}k\theta d\theta =\frac{k{\theta }_0^2}{2}$

By work energy method,

$\frac{1}{2}I{w}^2-0$ = work done = $\frac{k{\theta }_0^2}{2}$

$\therefore {\omega }^2=\frac{k{\theta }^2}{I}=\frac{k{\theta }_0^2}{m{L}^2}$

Now, from the free body diagram of the rod,

$T_2=\sqrt{(m{\omega }^{2}L{)}^2+(mg{)}^2}$

$=\sqrt{{(m\frac{k{\theta }_0^2}{m{L}^2}\times L)}^2}+m^2{g}^2$

$=\sqrt{\frac{k^2{\theta }_0^4}{L^2}+m^2{g}^2}$