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 θ₀ and released. Find the force exerted by the rod on one of the balls as the system passes through the mean position.

Figure

Answer 1

It is given that the mass of both the balls is m and they are connected to each other with the help of a light rod of length L.

Moment of inertia of the two-ball system $\left(I\right)$ is given by,
$I=2m{\left(\frac{L}{2}\right)}^2=\frac{m{L}^2}{2}$


Torque $\left(\tau \right)$, produced at any given position θ is given as:
$\tau$ = kθ
$\Rightarrow$ Work done during the displacement of system from 0 to θ₀ will be,
$W={\int }_0^{{\mathrm{\theta }}_0}k\mathrm{\theta }\mathrm{d\theta }=\frac{k{\mathrm{\theta }}_0^2}{2}$
On applying work-energy theorem, we get:
$$\begin{gathered} \frac{1}{2}I{\omega }^2-0=\mathrm{work}\mathrm{done}=\frac{k{\theta }_0^2}{2} \\ \therefore {\omega }^2=\frac{k{\theta }_{\mathit{0}}^{\mathit{2}}}{I}=\frac{k{\theta }_0^2}{m{L}^2} \end{gathered}$$



From the free body diagram of the rod, we can write:
$$\begin{gathered} \mathrm{Force},T_2=\sqrt{{\left(m{\omega }^{2}L\right)}^2+{\left(mg\right)}^2} \\ =\sqrt{{\left(m\frac{k{\theta }_0^2}{m{\mathrm{L}}^2}\times \mathrm{L}\right)}^2+m^2{g}^2} \\ =\sqrt{\frac{k^2{\theta }_0^4}{{\mathrm{L}}^2}+m^2{g}^2} \end{gathered}$$