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.

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,

\[W = \int^{\theta_0 }_0 \text { k} \theta \  d \theta = \frac{k \theta_0^2}{2}\]

On applying work-energy theorem, we get:

\[\frac{1}{2}I \omega^2  - 0 = \text { work  done } = \frac{k \theta_0^2}{2}\] 

\[ \therefore  \omega^2  = \frac{k \theta_0^2}{I} = \frac{k \theta_0^2}{m L^2}\]

From the free body diagram of the rod, we can write:

\[\text {Force},    T_2  = \sqrt{\left( m \omega^2 L \right)^2 + \left( mg \right)^2}\] 

\[     = \sqrt{\left( m\frac{k \theta_0^2}{m L^2} \times L \right)^2 + m^2 g^2}\] 

\[   = \sqrt{\frac{k^2 \theta_0^4}{L^2} + m^2 g^2}\]