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

Answer:

I=2m(L2)2=mL22

Explanation:

Answer 2

Explanation:

• Let us consider the mass of the balls as m and a light rod of length L connect both the balls together.
• We can write the moment of inertia of the two ball system as $I=2 m\left(\frac{L}{2}\right)^{2}=\left(\frac{m L^{2}}{2}\right)$.
• Also, the torque produced at any random position θ can be written as  $\tau=k \theta$.  Therefore, the work done by the displacement of the system from 0 to  $\theta_0$  position is $W=\frac{k \theta_{0}^{2}}{2}$.
• Hence, by the application of work energy theorem rule, we can write, $\frac{1}{2} I \omega^{2}=\frac{k \theta_{0}^{2}}{2}.$ Thus, the force is  $T_{2}=\sqrt{\left[\left(m \omega^{2} L\right)^{2}+(m g)^{2}\right]}=\sqrt{\left[\frac{k^{2} \theta_{0}^{4}}{L^{2}}+m^{2} g^{2}\right]}.$