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 tension in the rod as the system passes through the mean position.

Answer

A, B, D

Explanation

The M.I. of the two ball system
`I=2m(L/2)^2=(mL^2)/2`

At any position `theta` during the oscillation
`torque=ktheta`
So, work done during the displacement `0 to theta_0`.
`omega=int_0^(theta_0)k theta dtheta =(ktheta_0^2)/2`
By work energy method
`1/2Iomega^2-0=work done =(ktheta_0^2)/2`
`:.=(ktheta^2)/I=(ktheta_0^2)/(mL^2)`

Now from the free body diagram of the rod
`T_2=sqrt((momega^2L)^2+(mg)^2)`
`=sqrt(m(ktheta_0^2)/(mL^2)xxL)^2)+m^2g^2`
`=(sqrt(k^2theta_0^2)/L^2+m^2g^2)`