Two small bails, 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.
let θ be the angle made by rod connecting the masses during oscillations.
We know torque on the rod τ 
θ
Therefore τ = Kθ where k is torsional constant
Now work done w =
= K
/2
From work energy theorem , change in kinetic enrgy is equal to work done
K
/2 where I =
=
Tension in the string is due to centripetal force and also weight and is resultant of both .
T= 
=
substituting ω
T = 
1