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 θ_{0} 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