Two small balls, each having a mass m , are joined by a light rigid...

The moment of inertia of the system about the suspension wire is given by the sum of the moments of inertia of the two balls.

$I = m \left(\dfrac{L}{2}\right)^2 + m \left(\dfrac{L}{2}\right)^2 = \dfrac{mL^2}{2}$

By the principle of conservation of mechanical energy, the maximum potential energy stored in the torsional wire is converted into kinetic energy as the system crosses the mean position.

$\dfrac{1}{2} k \theta_0^2 = \dfrac{1}{2} I \omega^2$

Substituting the value of $I$:

$k \theta_0^2 = \left(\dfrac{mL^2}{2}\right) \omega^2$

$\omega^2 = \dfrac{2k \theta_0^2}{mL^2}$

As the system crosses the mean position, the ball experiences two forces exerted by the rod: a vertical force to balance its weight and a horizontal force to provide the necessary centripetal acceleration.

The vertical component of the force is $F_y = mg$.

The horizontal component of the force is the centripetal force $F_x = m \omega^2 r$, where $r = \dfrac{L}{2}$.

$F_x = m \left(\dfrac{2k \theta_0^2}{mL^2}\right) \left(\dfrac{L}{2}\right) = \dfrac{k \theta_0^2}{L}$

The net force exerted by the rod on one of the balls is the resultant of these two mutually perpendicular components.

$F = \sqrt{F_x^2 + F_y^2}$

$F =$\left[ \left(\dfrac{k \theta_0^2}{L}\right)^2 + (mg)^2 \right]^{1/2}

$F = \left[ \dfrac{k^2 \theta_0^4}{L^2} + m^2 g^2 \right]^{1/2}$