Physics - Rotation Question with Solution | TestHub

Let the mass of the uniform rod be $m$.

The moment of inertia of the rod about the fixed horizontal axis passing through its end B is $I=\frac{m{l}^2}{3}$.

Let the horizontal plane passing through the pivot B be the reference level for potential energy.

In the initial unstable equilibrium position, the rod is vertically upright. The height of its center of mass from B is $h_i=\frac{l}{2}$.

When the rod has turned through an angle $\theta$, the height of its center of mass from B becomes $h_f=\frac{l}{2}\cos \theta$.

According to the principle of conservation of mechanical energy, the loss in gravitational potential energy is equal to the gain in rotational kinetic energy.

Loss in potential energy = $mg(h_i-h_f)=mg\left(\frac{l}{2}-\frac{l}{2}\cos \theta \right)=\frac{mgl}{2}(1-\cos \theta )$

Gain in rotational kinetic energy = $\frac{1}{2}I{\omega }^2=\frac{1}{2}\left(\frac{m{l}^2}{3}\right){\omega }^2=\frac{m{l}^2}{6}{\omega }^2$

Equating the energies:

$\frac{m{l}^2}{6}{\omega }^2=\frac{mgl}{2}(1-\cos \theta )$

${\omega }^2=\frac{3g}{l}(1-\cos \theta )$

Using the trigonometric identity $1-\cos \theta =2{\sin }^2\left(\frac{\theta }{2}\right)$:

${\omega }^2=\frac{3g}{l}\times 2{\sin }^2\left(\frac{\theta }{2}\right)$

${\omega }^2=\frac{6g}{l}{\sin }^2\left(\frac{\theta }{2}\right)$

$\omega =\sqrt{\frac{6g}{l}}\sin \left(\frac{\theta }{2}\right)$