Physics - Rotation Question with Solution | TestHub

By the principle of conservation of mechanical energy, the maximum rotational kinetic energy of the rod is equal to the maximum potential energy gained by its centre of mass.

The moment of inertia of a thin uniform rod about an axis passing through its end is $I=\frac{M{L}^2}{3}$.

Maximum rotational kinetic energy is given by:

$K=\frac{1}{2}I{\omega }^2=\frac{1}{2}\left(\frac{M{L}^2}{3}\right){\omega }^2=\frac{M{L}^2{\omega }^2}{6}$

Let $h$ be the maximum height to which the centre of mass rises. The maximum potential energy is:

$U=Mgh$

Equating kinetic and potential energies:

$Mgh=\frac{M{L}^2{\omega }^2}{6}$

$h=\frac{L^2{\omega }^2}{6g}$

Answer: $\frac{{\mathrm{L}}^2{\omega }^2}{6\mathrm{g}}$