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 maximum kinetic energy is given by

$K=\frac{1}{2}I{\omega }^2$

The moment of inertia of a uniform rod of mass $m$ and length $L$ about an axis passing through its end is

$I=\frac{m{L}^2}{3}$

Substituting $I$ into the kinetic energy expression, we get

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

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

$U=mgh$

Equating the maximum kinetic energy and the maximum potential energy:

$mgh=\frac{m{L}^2{\omega }^2}{6}$

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

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