Question

A rigid structure consists of a thin ring of radius $R=\frac{75}{290}$ m and a thin radial rod of equal mass and length $2R$. The structure is pivoted around a horizontal axis xx' in the plane of the ring, passing through its centre. The structure is released from rest and it rotates around the axis xx' from the initial upright orientation. Find its angular speed in rad/s about the axis when it is inverted. Assume no frictional losses.

Answer 1

When the structure becomes inverted, there is no decrease in the potential energy of the ring.
$\therefore$Decrease in PE of the rod $=$ Gain in rotational kinetic energy of
the structure
$\Rightarrow M.g.4R=\frac{1}{2}{I}_{sy}{\omega }^2$
(As COM of the rod comes down by distance 4R)
Now $I_{sy}=\frac{M{R}^2}{2}+[\frac{M.4R^2}{12}+M.4R^2]$
$=\frac{M{R}^2}{2}+\frac{13M{R}^2}{3}=\frac{29M{R}^2}{6}$
$\omega =\sqrt{\frac{8M.gR}{\left(\frac{29}{6}M{R}^2\right)}}=\sqrt{\frac{48g}{29R}}=8$ rad/s