A circular loop of radius R is made of a perfectly elastic wire and is rotating with a constant angular velocity $\mathrm{\omega }$ lying on a smooth horizontal table. The rotation axis is vertical passing through the centre. A small radial push given to the loop at a point P on the table causes a transverse pulse to propagate on it. Find the smallest time in which the pulse will be back to its originating point P on the table.

First let us calculate the tension in the wire loop. Consider a small element of angular width d$\mathrm{\theta }$ on the loop.
$\begin{array}{l}2\mathrm{Tsin}\frac{\mathrm{d\theta }}{2}=\left(\mathrm{\mu Rd\theta }\right){\mathrm{\omega }}^2\mathrm{R}\\ \therefore 2\mathrm{T}\frac{\mathrm{d\theta }}{2}={\mathrm{\mu Rd\theta \omega }}^2\mathrm{R}\Rightarrow \mathrm{T}={\mathrm{\mu R}}^2{\mathrm{\omega }}^2\end{array}$
Speed of transverse wave relative to the string is $\mathrm{V}=\sqrt{\frac{\mathrm{T}}{\mathrm{\mu }}}=\mathrm{\omega R}$
Relative to the ground, the pulse will travel with speed 2V (or angular speed 2$\mathrm{\omega }$)
$\therefore$Time required to reach back at P is $\mathrm{t}=\frac{\mathrm{\pi }}{\mathrm{\omega }}$
[Note: The pulse that starts travelling in direction opposite to the direction of rotation of the ring will remain static at P]