Question

A ring of radius $r=25cm$ made of lead wire is rotated about a stationary vertical axis passing through its centre and perpendicular to the plane of the ring. The angular speed in $rps$ at which the ring ruptures is (20+x). The value of x is:

Answer 1

Let us consider an element of the ring (figure shown below). From Newton's law $F_n=m{w}_n$ for this element, we get,
$Td\theta =\left(\frac{m}{2\pi }d\theta \right){\omega }^{2}r$
So, $T=\frac{m}{2\pi }{\omega }^{2}r$
Condition for the problem is:
$\frac{T}{\pi r^2}\le {\sigma }_m$ or, $\frac{m{\omega }^{2}r}{2{\pi }^2{r}^2}\le {\sigma }_m$
or, ${\omega }_{max}^2=\frac{2{\pi }^2{\sigma }_{m}r}{\pi r^2\left(2\pi r\rho \right)}=\frac{{\sigma }_m}{\rho r^2}$
Using the table of appendices $n=23rps$