Question

A ring of mass $m$ and radius $R$ rotates about an axis passing through its centre and perpendicular to its plane with angular velocity $\omega$. Find the velocity of transverse pulse in the ring.

• A. $\frac{R\omega }{2}$
• B. $2R\omega$
• C. $\frac{3}{2}R\omega$
• D. $R\omega$

Answer 1

The correct option is D $R\omega$

Let $T$ be the tension produce in the ring.
Take a elementary part of ring having mass $dm$, subtending an angle $d\theta$ at centre


Here, $\text{T sin}\left(\frac{d\theta }{2}\right)+\text{T sin}\left(\frac{d\theta }{2}\right)=2\text{T sin}\left(\frac{d\theta }{2}\right)$ will be responsible for coertripetal force
So $2\text{T sin}\left(\frac{d\theta }{2}\right)=dmR{\omega }^2$
$\Rightarrow 2T\frac{d\theta }{2}=\frac{m}{2\pi R}Rd\theta .R{\omega }^2$
$\Rightarrow T=\frac{mR{\omega }^2}{2\pi }$
and $\mu =\frac{m}{2\pi R}$
$\left[\begin{array}{c}\frac{d\theta }{2}\text{is very small}\text{and dm}=\frac{m}{2\pi R}Rd\theta \end{array}\right]$
So, velocity of transverse pulse
$v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{\left(\frac{mR{\omega }^2}{2\pi }\right)}{\left(\frac{m}{2\pi R}\right)}}=R\omega$