Question

A circular loop of string rotates about its axis on a frictionless horizontal plane at a uniform rate so that the tangential speed of any particle of the string is $V$. If a small transverse disturbance is produced at a point of the loop, with what speed (relative to the string) will this disturbance travel on the string?

Answer 1

Step 1 : Given data

$R=$ Radius of the loop

$m=$mass per unit length of the string $=mRd\theta$

$\omega$ $=$ angular velocity

$V=$ linear velocity of the string

Step 2 : To find the velocity

The force for a small portion in the ring $dF=\left(mRd\theta \right){\omega }^{2}R$

Centrifugal force $=\left(mRd\theta \right){\omega }^{2}R$

Element length $=Rd\theta$

$dF=2m{R}^{2}d\theta {\omega }^2\sin \theta$

$$\begin{gathered} F={\int }_0^{\frac{\pi }{2}}2m{R}^2{\omega }^2\sin \theta d\theta \\ =2m{R}^2{\omega }^2{\left[-\cos \theta \right]}^{\frac{\pi }{2}} \\ =2m{R}^2{\omega }^2\left[1-0\right] \\ 2T=2m{R}^2{\omega }^2 \\ T=m{R}^2{\omega }^2 \end{gathered}$$

Transverse vibration's velocity is given by ,

$$\begin{gathered} V=\sqrt{\frac{T}{m}} \\ V=\sqrt{\frac{m{R}^2{\omega }^2}{m}} \\ V=R\omega \end{gathered}$$

Then the speed of the disturbance is $R\omega$

Hence the answer is $R\omega$