Question

A circular loop of string rotates about its axis on a frictionless horizontal plane at 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

m = mass per unit length of the string.

R = Radius of the loop

w = angular velocity

V = Linear velocity of the string

Consider one half of the string as shown in figure.

The half loop experiences centrifugal force at every point, away from centre, which is balanced by tension 2T.

Consider an element of angular part $d\theta$ at angle $\theta$

Consider another element symmetric to this.

Centrifugal force experienced by the element $=\left(mRd\theta \right)w^{2}R$

(.. Length of element $=Rd\theta$, mass $=mRd\theta$ )Resolving in to rectangular components, net force on the two symmetric elements,

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

[Horizontal components cancels each other]

So, Total $F={\int }_0^{\frac{\pi }{2}}2m{R}^2{\omega }^{2}sin\theta d\theta$

$=2m{R}^2{\omega }^2[-cos\theta ]$

$=2m{R}^2{\omega }^2$

Again, $2T=2m{R}^2{\omega }^2$

$\Rightarrow T=m{R}^2{\omega }^2$

Velocity of transverse vibration,

$V=\sqrt{\left(\frac{T}{m}\right)}$

$=\sqrt{\left(m{R}^2{\omega }^2\right)}=\omega R=V$

So, the speed of the disturbance will be V.