Question

A circular loop of string rotates about its axis on a frictionless horizontal place at a uniform rate so that the tangential speed of any particle of the string is . 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

Let,
V = Linear velocity of the string
m = Mass per unit length of the the string.
R = Radius of the loop
ω = Angular velocity



Consider one half of the string, as shown in the figure.
The half loop experiences centrifugal force at every point (away from the centre) balanced by tension 2T.
Consider an element of angular part dθ at angle θ.
So,
Length of the element$=Rd\mathrm{\theta },\mathrm{mass}=mRd\theta$
Centrifugal force experienced by the element$=\left(mRd\theta \right){\omega }^{2}R$
Resolving the centrifugal force into rectangular components,
Since the horizontal components cancel each other, the net force on the two symmetric elements is given as
$dF=2m{R}^{2}d\theta {\omega }^2\mathrm{sin\theta }$
$$\begin{gathered} \mathrm{Total}\mathrm{force},F={\int }_0^{\pi /2}2m{R}^2{\omega }^2\sin \theta d\theta \\ =2m{\mathrm{R}}^2{\omega }^2\left[-\cos \mathrm{\theta }\right] \\ =2m{\mathrm{R}}^2{\omega }^2 \\ \mathrm{And}, \\ 2T=2m{R}^2{\omega }^2 \\ \Rightarrow T=m{R}^2{\omega }^2 \end{gathered}$$
Velocity of the transverse vibration is given as
$$\begin{gathered} V\text{'}=\sqrt{\left(\frac{T}{m}\right)} \\ V\text{'}=\sqrt{\left(\frac{m{R}^2{\omega }^2}{m}\right)}=\omega R \end{gathered}$$
Linear velocity of the string, V = $\omega R$
∴ Speed of the disturbance, V' = V