Question

A loop of a string of mass per unit length $\mu$ and radius $R$ is rotated about an axis passing through its centre and perpendicular to its plane with an angular velocity $\omega$. A small disturbance is created in the loop having the same sense of rotation w.r.t the string. The linear speed of the disturbance for a stationary observer is

• A. $\omega R$
• B. $2\omega R$
• C. $3\omega R$
• D. Zero

Answer 1

The correct option is B $2\omega R$

For equilibrium of elementary mass $dm$, we use
$dm.{\omega }^{2}R=2T\sin \frac{d\theta }{2}$


As $d\theta$ is small, $\sin \frac{d\theta }{2}\approx \frac{d\theta }{2}$
$\Rightarrow \mu Rd\theta {\omega }^{2}R=2T\frac{d\theta }{2}$
[$\because \text{length of segment}=Rd\theta$ and mass of segment $=\mu \times Rd\theta$]
$\Rightarrow \mu {\omega }^2{R}^2=T$
Hence, velocity of wave w.r.t the string is
$v_w=\sqrt{\frac{T}{\mu }}=\sqrt{{\omega }^2{R}^2}=\omega R$

Also, speed of string is $\omega R$
Therefore, velocity of the disturbance w.r.t. ground $=\omega R+\omega R=2\omega R$.