Question

A chain in the shape of circular loop of mass $m$ and radius $R$ is placed on a frictionless surface. Loop is given an angular velocity of $\omega$ as shown in the figure. Tension in the chain is

• A. $m{\omega }^{2}R$
• B. $\frac{m{\omega }^{2}R}{2\pi }$
• C. $\frac{m{\omega }^{2}R}{2}$
• D. $\frac{m{\omega }^{2}R}{\pi }$

Answer 1

The correct option is B $\frac{m{\omega }^{2}R}{2\pi }$


$dm=\frac{m}{2\pi R}dl=\frac{m}{2\pi R}\left(2R\theta \right)$
For equilibrium
$2T\sin \theta =2T\theta =dm{\omega }^{2}R\Rightarrow 2T\mathrm{\theta /}=\frac{m\left(2\mathrm{\theta /}\right)}{2\pi }{\omega }^{2}R\Rightarrow T=\frac{m{\omega }^{2}R}{2\pi }$