Question

A ring of mass $7.2\text{kg}$ and of radius $25\text{cm}$ is experiencing a tension of $880\text{N}$ while making rotation about its own axis is such a way that each part of the ring has uniform speed. The frequency of rotation of the ring (in $\text{rev/min}$) is (assume surfaces of contact are frictionless)

• A. $10\text{rev/min}$
• B. $60\text{rev/min}$
• C. $520\text{rev/min}$
• D. $529\text{rev/min}$

Answer 1

The correct option is D $529\text{rev/min}$

Tension in a ring of mass $m$ rotating with angular velocity $\omega$ about its axis is given by
$T=\frac{mR{\omega }^2}{2\pi }$

Since, $\omega =2\pi f$
$\Rightarrow T=\frac{mR(2\pi f{)}^2}{2\pi }$
$\Rightarrow f=\sqrt{\frac{T}{2\pi mR}}$

Given: $T=880\text{N};m=7.2\text{kg};R=25\text{cm}=0.25\text{m}$

We get $f=\sqrt{\frac{880}{2\pi \times 7.2\times 0.25}}=\sqrt{\frac{880\times 7}{2\times 22\times 7.2\times 0.25}}$
$=8.82\text{rev/s}=529\text{rev/min}$