Question

A ring and a disc of different masses are rotating with the same kinetic energy. If we apply a retarding torque $\tau$ on the ring, it stops after making $n$ revolutions. After how many revolutions will the disc stop if the retarding torque on it is also $\tau$?

• A. $\frac{n}{2}$
• B. $n$
• C. $2n$
• D. $4n$

Answer 1

The correct option is C $2n$

Ring $=r$

Disc $=d$

$\therefore \frac{1}{2}{I}_r{\omega }_r^2=KE$

And $\frac{1}{2}{I}_d{\omega }_d^2=KE$

$\gamma =I_r{\alpha }_r$ or, $\gamma =I_d{\alpha }_d$

${\omega }^2={\omega }_o^2+2{\alpha }_r{\theta }_r$

or, $0={\omega }_r^2-2\left(\frac{\tau }{I_r}\right)\times \left(2\pi n\right)$

$\therefore I_r{\omega }_r^2=4\pi n\tau$

Now, $0={\omega }_d^2-2\frac{\tau }{I_d}\times {\theta }_d$

$\therefore {\theta }_d=\frac{1}{2}{I}_d{\omega }_d^2\times \frac{1}{\tau }=4\pi n=2nrevolution$