Question

A circular disc is rotating about its own natural axis. A constant opposing torque $2.75Nm$ is applied on the disc due to which it comes to rest in $28$ rotations. If moment of inertia of disc is $0.5kg{m}^2$, the initial angular velocity of disc is:

• A. $210rpm$
• B. $280rpm$
• C. $360rpm$
• D. $420rpm$

Answer 1

The correct option is D $420rpm$

we know that $\tau =I\alpha$
$\Rightarrow \alpha =\frac{\tau }{I}=\frac{2.75Nm}{0.5}=5.5rad{s}^{-2}$
given no of rotations $=28$
$\therefore Totalangle\theta =28\left(2\pi \right)=56\pi$
${\omega }^2=2\alpha \theta$
$\Rightarrow {\omega }^2=2\times 5.5\times 56\pi$
$\omega =43.99rad/s$
$=43.99\times \frac{60}{2\pi }rpm=420rpm$