Question

A wheel has moment of inertia 5 x ${10}^{-3}kg{m}^2$ and is making 20 rev/sec. The torque needed to stop it in 10 sec is ................ x ${10}^{-2}N-m$ :-

• A. $2\pi$
• B. $2.5\pi$
• C. $4\pi$
• D. $4.5\pi$

Answer 1

The correct option is A $2\pi$

Initial angular velocity is ${\omega }_0=20\times 2\pi rad/s=40\pi$ $rad/s$

The angular $retardation$ needed to stop it in $t=10sec$ will be given by $\alpha =\frac{{\omega }_0}{t}=40\pi /10=4\pi$

So the toque needed will be

$\tau =I\times \alpha =5\times {10}^{-3}\times 4\pi =2\pi \times {10}^{-2}N-m$,

$I$ is the moment of inertia.

Option a is correct.