Question

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

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

Answer 1

The correct option is A $2\pi$

Given,

${\omega }_i=20rev/s=20\times 2\pi =40\pi rad/s$

$I=5\times {10}^{-3}kg{m}^2$

${\omega }_f=0rad/s$

$t=10sec$

1st equation of motion in rotation,

${\omega }_f={\omega }_i+\alpha t$

$0=40\pi +10\alpha$

$\alpha =-\frac{40\pi }{10}=-4\pi rad/s^2$

Torque, $\tau =I\alpha$

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

The correct option is A.