Question

A wheel of moment of inertia $5\times {10}^{-3}$ $kg-m^2$ is making $20$ $rev/s$. For the torque required to step it in 10 s is

• A. $2\pi \times {10}^{-2}$ N.m
• B. $2\pi \times {10}^2$ N.m
• C. $4\pi \times {10}^{-2}$ N.m
• D. $4\pi \times {10}^2$ N.m

Answer 1

The correct option is A $2\pi \times {10}^{-2}$ N.m

$\begin{array}{c}\text{Given-}\ast \text{Moment of Inertia}I=5\times {10}^{-3}kg{m}^2\\ \text{* Angular velocity}\omega =20\text{rev}18=40\pi rad/s\\ \text{Initial angular momentum L}\\ L=I\omega =5\times {10}^{-3}\times 40\pi =0.2\pi \text{units}\\ \text{torque required to stop it in los (t)}\\ t=\frac{\Delta L}{\Delta t}=\frac{0.2\pi -0}{10}=0.02\pi Nm\\ \text{Hence, option (A) is correct.}\end{array}$