Question

A wheel having moment of inertia $2kg{m}^2$ about its vertical axis, rotates at the rate of $60rpm$ about this axis. The torque which can stop the wheels rotation in one minute would be

• A. $\frac{2\pi }{15}Nm$
• B. $\frac{\pi }{12}Nm$
• C. $\frac{\pi }{15}Nm$
• D. $\frac{\pi }{18}Nm$

Answer 1

The correct option is C $\frac{\pi }{15}Nm$

Let angular acceleration is $\alpha$.

Given ${\omega }_o=60rpm=\frac{2\pi 60}{60}=2\pi$ $rad/s$, time $t=1minute=60s$, MOI $I=2kg{m}^2$

Using $\omega ={\omega }_o+\alpha t$ $\Rightarrow 0=2\pi -\alpha 60$ $\Rightarrow \alpha =\frac{\pi }{30}$

$\therefore$ Torque $=I\alpha =2\times \frac{\pi }{30}=\frac{\pi }{15}Nm$