Question

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

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

Answer 1

Answer: (C)

Given, $I=2 \,kg - m ^{2}, \omega_{0}=\frac{60}{60} \times 2 \pi\, rad / s$
$\omega=0, t=60\, s$
The torque required to stop the wheel's rotation is
$\tau=I \alpha=I\left(\frac{\omega_{0}-\omega}{t}\right) $
$\therefore \tau=\frac{2 \times 2 \pi \times 60}{60 \times 60}$
$=\frac{\pi}{15} N - m$