Question

A grindstone has a moment of inertia of $6kg{m}^2$. A constant torque is applied and the grindstone is found to have a speed of $150rpm,10$ seconds after starting from rest. The torque is

• A. $3\pi Nm$
• B. $3Nm$
• C. $\frac{\pi }{3}Nm$
• D. $4\pi Nm$

Answer 1

Answer: (A)

Here, $I=6kg{m}^2,t=10s,{\omega }_0=0$
$\upsilon =150rpm=\frac{150}{60}rps=\frac{5}{2}rps$
$\omega =2\pi \upsilon =2\pi \times \frac{5}{2}=5\pi rad{s}^{-1}$
$\alpha =\frac{\omega -{\omega }_0}{t}=\frac{5\pi -0}{10}=\frac{\pi }{2}rad{s}^{-2}$
$\therefore$ Torque, $\tau =I\alpha =6\times \frac{\pi }{2}=3\pi Nm$