Question

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

• A. $3 \pi \,N \,m$
• B. $3 \,N\,m$
• C. $\frac{\pi}{3} \,N \,m$
• D. $4 \pi \,N \,m$

Answer 1

Answer: (A)

$3 \pi \,N \,m$

Answer 2

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