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  4. A grindstone has a moment of inertia of 6 kg m2. 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

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Q. 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

System of Particles and Rotational Motion

Solution:

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, $\tau = I \alpha = 6 \times\frac{\pi}{2} = 3\pi N m$