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 $150rpm, 10s$. after starting from rest. Calculate the torque.
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Here, Moment of inertia of grindstone, $I = 6kgm^2$ Initial angular velocity $\omega_1=0$ Final angular velocity, $\omega_2=2\pi\text{n}=2\pi\times\frac{150}{60}$
$=5\pi\text{ rad/sec}^2$ Time for which torque acts, $\text{t}=10\text{\sec.}$
$\therefore$ Angular acceleration $(\alpha)=\frac{\omega_2-\omega_1}{\text{t}}$
$=\frac{5\pi-0}{10}=\frac{\pi}{2}\text{rad/sec}^2$ As $\tau=\text{I}\alpha$
$\therefore\tau=6\times\frac{\pi}{2}=3\pi\text{Ns}.$
$=5\pi\text{ rad/sec}^2$ Time for which torque acts, $\text{t}=10\text{\sec.}$
$\therefore$ Angular acceleration $(\alpha)=\frac{\omega_2-\omega_1}{\text{t}}$
$=\frac{5\pi-0}{10}=\frac{\pi}{2}\text{rad/sec}^2$ As $\tau=\text{I}\alpha$
$\therefore\tau=6\times\frac{\pi}{2}=3\pi\text{Ns}.$
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