Question

Sol. [Key Concept: For a solid sphere $I = \frac{2}{5}\ {MR}^{2}$; find $\alpha$ from $\omega = \omega_{0} + \alpha\ t$, then $\tau = I\alpha$ and revolutions $= \frac{\omega_{0}^{2}/(2\alpha)}{2\pi}$.]

$${\ \omega_{i} = 1200rpm = 1200 \times \frac{2\pi}{60} = 40\pi rad/sec }{\rightarrow \omega_{f} = \omega_{i} + \alpha t \Rightarrow 0 = 40p + \alpha(10) }{\alpha = - 4\pi rad/\sec^{2} }{\rightarrow I = \frac{2}{5}{MR}^{2} = \frac{2}{5}(5)(0.04)^{2} = 2 \times 0.0016 }{I = 0.0032Kg - m^{2} }{\rightarrow \tau = I\alpha = 0.0032 \times 4\pi }{\tau = 0.0128\pi\text{ }N - m }{\rightarrow \omega_{f}^{2} = \omega_{i}^{2} + 2\alpha\theta }{0 = (40\pi)^{2} + 2(4\pi)\theta }{\theta = \frac{1600\pi^{2}}{8\pi} = 200\pi }$$No. of revolution $= \frac{\theta}{2\pi} = \frac{200\pi}{2\pi}$

No. of revolution $= 100$