A fly wheel of moment of inertia $3 imes 10^{2} \mathrm{kgm}^{2}$ is rotating with uniform angular

Question

A fly wheel of moment of inertia $3 	imes 10^{2} \mathrm{kgm}^{2}$ is rotating with uniform angular speed of $4.6 \mathrm{rads}^{-1}$. If a torque of $6.9 	imes 10^{2} \mathrm{Nm}$ retards the wheel, then the time is which the wheel comes to rest is\

Professor PC Thomas Classes · Answer

Here, Momentum of inertia, $\mathrm{I}=3 	imes 10^{2} \mathrm{kgm}^{2}$\ Torque, $	au=6.9 	imes 10^{2} \mathrm{Nm}$\ Initial angular speed, $\omega_{0}=4.6 \mathrm{rads}^{-1}$\ Final angular speed, $\omega=0 \mathrm{rad} \mathrm{s}^{-1}$\ As $\omega=\omega_{0}+\alpha \mathrm{t}$\ $\alpha=\frac{\omega-\omega_{0}}{\mathrm{t}}=\frac{0-4.6}{\mathrm{t}}=\frac{4.6}{\mathrm{t}} \mathrm{rads}^{-2}$ Negative sign is for deceleration\ Torque, $	au=\mathrm{I} \alpha$\ $6.9 	imes 10^{2}=3 	imes 10^{2} 	imes \frac{4.6}{t}$\ $\mathrm{t}=\frac{3 	imes 10^{2} 	imes 4.6}{6.9 	imes 10^{2}}=2 \mathrm{~s}$\