A wheel of moment of inertia $2.5 \mathrm{~kg}-\mathrm{m}^{2}$ has an initial angular velocity of $4

Question

A wheel of moment of inertia $2.5 \mathrm{~kg}-\mathrm{m}^{2}$ has an initial angular velocity of $40 \mathrm{rad} \mathrm{s}^{-1}$. A constant torque of 10 Nm acts on the wheel. The time during which the wheel is accelerated to $60 \mathrm{rad} \mathrm{s}^{-1}$ is\

Professor PC Thomas Classes · Answer

Here, $\mathrm{I}=2.5 \mathrm{kgm}^{2}, \omega_{0}=40 \mathrm{rads}^{-1}$\ $	au=10 \mathrm{Nm}, \omega=60 \mathrm{rads}^{-1}$\ or $\alpha=\frac{	au}{\mathrm{I}}=\frac{10}{2.5}=4 \mathrm{rads}^{-2}$\ Using $\omega=\omega_{0}+\alpha \mathrm{t}$\ $\mathrm{t}=\frac{\omega-\omega_{0}}{\alpha}$\ Substituting the values, we get\ $\mathrm{t}=\frac{60-40}{4}=5 \mathrm{~s}$\