Work Done On A Rotating Wheel

If the torque applied is $Q$ Nm and the moment of inertia of the wheel is $I=0.2$ kg m $^2$ , then $I\dot\omega \; = \; -Q$ So $\dot\omega$ is constant, and the wheel is reduced from $\omega=20$ to $\omega=0$ in $4$ seconds. Thus $\dot\omega = -5$ s $^{-2}$ , and hence $Q = 1$ Nm. Thus $\omega \; = \; 20 - 5t$ The work done by the torque is $\int Q\,d\theta \; = \; \int Q \dot\theta\,dt \; = \; \int Q\omega\,dt$ and so the work done in the first $2$ seconds is $\int_0^2 (20 - 5t)\,dt \; = \; \Big[20t - \tfrac52t^2\Big]_0^2 \; = \; 30 \mbox{ J}$

As the wheel was rotating with an angular speed of 20 rad/s. and It is stopped to rest by applying a constant torque in 4 seconds.so retarding angular acceleration will be -5rad / $s^{2}$ . so initial angular kinetic energy will be $\frac{1}{2}$ I $w^{2}$ = 40 whereas as the angular velocity will be 10rad/s after 2 secs so final angular kinetic energy will be 10 . by work energy theorem work=40-10=30J

ω₀=20 rad/s. To come to rest ,time taken is 4s. Therefore,from the first equation of rotational motion, ω=ω₀+αt 0=20+α 4 α=-5 rad/s² In the first two seconds ω attained will be, ω=ω₀+αt ω=20-4 2 ω=12rad/s. Using Work Energy theorem for rotation, Work done=1/2 I (ω²-ω₀²) =0.5 0.20 (10²-20²) =-30J .