Work-energy Theorem For Rotation

The rotational work-energy theorem states that the work done by a torque equals the change in rotational kinetic energy: $\tau \theta =\frac{1}{2}I{\omega }^2-\frac{1}{2}I{\omega }_0^2$. Starting from rest, the right side is simply $\frac{1}{2}I{\omega }^2=\frac{1}{2}(5)(120{)}^2=2.5\times 14400=36000\text{ J}$. The angular displacement must be converted to radians: $\theta =30\text{ rev}=30\times 2\pi =60\pi \approx 188.5\text{ rad}$. Therefore $\tau =\frac{36000}{188.5}\approx 191\text{ N m}$. The value 1200 N m forgets to convert revolutions to radians and divides by 30 instead. The value 95 N m omits the factor of one-half on the energy, halving the torque. The value 300 N m results from using $\omega$ instead of ${\omega }^2$ in the energy. This is the NCERT rotational analogue of the linear work-energy theorem. As a unit check, joules divided by radians (a dimensionless angle) yields newton-metres, and roughly 190 N m is reasonable for storing 36 kJ over about 188 radians. Equivalently one could extract the constant angular acceleration from ${\omega }^2=2\alpha \theta$ and then apply $\tau =I\alpha$, but the work-energy route is quicker here because it bypasses the explicit calculation of angular acceleration and works directly with the energy change.