Work-energy Theorem For Rotation

Question

A flywheel of moment of inertia 5 kg m^2, starting from rest, is brought to an angular speed of 120 rad/s by a constant torque acting through 30 complete revolutions. Neglecting friction, what is the magnitude of the applied torque?

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The rotational work-energy theorem states that the work done by a torque equals the change in rotational kinetic energy: $	au 	heta = \frac{1}{2}I\omega^2 - \frac{1}{2}I\omega_0^2$. Starting from rest, the right side is simply $\frac{1}{2}I\omega^2 = \frac{1}{2}(5)(120)^2 = 2.5 	imes 14400 = 36000	ext{ J}$. The angular displacement must be converted to radians: $	heta = 30	ext{ rev} = 30 	imes 2\pi = 60\pi \approx 188.5	ext{ rad}$. Therefore $	au = \frac{36000}{188.5} \approx 191	ext{ 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 $\omega^2 = 2\alpha	heta$ and then apply $	au = 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.

Work-energy Theorem For Rotation

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