Rotational Kinetic Energy

Question

An industrial flywheel with a moment of inertia of 0.5 kg m^2 spins steadily at 600 revolutions per minute about its central axis. What is the rotational kinetic energy stored in the flywheel?

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Rotational kinetic energy is stored in a spinning body according to $KE = \frac{1}{2}I\omega^2$, the rotational analogue of $\frac{1}{2}mv^2$ for linear motion. The first essential step is to convert the spin rate to SI units: $\omega = 600	ext{ rpm} = \frac{600 	imes 2\pi}{60} = 20\pi \approx 62.83	ext{ rad/s}$. Then $KE = \frac{1}{2}(0.5)(62.83)^2 = 0.25 	imes 3947.8 \approx 987	ext{ J}$. The value 90 J results from using the spin rate in rpm directly without converting to rad/s, a serious unit error. The value 1974 J double-counts the energy by omitting the factor of one-half. The value 500 J ignores the squaring of the angular velocity altogether. This is the standard NCERT expression for energy stored in a rotating flywheel, a principle exploited in engines to smooth out uneven power delivery. As a unit check, kg m$^2$ multiplied by (rad/s)$^2$ yields joules, and roughly one kilojoule is reasonable for a flywheel spinning at ten revolutions per second. This stored energy is what lets a flywheel deliver a sudden burst of power in a punch press or keep an engine turning smoothly between firing strokes, and because it grows with the square of angular speed, doubling the rpm would quadruple the energy held in the wheel.

Rotational Kinetic Energy

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