A flywheel spinning in place goes nowhere, yet it can light a bulb or lift a weight: it stores kinetic energy. This topic finds how much, using rotational inertia, and extends the work-energy theorem to turning.
Why it works
Split a spinning object into small pieces. A piece of mass $m$ at radius $r$ moves at $v = r\omega$, so its kinetic energy is $\tfrac{1}{2}m(r\omega)^2$. Every piece shares the same $\omega$, so adding them gives $$K = \tfrac{1}{2}\left(\Sigma mr^2\right)\omega^2 = \tfrac{1}{2}I\omega^2.$$ For work: a force $F$ at radius $r$ pushing a piece along an arc $s = r\Delta\theta$ does work $Fs = (rF)\Delta\theta = \tau\Delta\theta$.
Key idea
A spinning object stores $K_{\text{rot}} = \tfrac{1}{2}I\omega^2$ ($\omega$ in rad/s). A torque turning it through $\Delta\theta$ (in radians) does work $W = \tau\Delta\theta$, and the net work equals the change in rotational kinetic energy: $\Sigma\tau\,\Delta\theta = \tfrac{1}{2}I\omega_f^2 - \tfrac{1}{2}I\omega_i^2$.
Worked examples
Worked example: A spinning flywheel
A solid-disk flywheel ($50\ \text{kg}$, radius $0.30\ \text{m}$) spins at $600$ rpm. How much kinetic energy does it store?
- $I = \tfrac{1}{2}(50)(0.30)^2 = 2.25\ \text{kg·m}^2$.
- $\omega = 600 \times \dfrac{2\pi}{60} = 62.8\ \text{rad/s}$.
- $K = \tfrac{1}{2}(2.25)(62.8)^2 = 4440\ \text{J}$, about the energy needed to lift a $45\ \text{kg}$ person $10\ \text{m}$.
Answer: about $4.4 \times 10^{3}\ \text{J}$
Worked example: Spinning up with a torque
A motor applies a net torque of $4.0\ \text{N·m}$ to a wheel ($I = 0.50\ \text{kg·m}^2$) at rest while it turns $5$ revolutions. Find its final $\omega$.
- $\Delta\theta = 5 \times 2\pi = 31.4\ \text{rad}$, so $W = (4.0)(31.4) = 126\ \text{J}$.
- $W = \tfrac{1}{2}I\omega^2$: $\omega = \sqrt{\dfrac{2(126)}{0.50}} = 22.4\ \text{rad/s}$.
Answer: $22\ \text{rad/s}$
Common mistake
Using $\tfrac{1}{2}mv^2$ with the rim speed for an object that only spins. Only the rim moves at that speed; everything inside moves more slowly. For a solid disk, $\tfrac{1}{2}I\omega^2 = \tfrac{1}{2}\left(\tfrac{1}{2}MR^2\right)\left(\tfrac{v}{R}\right)^2 = \tfrac{1}{4}Mv^2$, half of $\tfrac{1}{2}Mv^2$. Another trap is $\omega$ in rpm: $\tfrac{1}{2}I\omega^2$ needs rad/s. Check: a spinning object's $K$ can never be more than $\tfrac{1}{2}Mv_{\text{rim}}^2$, which only a thin hoop reaches.
Rotational kinetic energy is the second kind of kinetic energy. An object can have both: a rolling ball moves and spins at once, which is the next topic.