Rotational Kinetic Energy

Substitute $I$ for $m$ and $\omega$ for $v$ in $\tfrac{1}{2}mv^2$ and you have the energy stored in anything that spins: $KE_{rot} = \tfrac{1}{2}I\omega^2$. The substitution is not a mnemonic: it is what falls out of adding up $\tfrac{1}{2}m_i v_i^2$ over every particle in the body, using $v_i = \omega r_i$, and collecting the $\sum m_i r_i^2$ into a single symbol called $I$. The moment of inertia is defined precisely so that this works.

A flywheel with $I = 40$ kg·m² turning at 3000 rpm, that is $\omega = 314$ rad/s, stores $\tfrac{1}{2} \times 40 \times 314^2 \approx 1.97$ MJ, roughly the energy in half a litre of petrol, and it can be given back in seconds. That is a real technology, not an illustration: grid-scale flywheel installations store megajoules in steel or carbon-fibre rotors and use them to smooth demand, and the same principle in miniature is what carries a single-cylinder engine between power strokes.

A rolling object carries both kinds of kinetic energy at once, translational $\tfrac{1}{2}mv^2$ and rotational $\tfrac{1}{2}I\omega^2$, and the split between them decides races down a ramp. A solid cylinder puts a smaller fraction of the available energy into spinning than a hollow hoop of the same mass and radius does, so more is left over for going forward, and the cylinder wins: regardless of mass, regardless of radius, which is the counter-intuitive and testable part.

The $\omega^2$ makes the rpm-to-rad/s conversion twice as expensive as usual. Feed 3000 straight in where 314 belongs and the answer is not 9.55 times too large but $9.55^2 = 91$ times too large: the flywheel above would appear to store 180 MJ. An answer that absurd is at least visible, but the same error on a smaller rotor produces a number that merely looks generous. The other error is one of omission: for anything that rolls rather than merely spins in place, the rotational energy is only part of the total, and an energy balance that counts $\tfrac{1}{2}I\omega^2$ alone will not close. A solid disk rolling without slipping carries exactly a third of its kinetic energy in rotation and two-thirds in translation. And as always, $I$ must be taken about the axis the body is actually turning about: for a rolling wheel analysed about its contact point rather than its centre, the parallel-axis theorem applies and the number changes.