Push a box and it resists you. That resistance is its mass. Now try to spin something, and you meet a different kind of stubbornness — one that depends not just on how much mass there is, but on where that mass is.
That quantity is the moment of inertia, written $I$.
The definition
For a collection of point masses, each a distance $r_i$ from the axis of rotation:
$$I = \sum_i m_i r_i^2$$For a continuous body, the sum becomes an integral:
$$I = \int r^2 \, dm$$Notice the $r^2$. This is the whole story. Mass twice as far from the axis contributes four times as much. Distance dominates.
Why a hoop beats a disc
Take a hoop and a solid disc, both of mass $M$ and radius $R$. Identical mass, identical size.
- Hoop: all mass sits at radius $R$, so $I = MR^2$
- Disc: mass is spread from centre to edge, so $I = \tfrac{1}{2}MR^2$
The hoop is twice as hard to spin up. Roll them down a ramp together and the disc wins every time — it spends less energy on rotation and more on moving forward.
This is not a trick of the equations. It is why flywheels put their mass at the rim, and why a figure skater pulls their arms in to spin faster.
Common results worth memorising
| Body (axis through centre) | Moment of inertia |
|---|---|
| Hoop / thin cylindrical shell | $MR^2$ |
| Solid disc / cylinder | $\tfrac{1}{2}MR^2$ |
| Solid sphere | $\tfrac{2}{5}MR^2$ |
| Thin spherical shell | $\tfrac{2}{3}MR^2$ |
| Rod, axis through middle | $\tfrac{1}{12}ML^2$ |
Rotational kinetic energy
Everything you know about linear motion has a rotational twin:
$$K_{\text{linear}} = \tfrac{1}{2}mv^2 \qquad\qquad K_{\text{rot}} = \tfrac{1}{2}I\omega^2$$Mass becomes moment of inertia; velocity becomes angular velocity. The structure is identical — which is the real reason $I$ is worth understanding properly.
Check yourself
A solid sphere and a hollow sphere of equal mass and radius roll from rest down the same incline. Which reaches the bottom first, and why?
Hint: compare $\tfrac{2}{5}MR^2$ with $\tfrac{2}{3}MR^2$ and think about where the energy goes.