Mass tells you how hard it is to change straight-line motion. For turning, what matters is not only how much mass there is, but how far it sits from the axis. That is rotational inertia, built from the torque needed to spin something up and the idea of a system of masses.
Why it works
Picture one small mass $m$ on a light rod, a distance $r$ from the axis. Push it along its circle with force $F$: $F = ma_t = m(r\alpha)$. The torque of that push is $\tau = rF = mr^2\alpha$. So the "rotational mass" of this one bead is $mr^2$. The same bead twice as far out is four times as hard to spin up: for the same $\alpha$ it needs twice the acceleration, so twice the force, and that force acts at twice the distance, which doubles the torque again. For many pieces, add them up.
Key idea
Rotational inertia: $I = \Sigma m r^2$, with $r$ measured from the axis, in $\text{kg·m}^2$. Mass far from the axis counts much more than mass near it. For common shapes of mass $M$ and radius $R$: hoop $MR^2$, solid disk or cylinder $\tfrac{1}{2}MR^2$, solid sphere $\tfrac{2}{5}MR^2$, thin hollow sphere $\tfrac{2}{3}MR^2$, thin rod about its center $\tfrac{1}{12}ML^2$, about one end $\tfrac{1}{3}ML^2$.
Worked examples
Worked example: Beads on a rod
A light rod spins about its middle. It carries $2.0\ \text{kg}$ at $0.30\ \text{m}$ on one side and $1.0\ \text{kg}$ at $0.50\ \text{m}$ on the other. Find $I$.
- $$I = (2.0)(0.30)^2 + (1.0)(0.50)^2 = 0.18 + 0.25 = 0.43\ \text{kg·m}^2.$$
- The lighter bead contributes more because it is farther out.
Answer: $0.43\ \text{kg·m}^2$
Worked example: Same object, different axis
The same two beads, but now the axis passes through the $2.0\ \text{kg}$ bead, so the other bead is $0.80\ \text{m}$ away. Find $I$.
- The bead on the axis has $r = 0$ and adds nothing.
- $I = (1.0)(0.80)^2 = 0.64\ \text{kg·m}^2$, larger than before: $I$ depends on the axis.
Answer: $0.64\ \text{kg·m}^2$
Worked example: A disk with a lump
A $4.0\ \text{kg}$ solid disk of radius $0.25\ \text{m}$ has $0.50\ \text{kg}$ of clay stuck to its rim. Find $I$ about its center.
- Disk: $\tfrac{1}{2}(4.0)(0.25)^2 = 0.125\ \text{kg·m}^2$. Clay: $(0.50)(0.25)^2 = 0.031\ \text{kg·m}^2$.
- $I = 0.156\ \text{kg·m}^2$. The clay is one ninth of the mass but adds a quarter as much again.
Answer: $0.16\ \text{kg·m}^2$
Common mistake
Thinking rotational inertia depends only on the mass. A hoop and a disk of the same mass and radius have different $I$ ($MR^2$ against $\tfrac{1}{2}MR^2$), because the hoop keeps all its mass at the rim. Moving mass outward raises $I$ even though the mass doesn't change. Check: every term in $I$ has $r^2$ in it; if your answer doesn't use distances from the axis, it can't be right.