Rotational Inertia defined $I = \sum mr^2$ for point masses: mass far from the axis counts more. For a solid object, slice it into tiny pieces as in Center of Mass by Integration and add up $r^2\,dm$.
Key idea
$$I = \int r^2\,dm,$$ where $r$ is each piece's distance from the axis of rotation. Results worth knowing: thin rod about its center $\tfrac{1}{12}ML^2$, about one end $\tfrac{1}{3}ML^2$; thin hoop $MR^2$; solid disk or cylinder about its axle $\tfrac{1}{2}MR^2$; solid sphere $\tfrac{2}{5}MR^2$; thin spherical shell $\tfrac{2}{3}MR^2$.
Why it works
Kinetic energy explains the $r^2$. A piece at distance $r$ moves at $v = r\omega$, so its energy is $\tfrac{1}{2}dm\,(r\omega)^2$. Adding over the body: $K = \tfrac{1}{2}\left(\int r^2\,dm\right)\omega^2 = \tfrac{1}{2}I\omega^2$. Doubling a piece's distance from the axis doubles its speed and quadruples its energy at the same $\omega$.
Worked example: A uniform rod
Find $I$ for a uniform $2\ \text{kg}$ rod of length $1.2\ \text{m}$ about an axis through one end, perpendicular to the rod.
- $dm = \dfrac{M}{L}dx$ at distance $x$ from the end.
- $I = \dfrac{M}{L}\int_0^L x^2\,dx = \dfrac{M}{L}\cdot\dfrac{L^3}{3} = \tfrac{1}{3}ML^2$.
- $I = \tfrac{1}{3}(2)(1.2)^2 = 0.96\ \text{kg·m}^2$. (About the center it would be $\tfrac{1}{12}ML^2 = 0.24\ \text{kg·m}^2$, a quarter as much.)
Answer: $0.96\ \text{kg·m}^2$.
Worked example: A solid disk from rings
Show that a uniform disk of mass $M$ and radius $R$ has $I = \tfrac{1}{2}MR^2$ about its axle.
- Use rings: a ring of radius $r$ and width $dr$ has mass $dm = \sigma\,2\pi r\,dr$, with $\sigma = M/(\pi R^2)$, and all of it is at distance $r$.
- $$I = \int_0^R r^2\,\sigma\,2\pi r\,dr = 2\pi\sigma\dfrac{R^4}{4} = \dfrac{M}{\pi R^2}\cdot\dfrac{\pi R^4}{2} = \tfrac{1}{2}MR^2$$
Answer: $I = \tfrac{1}{2}MR^2$.
Two objects with the same mass and size can have very different $I$: a hoop puts all its mass at the rim ($MR^2$), a solid disk spreads it inward ($\tfrac{1}{2}MR^2$). That difference decides races down ramps in Rolling Dynamics. For axes that don't pass through the center of mass, The Parallel-Axis Theorem saves you from integrating again.
Common mistake
Using each piece's distance from the center of mass, or from the middle of the object, when the axis is somewhere else. For a rod spinning about its end, the far end is a distance $L$ from the axis, not $L/2$. Check: draw the axis first and measure every $r$ from it. Also check the ranking: the same mass gives the largest $I$ when it sits farthest from the axis.