Moment of Inertia

In rotational motion, mass alone doesn't determine how hard it is to spin something — where the mass sits matters too. Two objects of the same total mass can have very different rotational inertia. This rotational analogue of mass is the moment of inertia.

Concept

For a system of discrete particles, the moment of inertia about a fixed axis is

$I = \sum_i m_i r_i^2$

where $r_i$ is the perpendicular distance of $m_i$ from the axis.

For a continuous body,

$I = \int r^2\,dm$

SI unit: kg·m².

Properties:

• Depends on the axis chosen.
• Mass farther from the axis contributes much more (quadratic dependence).
• Used in $L = I\omega$ and $\tau = I\alpha$ for fixed-axis rotation.
• Rotational KE: $K_{\text{rot}} = \tfrac{1}{2}I\omega^2$.

Radius of gyration $k$: A representative distance such that $I = Mk^2$, where $M$ is the total mass. So

$k = \sqrt{\frac{I}{M}}$

Moments of inertia of standard bodies (about specified axes):

Derivation

Thin uniform rod about perpendicular axis through centre. Linear density $\lambda = M/L$. Take an element $dm = \lambda\,dx$ at position $x$ from centre:

$I = \int_{-L/2}^{L/2} x^2 \cdot \lambda\,dx = \lambda\cdot\frac{x^3}{3}\Big|_{-L/2}^{L/2} = \lambda \cdot \frac{L^3}{12} = \frac{M L^2}{12}$

Thin ring about perpendicular axis through centre. All mass is at distance $R$ from the axis:

$I = \int R^2\,dm = R^2 \int dm = MR^2$

Solid disc about axis. Use thin rings of radius $r$, thickness $dr$, mass $dm = (M/(\pi R^2))\cdot 2\pi r\,dr$:

$I = \int_0^R r^2\,dm = \frac{2M}{R^2}\int_0^R r^3\,dr = \frac{2M}{R^2}\cdot\frac{R^4}{4} = \frac{MR^2}{2}$

Worked Example

A thin uniform rod of mass 2 kg and length 1 m rotates about a perpendicular axis through one end. Find the moment of inertia and the radius of gyration.

Solution:

About the end, $I = \tfrac{1}{3}ML^2 = \tfrac{1}{3}(2)(1)^2 = \tfrac{2}{3} \approx 0.667 \, \text{kg·m}^2$.

Radius of gyration: $k = \sqrt{I/M} = \sqrt{(2/3)/2} = \sqrt{1/3} = 1/\sqrt{3} \approx 0.577 \, \text{m}$.

So all the mass concentrated at distance $L/\sqrt{3}$ from the end gives the same $I$.

Common Confusions

• MOI depends on axis. A rod has $ML^2/12$ about its centre but $ML^2/3$ about its end — same mass, four times the MOI.
• MOI ≠ mass. It includes mass distribution.
• Quadratic in distance. Doubling distance from axis quadruples a particle's contribution.
• Radius of gyration is a defined distance, not a measured one. It need not coincide with any physical feature of the body.

Key Takeaways

• $I = \sum m_i r_i^2$ or $\int r^2 dm$.
• Standard bodies: $ML^2/12$ (rod, centre), $MR^2$ (ring), $\tfrac{1}{2}MR^2$ (disc), $\tfrac{2}{5}MR^2$ (solid sphere).
• Radius of gyration $k = \sqrt{I/M}$, with $I = Mk^2$.
• Mass farther from axis contributes more — distribution matters.
• Required for $L = I\omega$, $\tau = I\alpha$, $K_{\text{rot}} = \tfrac{1}{2}I\omega^2$.