Classical angular momentum

Angular momentum $L$ is the rotational analogue of linear momentum $p$ . For a particle rotating in a plane at radius $r$ about $O$ (see diagram below), its velocity component $v_\perp$ that is perpendicular to $r$ is

$v_\perp=\frac{2\pi r}{T}=r\omega\; \; \; \; \; \; \; \; 58$

where $T$ is the period (time taken to complete a revolution) and $\omega=\frac{2\pi}{T}$ is the angular velocity.

The rotational kinetic energy $KE_{rot}$ of the particle is

$KE_{rot}=\frac{1}{2}mv_{\perp}^{2}=\frac{1}{2}mr^{2}\omega^{2}=\frac{1}{2}I\omega^{2}$

where $I=mr^{2}$ is the moment of inertia.

Question

What is moment of inertia and why is it equal to $mr^{2}$ ?

Answer

The moment of inertia is the rotational equivalent of a particle’s inertia in linear motion. For a particle in linear motion, its inertia is quantified by its mass. For a particle in rotational motion, its inertia is dependent on both its mass and the distribution of that mass relative to $O$ (i.e. dependent on $m$ and $r$ for a point mass). Since $\frac{1}{2}mv_{\perp}^{\, \, 2}=\frac{1}{2}mr^{2}\omega^{2}$ , where $\omega$ is the angular velocity of the rotating particle, we define $I=mr^{2}$ such that there is a correspondence between $\omega$ and $v_{\perp}$ , and between $I$ and $m$ .