Angular momentum of rigid body in plane motion

Angular momentum of a rigid body in plane motion is the sum of the moments of the momenta of all its particles about a point. For a body of mass $m$ in plane motion with angular velocity $\omega$, the momenta of the particles are equivalent to the linear momentum $m\bar{\vec v}$ at the mass centre $G$ together with a couple $\bar I\omega$.

About the mass centre

$$H_G = \bar I\,\omega$$

where $\bar I$ is the moment of inertia about the axis through $G$ perpendicular to the plane. This is true whether or not $G$ is moving.

About any other point O

$$H_O = \bar I\omega + (\bar{\vec r}\times m\bar{\vec v}) = \bar I\omega + m\bar v\,d$$

where $d$ is the perpendicular distance from $O$ to the line of action of $m\bar{\vec v}$.

Special cases

The rate of change of angular momentum equals the sum of moments: $\sum M_G = \dot H_G = \bar I\alpha$.

Examples

1. A flywheel of mass 20 kg and radius of gyration 0.3 m turning at 100 rad/s about its axis: $H = mk^2\omega = 20(0.3)^2(100) = 180$ kg m²/s.
2. A uniform disk (mass 10 kg, radius 0.5 m) rolling at $\bar v = 2$ m/s: $\omega = 4$ rad/s, $H_G = \tfrac12(10)(0.5)^2(4) = 5$ kg m²/s; about the contact point $H_C = 5 + 10(2)(0.5) = 15$ kg m²/s, which equals $I_C\omega = \tfrac32(10)(0.25)(4)= 15$ kg m²/s.