Animated Solution for Physics - Rotational Motion: A disc of mass $M$ and radius $R$ is rolling with angular speed $\omega$ on a horizontal plane as shown. The magnitude of angular momentum of the disc about the origin $O$ is

When a rigid body like a disc is both translating and rotating, its total angular momentum about any fixed point in space is not just a single simple term. It is a beautiful vector sum of two distinct physical phenomena.

Imagine you are standing at the origin $O$, watching this disc roll past you. You see two things happening simultaneously: the disc is spinning around its own center, and the center of the disc is moving linearly past you.

Mathematically, this is expressed by the master equation:

Here, ${\vec{L}}_{CM}$ is the spin angular momentum (the rotation of the disc about its own center of mass), and $\vec{r}\times {\vec{p}}_{CM}$ is the orbital angular momentum (the angular momentum of the center of mass itself, treated as a point particle, about the origin).

Let's break down the first term, the spin angular momentum. For a symmetric body rotating about its center of mass, this is simply the moment of inertia multiplied by the angular velocity:

For a uniform solid disc, the moment of inertia about its central axis is $I_{CM}=\frac{1}{2}M{R}^2$.

The disc is rolling forward, which means it is rotating clockwise. If you curl the fingers of your right hand in a clockwise direction, your thumb points directly into the page (or screen). Therefore, the magnitude of the spin angular momentum is $\frac{1}{2}M{R}^2\omega$, and its direction is into the page.

Now, let's look at the orbital term, $\vec{r}\times {\vec{p}}_{CM}$. The center of mass of the disc is moving in a straight horizontal line. Because the disc is undergoing pure rolling, the velocity of the center of mass is intimately tied to its angular speed by the relation $v=R\omega$.

The linear momentum of the center of mass is ${\vec{p}}_{CM}=Mv=MR\omega$.

To compute the cross product $\vec{r}\times {\vec{p}}_{CM}$, we don't need to do complicated matrix math. We can use the geometric definition: the magnitude is simply the momentum multiplied by the perpendicular distance from the origin to the line of motion.

The center of mass moves along a horizontal line at a constant height $R$ above the x-axis. Therefore, the perpendicular distance is exactly $R$.

What about the direction? Using the right-hand rule for the cross product (fingers from $\vec{r}$ curling towards ${\vec{p}}_{CM}$), your thumb again points directly into the page.

We have found our two pieces of the puzzle. Both the spin angular momentum and the orbital angular momentum point in the exact same direction (into the page). Because they are collinear, we can simply add their magnitudes together algebraically to find the total angular momentum about the origin:

This elegant result shows that the translational motion of the rolling disc contributes twice as much to the total angular momentum about the origin as its own internal spinning does!