Animated Solution for Physics - Rotational Motion: Angular momentum of a single particle moving with constant speed along circular path

Imagine you are standing at the center of a giant merry-go-round, watching a single point on the edge spin around you. This simple, mesmerizing motion holds some of the most profound secrets of physics.

When a particle moves in a circular path with a constant speed, it is in a state of uniform circular motion. But to truly understand the dynamics of this particle, we need to look beyond just its speed and path. We need to explore its angular momentum.

Angular momentum is the rotational equivalent of linear momentum. It tells us how much "spin" an object has and how hard it would be to stop that spin.

Let's break down the physical reality of our particle. At any given instant, the particle has a specific position relative to the center of the circle. We represent this with the position vector, $\vec{r}$.

Simultaneously, the particle is moving. Its velocity is always directed along the tangent to the circular path. We represent this with the velocity vector, $\vec{v}$.

Because the particle is constrained to a circular path, the radius (position vector) and the tangent (velocity vector) are always perfectly perpendicular to each other. This geometric fact is the cornerstone of our solution.

To find the angular momentum, we use its fundamental definition. The angular momentum $\vec{L}$ of a particle is defined as the cross product of its position vector $\vec{r}$ and its linear momentum $\vec{p}$.

Since linear momentum is simply mass times velocity ($\vec{p}=m\vec{v}$), we can rewrite our master equation as:

This equation is a vector equation, meaning it contains information about both the magnitude (how much angular momentum there is) and the direction (where the axis of rotation points). Let's evaluate them one by one.

The magnitude of a cross product depends on the angle between the two vectors. Mathematically, the magnitude of the angular momentum is given by:

Here, $\theta$ is the angle between the position vector $\vec{r}$ and the velocity vector $\vec{v}$.

Recall our earlier observation: in a circular path, the radius and the tangent are always perpendicular. Therefore, $\theta =90^{\circ }$.

Since $\sin (90^{\circ })=1$, our equation simplifies beautifully:

Now, let's look at the variables. The mass $m$ of the particle is constant. The radius $r$ of the circular path is constant. And the problem explicitly states that the particle is moving with a constant speed $v$.

Because $m$, $v$, and $r$ are all constant, their product must also be constant. Therefore, the magnitude of the angular momentum remains perfectly constant.

Now, what about the direction? This is where the magic of 3D space comes into play. The direction of a cross product is determined by the Right-Hand Rule.

If you point the fingers of your right hand in the direction of the position vector $\vec{r}$, and then curl them towards the velocity vector $\vec{v}$, your thumb will point in the direction of the angular momentum vector $\vec{L}$.

Because $\vec{r}$ and $\vec{v}$ both lie in the plane of the circle, your thumb will point perfectly perpendicular to that plane.

As the particle moves around the circle, the vectors $\vec{r}$ and $\vec{v}$ continuously change their direction. However, they always remain within the exact same 2D plane. Consequently, the perpendicular vector $\vec{L}$ always points in the exact same direction—straight along the axis of rotation.

Therefore, the direction of the angular momentum remains perfectly constant.

We have meticulously analyzed both components of the angular momentum vector.

The constant speed and fixed radius guarantee that the magnitude never changes. The fixed plane of motion guarantees that the direction never changes.

Thus, for a single particle moving with constant speed along a circular path, its angular momentum remains the same in both magnitude and direction.

What if we tweaked the problem? Imagine the particle is still moving in a circle, but it is accelerating—its speed $v$ is increasing. This is known as non-uniform circular motion.

In this scenario, the plane of motion is still fixed, so the direction of $\vec{L}$ would remain constant. However, because $v$ is increasing, the magnitude $|\vec{L}|=mvr$ would also increase.

Understanding these subtle distinctions is what transforms a good physics student into a master problem solver. Always ask yourself: "What if?"