Animated Solution for Physics - Rotational Motion: A bob of mass $m$ attached to an inextensible string of length $l$ is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed $\omega$ rad/s about the vertical support. About the point of suspension,

Visualizing the Conical Pendulum

Imagine a bob tied to an inextensible string, whirling around in a horizontal circle. This beautiful setup is known as a conical pendulum. The string traces out a cone in space, and the bob moves with a constant angular speed $\omega$.

Our goal is to analyze the angular momentum of this bob specifically about the point of suspension, let's call it $O$.

Angular Momentum

The fundamental definition of angular momentum $\mathbf{L}$ of a particle about a specific origin is given by the cross product of its position vector $\mathbf{r}$ and its linear momentum $\mathbf{p}$:

Here, $\mathbf{r}$ is the vector pointing from the suspension point $O$ to the bob, and $\mathbf{v}$ is the tangential velocity of the bob along its circular path.

A Perfect Perpendicularity

To find the magnitude of the angular momentum, we need to look at the angle between the position vector $\mathbf{r}$ and the velocity vector $\mathbf{v}$.

Because the bob is moving in a horizontal circle, its velocity $\mathbf{v}$ is purely tangential to the circle. The position vector $\mathbf{r}$, originating from the suspension point, has a downward vertical component and an outward radial component.

If you take the dot product $\mathbf{r}\cdot \mathbf{v}$, you will find it is exactly zero! The tangential velocity is perpendicular to both the radial direction and the vertical direction. Therefore, $\mathbf{r}$ and $\mathbf{v}$ are always perfectly perpendicular ($\theta =90^{\circ }$).

The magnitude of the angular momentum is:

Since the mass $m$, the string length $l$, and the speed $v$ are all constant, the magnitude of the angular momentum remains perfectly constant.

The Sweeping Cone

Now, what about the direction? The direction of $\mathbf{L}$ is given by the right-hand rule for the cross product $\mathbf{r}\times \mathbf{v}$.

Because $\mathbf{L}$ must be perpendicular to both $\mathbf{r}$ and $\mathbf{v}$, it points upwards and outwards, perpendicular to the string. As the bob revolves around the vertical axis, the plane containing $\mathbf{r}$ and $\mathbf{v}$ continuously tilts and rotates.

Consequently, the perpendicular vector $\mathbf{L}$ cannot stay still. It rotates along with the bob, sweeping out its own cone in space! Therefore, the direction of the angular momentum is continuously changing.

Combining our findings, the angular momentum changes in direction but not in magnitude.

A Crucial Catch: What if the question asked for the angular momentum about the center of the horizontal circle instead of the suspension point? In that case, the position vector would be purely radial, and $\mathbf{r}\times \mathbf{v}$ would point strictly vertically upwards. Both the magnitude and direction would be constant! This highlights a profound truth in rotational dynamics: Angular momentum is entirely dependent on your choice of reference point.