Animated Solution for Physics - Rotational Motion: A particle moves in a circular path with decreasing speed. Choose the correct statement.

Imagine you are watching a particle moving along a circular track, but it's running out of energy—its speed is continuously decreasing. This simple setup is a classic trap in rotational mechanics, testing your ability to separate vectors into their magnitude and direction.

Let's break down the physics step by step.

The angular momentum $\mathbf{L}$ of a particle about the center of its circular path is defined by the cross product $\mathbf{L}=m(\mathbf{r}\times \mathbf{v})$.

Because the particle is confined to a flat, circular plane, the position vector $\mathbf{r}$ and the velocity vector $\mathbf{v}$ always lie in this 2D plane. If you apply the right-hand rule—curling your fingers from $\mathbf{r}$ to $\mathbf{v}$—your thumb will point perfectly perpendicular to the plane. Since the plane doesn't tilt or shift, this direction of angular momentum remains absolutely constant.

But what about its magnitude? The magnitude of angular momentum is given by $|\mathbf{L}|=mrv$. The mass $m$ and the radius $r$ are constant, but the problem explicitly states that the speed $v$ is decreasing. Therefore, the magnitude of $\mathbf{L}$ is shrinking. Because the magnitude changes, the angular momentum vector as a whole is not constant. This immediately eliminates option (a).

Now, let's look at the acceleration. It is a common misconception to assume that any particle in a circle has an acceleration pointing straight to the center.

That is only true for uniform circular motion. Here, the speed is decreasing. This means there are two distinct components of acceleration at play: 1. Centripetal Acceleration (${\mathbf{a}}_c$): This points towards the center and is responsible for changing the direction of the velocity to keep the particle in a circle. 2. Tangential Acceleration (${\mathbf{a}}_t$): Because the particle is slowing down, there must be an acceleration component pointing directly opposite to the velocity vector.

The net acceleration $\mathbf{a}$ is the vector sum of these two components ($\mathbf{a}={\mathbf{a}}_c+{\mathbf{a}}_t$). Because of the tangential component, the net acceleration vector tilts away from the center. Thus, option (b) is incorrect.

What about option (c)? The problem explicitly defines the trajectory as a "circular path." A circle has a fixed radius. Even though the particle is slowing down, it is forced to stay on this circular track, so it does not spiral inwards.

This leaves us with the beautiful geometric truth of option (d): The direction of angular momentum remains constant.