Animated Solution for Physics - Rotational Motion: A particle undergoes uniform circular motion. About which point on the plane of the circle, will the angular momentum of the particle remain conserved ?

The beauty of physics often lies in its conservation laws. These laws act as the absolute truths of the universe, providing us with powerful shortcuts to solve complex problems without getting bogged down in the messy details of forces and accelerations at every single instant. One of the most profound of these is the Conservation of Angular Momentum. But conservation laws don't just happen magically; they are deeply tied to the symmetries of the system and the specific reference points we choose.

In this problem, we are asked to find the point about which the angular momentum of a particle undergoing uniform circular motion remains conserved. Let's embark on a journey to understand not just the 'what', but the deep 'why' behind this phenomenon.

Imagine a particle of mass $m$ tied to a string, whirling around in a perfect circle on a frictionless table. The speed of the particle, $v$, is constant. This is the hallmark of uniform circular motion.

Even though the speed is constant, the velocity is constantly changing because the direction of motion is continuously altering. According to Newton's First Law, a change in velocity requires a net external force. In uniform circular motion, this force is the centripetal force, denoted as ${\vec{F}}_c$.

The defining characteristic of the centripetal force is its direction: it always points radially inward, directly towards the center of the circular path. If the string were to suddenly snap, this force would vanish, and the particle would fly off in a straight line tangent to the circle.

To determine if angular momentum is conserved, we must look at its rotational counterpart to force: Torque ($\vec{\tau }$). Just as a net linear force causes a change in linear momentum, a net torque causes a change in angular momentum. This relationship is elegantly captured by Newton's Second Law for rotation:

This equation tells us a simple but profound truth: Angular momentum $\vec{L}$ is conserved (meaning its rate of change is zero) if and only if the net torque ${\vec{\tau }}_{net}$ acting on the system is zero.

So, our quest to find the point of conserved angular momentum translates directly into a quest to find the point about which the net torque is zero.

Torque is not an absolute quantity; it is always defined relative to a specific reference point. Mathematically, the torque produced by a force $\vec{F}$ about a point $O$ is given by the cross product of the position vector $\vec{r}$ (drawn from $O$ to the point of application of the force) and the force vector itself:

The magnitude of this cross product is given by:

where $\theta$ is the angle between the position vector $\vec{r}$ and the force vector $\vec{F}$ when they are placed tail-to-tail.

Let's test the center of the circle as our reference point.

We draw the position vector $\vec{r}$ from the center of the circle to the particle. By definition, this vector points radially outward.

Now, consider the only force acting on the particle: the centripetal force ${\vec{F}}_c$. As we established earlier, this force points radially inward, directly towards the center.

Notice the beautiful geometric relationship here. The position vector $\vec{r}$ and the force vector ${\vec{F}}_c$ lie along the exact same line, but they point in exactly opposite directions. They are anti-parallel.

Because they are anti-parallel, the angle $\theta$ between them is exactly $180^{\circ }$. Let's plug this into our torque magnitude equation:

Since $\sin (180^{\circ })=0$, the entire expression collapses:

The torque about the center of the circle is identically zero. Returning to our master equation, if ${\vec{\tau }}_{net}=0$, then $\frac{d\vec{L}}{dt}=0$.

To truly appreciate this result, let's consider what happens if we choose a different reference point, say, a point $P$ on the circumference of the circle.

We draw a new position vector, ${\vec{r}}^{\prime }$, from point $P$ to the particle. The force acting on the particle is still the same centripetal force ${\vec{F}}_c$ pointing towards the center $O$.

However, look at the geometry now. The vector ${\vec{r}}^{\prime }$ and the vector ${\vec{F}}_c$ are no longer collinear. They form some arbitrary angle $\theta$ that is neither $0^{\circ }$ nor $180^{\circ }$.

Because $\sin\theta eq 0$, the cross product ${\vec{r}}^{\prime }\times {\vec{F}}_c$ will yield a non-zero value. A net torque exists about point $P$. Consequently, the angular momentum of the particle calculated about point $P$ will continuously change as the particle moves around the circle. It is not conserved.

This logic holds true for any point inside or outside the circle, except for the exact geometric center. The center is the unique point of symmetry where the line of action of the centripetal force always passes through the reference point, rendering the lever arm (and thus the torque) zero at all times. This is why central forces, like gravity acting on planets, always lead to the conservation of angular momentum about the force center, giving rise to phenomena like Kepler's Law of Equal Areas.