Angular momentum about a point

The angular momentum of a particle about point $P$ is $\vec L = \vec r \times \vec p = \vec r \times m\vec v$ where $\vec r$ is the position vector of the particle from point $P$.

Condition for zero angular momentum

If the particle moves along a straight line that passes through point $P$, then $\vec r$ is always along the same line as $\vec v$.

Hence, the angle between $\vec r$ and $\vec v$ is either $0^\circ$ or $180^\circ$.

Therefore, $L = mrv\sin\theta = mrv\sin 0^\circ = 0$ or $L = mrv\sin 180^\circ = 0$

Using perpendicular distance form

Magnitude of angular momentum about a point can also be written as $L = mv \times (\text{perpendicular distance of point } P \text{ from line of motion})$

Since the particle moves along line $PC$, the line of motion passes through $P$ itself. So the perpendicular distance is zero.

Thus, $L = mv \times 0 = 0$

Option check

• A: $mvL$ — incorrect
• B: $mvl$ — incorrect
• C: $mvr$ — incorrect
• D: zero — correct

Final answer

$\boxed{\text{Angular momentum about } P = 0}$