Q.Show that the angular momentum about any point of a single particle moving with constant velocity remains constant throughout the motion.

Angular momentum about a fixed point $O$: $\vec{L}=\vec{r}\times m\vec{v}$. Differentiating,

$\frac{d\vec{L}}{dt} = \vec{v}\times m\vec{v} + \vec{r}\times m\frac{d\vec{v}}{dt} = \vec{0} + \vec{0} = \vec{0}$ …

The figure shows a particle moving along a straight horizontal line with constant velocity $\mathbf{v}$. A fixed point O is marked below this line. At the instant shown, the particle is at point P on the line. A position vector $\mathbf{r}$ is drawn from O to P, making an angle $\theta$ with the horizontal line. A dashed vertical line from O meets the horizontal line at point M, so that OM is perpendicular to the direction of motion. The length OM is labelled as $r \sin \theta$.

The key physical idea is that even though the particle moves in a straight line with constant velocity, its angular momentum about point O is constant. The figure makes this obvious geometrically. The angular momentum of the particle about O is $\mathbf{l} = \mathbf{r} \times m\mathbf{v}$. Its magnitude is $l = m v r \sin \theta$. But $r \sin \theta$ is exactly the perpendicular distance from O to the line of motion — that is, the length OM. Since OM does not change as the particle moves along the line, $r \sin \theta$ is constant. Therefore $l$ is constant.

Here $m$ is the particle's mass, $v$ is its speed, $r$ is the distance from O to the particle, and $\theta$ is the angle between $\mathbf{r}$ and $\mathbf{v}$. The quantity $r \sin \theta$ is the lever arm — the perpendicular distance from the reference point O to the line along which the particle moves.

The figure also shows that the direction of $\mathbf{l}$ is perpendicular to the plane containing $\mathbf{r}$ and $\mathbf{v}$. For the motion shown, that direction is into or out of the page, depending on the orientation of the velocity relative to O. Since the particle's path is a straight line and O is fixed, this direction also remains constant.

The textbook uses this figure to ground the definition of angular momentum for a particle before extending it to systems of particles and rigid bodies. The constancy of angular momentum for a particle moving with constant velocity about any fixed point not on its path is a direct consequence of the cross product geometry. The dashed perpendicular OM is the visual key: it is the same length no matter where the particle is on the line.

Setting up

Let a particle of mass $m$ move with constant velocity $\vec{v}$. Pick any fixed point $O$ as the origin. At time $t$, the particle's position relative to $O$ is

where $\vec{r}_0$ is its position at $t=0$. Its angular momentum about $O$ is

Differentiating directly

Differentiate $\vec{L}$ with respect to time using the product rule:

$\frac{d\vec{L}}{dt} = \frac{d\vec{r}}{dt}\times m\vec{v} + \vec{r}\times m\frac{d\vec{v}}{dt} = \vec{v}\times m\vec{v} + \vec{r}\times m(\vec{0})$ …

Concept: Angular Momentum of a Particle Moving with Constant Velocity

Step 1: Write the angular momentum about a fixed point $O$

Step 2: Differentiate with respect to time using the product rule

Step 3: Evaluate each term

• $\vec v\times m\vec v = 0$, since the cross product of any vector with a scalar multiple of itself is zero.
• $\dfrac{d\vec v}{dt}=0$, since the particle moves with constant velocity.

Step 4: Conclude

Step 5: Confirm by direct substitution …