Angular momentum of a particle

Just as the moment of a force is the rotational analogue of force in linear motion, the quantity angular momentum is the rotational analogue of linear momentum. We shall first define angular momentum for the special case of a single particle and look at its usefulness in the context of single particle motion. We shall then extend the definition of angular momentum to systems of particles including rigid bodies.

Like moment of a force, angular momentum is also a vector product. It could also be referred to as moment of (linear) momentum. From this term one could guess how angular momentum is defined.

Consider a particle of mass m and linear momentum p at a position r relative to the origin O. The angular momentum l of the particle with respect to the origin O is defined to be

l = r × p

The magnitude of the angular momentum vector is

l = rpsin θ

where p is the magnitude of p and θ is the angle between r and p. We may write

l = rp⊥ or r⊥p

where r⊥ (= r sinθ) is the perpendicular distance of the directional line of p from the origin and p⊥(p sin θ) is the component of p in a direction perpendicular to r. We expect the angular momentum to be zero (l = 0), if the linear momentum vanishes (p = 0), if the particle is at the origin (r = 0), or if the directional line of p passes through the origin θ = $0^0$ or 180$^0$.

The physical quantities, moment of a force and angular momentum, have an important relation between them. It is the rotational analogue of the relation between force and linear momentum. For deriving the relation in the context of a single particle, we differentiate l = r × p with respect to time,

$\frac{d\mathbf{l}}{dt}=\frac{d}{dt}(\mathbf{r} \times \mathbf{p})$

Applying the product rule for differentiation to the right hand side,

$\frac{d}{dt}(\mathbf{r} \times \mathbf{p})=\frac{d\mathbf{r}}{dt}\times \mathbf{p}+\mathbf{r} \times \frac{d\mathbf{p}}{dt}$

Now, the velocity of the particle is v = dr/dt and p = mv

Because of this $\frac{d\mathbf{r}}{dt}\times \mathbf{p}=\mathbf{v} \times m \mathbf{v}=0$

as the vector product of two parallel vectors vanishes. Further, since dp/dt = F,

$\mathbf{r} \times \frac{d\mathbf{p}}{dt}= \mathbf{r} \times \mathbf{F}=$ τ

Hence $\frac{d}{dt}(\mathbf{r} \times \mathbf{p})=$ τ

Thus, the time rate of change of the angular momentum of a particle is equal to the torque acting on it. This is the rotational analogue of the equation F = dp/dt, which expresses Newton’s second law for the translational motion of a single particle.

Example 1

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

Answer

Let the particle with velocity v be at point P at some instant t. We want to calculate the angular momentum of the particle about an arbitrary point O.

Fig.1

The angular momentum is l = r × mv. Its magnitude is mvr sinθ, where θ is the angle between r and v as shown in Fig. 1. Although the particle changes position with time, the line of direction of v remains the same and hence OM = r sin θ. is a constant.

Further, the direction of l is perpendicular to the plane of r and v. It is into the page of the figure.This direction does not change with time.

Thus, l remains the same in magnitude and direction and is therefore conserved. Is there any external torque on the particle?