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 ( .25a) The magnitude of the angular momentum vector is sin l r p ( .26a) where p is the magnitude of p and θ is the angle between r and p . We may write l r p ⊥ or r p ⊥ ( .26b) where r ⊥ (= r sin θ ) is the perpendicular distance of the directional line of p from the origin and sin ) ⊥ = is the component of p in a direction perpendicular to r .
We expect the angular momentum to be zero ( l = ), if the linear momentum vanishes ( p = ), if the particle is at the origin ( r = ), or if the directional line of p passes through the origin θ = or . 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, d ( l Applying the product rule for differentiation to the right hand side, Now, the velocity of the particle is v = d r / dt and p = m v Because of this d , t × as the vector product of two parallel vectors vanishes.
Further, since d p