9 ANGULAR MOMENTUM 9.2 Angular momentum of a point particle

×

l = r p sin (^) 
p
l
origin r
O




particle

Figure 85: Angular momentum of a point particle about the origin.

Let us differentiate Eq. (9.3) with respect to time. We obtain

dl

dt

= ̇r × p + r × p ̇. (9.5)

Note that the derivative of a vector product is formed in much the same manner
as the derivative of an ordinary product, except that the order of the various

terms is preserved. Now, we know that ̇r = v = p/m and p ̇ = f. Hence, we obtain
dl

p × p

+ r f. (9.6)

dt m

However, p × p = 0 , since the vector product of two parallel vectors is zero. Also,

r × f = τ, (9.7)

where τ is the torque acting on the particle about an axis passing through the

origin. We conclude that
dl
= τ. (9.8)
dt
Of course, this equation is analogous to Eq. (9.2), which suggests that angular

momentum, l, plays the role of linear momentum, p, in rotational dynamics.

For the special case of a particle of mass m executing a circular orbit of ra-

dius r, with instantaneous velocity v and instantaneous angular velocity ω, the
magnitude of the particle’s angular momentum is simply

l = m v r = m ω r^2. (9.9)