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Angular momentum of a point particle
Consider a particle of mass , position vector , and instantaneous velocity ,
which rotates about an axis
passing through the origin of our coordinate system. We know that the
particle's linear momentum is written

(414) 
and satisfies

(415) 
where is the force acting on the particle. Let us search for the rotational
equivalent of .
Consider the quantity

(416) 
This quantitywhich is known as angular momentumis a vector of magnitude

(417) 
where is the angle subtended between the directions of and .
The direction of is defined to be mutually perpendicular to the directions
of and , in the sense given by the righthand grip rule. In other
words, if vector rotates onto vector (through an angle less
than ), and the fingers of the righthand are aligned with this rotation, then
the thumb of the righthand indicates the direction of . See Fig. 85.
Figure 85:
Angular momentum of a point particle about the origin.

Let us differentiate Eq. (416) with respect to time. We obtain

(418) 
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
and
.
Hence, we obtain

(419) 
However,
, since the vector product of two parallel
vectors is zero. Also,

(420) 
where
is the torque acting on the particle about an axis passing through the
origin. We conclude that

(421) 
Of course, this equation is analogous to Eq. (415), which suggests that angular
momentum, , plays the role of linear momentum, , in rotational
dynamics.
For the special case of a particle of mass executing a circular orbit of radius ,
with instantaneous velocity and instantaneous angular velocity , the
magnitude of the particle's angular momentum is simply

(422) 
Next: Angular momentum of an
Up: Angular momentum
Previous: Introduction
Richard Fitzpatrick
20060202