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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 quantity--which is known as angular momentum--is 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 right-hand grip rule. In other words, if vector rotates onto vector (through an angle less than ), and the fingers of the right-hand are aligned with this rotation, then the thumb of the right-hand indicates the direction of . See Fig. 85.

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 2006-02-02