Mathematics of Physics and Engineering

Newton's Laws of Motion and Gravitation 41

Recall that a force F acting on the point mass m has a torque, or
moment, about O equal to

T 0 = rxF.

Applying the rule (1.3.6), page 26, to formula (2.1.4) we find

dLQ

(2.1.5)

dt

m(rxr + rxr) = mrxf.

If the frame O is inertial, then r — F/m and

dLc

dt

= r x F = T 0. (2.1.6)

Relation (2.1.6) describes the rotational motion just as (2.1.3) describes the
translational motion.
As an example, consider the SIMPLE RIGID PENDULUM, which is a
massless thin rigid rod of length £ connected to a point mass m at one
end. The other end of the rod is connected to a frictionless pin-joint at a
point O, which is a zero-diameter bearing that permits rotation in a fixed
plane. We select a cartesian coordinate system (i, j, K) with center at O
and i, j fixed in the plane of the rotation (Figure 2.1.1), and assume that
the corresponding frame is inertial.

Fig. 2.1.1 Simple Rigid Pendulum

The motion of m is a circular rotation in the (i, j) plane, and is best

described using polar coordinates (r,6); see page 35. The rigidity as-

sumption implies r(t) = £ for all t. Denote by 0 = 6{t) the angle from

% to the rod at time t. Let r(t) be the position vector of the point

mass in the frame O. Then r(t) = £r(0(t)), and by (1.3.24) on page 35,

r — £80. Hence, the angular momentum of the point mass about O is