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13.4 Torque and Angular Momentum as Vectors 413

side of the axis, and when we add the angular-momentum vectors contributed by these If a body is symmetric about
z
two particles (or any other pair of particles), the resultant lies along the axis of rotation axis of rotation, resultant
angular momentum will be
(see Fig. 13.19).
along axis of rotation.
Since Newton’s Second Law for translational motion states that the rate of change
of the momentum equals the force, the analogy between the equations for transla- L 2 L 1
tional and rotational motion suggests that the rate of change of the angular momen-
tum should equal the torque. It is easy to verify this for the case of a single particle.With
the usual rule for the differentiation of a product,
r r 2 2 r 1 y
O O
d d
L (r p)
dt dt
(13.43)
dr dp x
p r
dt dt
The first term on the right side is FIGURE 13.19 For a rotating symmetric
body, the angular momentum is always
dr along the axis of rotation.
p v (mv) m(v v) 0 (13.44)
dt
This is zero because the cross product of a vector with itself is always zero. According
to Newton’s Second Law, the second term on the right side of Eq. (13.43) is
dp
r r F (13.45)
dt
where F is the force acting on the particle. Therefore, Eq. (13.43) becomes

d L
r F (13.46)
dt
In the case of a rigid body, the angular momentum is the sum of all the angular
momenta of the particles in the body, and the rate of change of this total angular
momentum can be shown to equal the net external torque:

d L equation of rotational motion for
(13.47)
dt vector angular momentum

This equation for the rate of change of the angular momentum of a rigid body is
analogous to the equation dp dt F for the rate of change of the translational momen-
tum of a particle.
To compare the vector equation (13.47) with our earlier equation I ,we
must focus our attention on the component of the angular momentum along the
axis of rotation, that is, the z axis. Figure 13.20 shows an arbitrary rigid body z
Angular momentum Component of
rotating about a fixed axis, which coincides with the z axis. As in Example
makes an angle with angular momentum
11, the angular-momentum vector of this body makes an angle with the axis z axis. along z axis is L .
z
of rotation. However, as we discussed in Example 11, the z component of
L
the angular momentum of each particle in the rotating body is simply equal to its
moment of inertia about the z axis multiplied by the angular velocity [see Eq. (13.42)].
L z
Hence, when we sum the contributions of all the particles in the rotating body, we
find that the z component of the net angular momentum of the entire rotating body y
O
equals the net moment of inertia of the entire body multiplied by the angular veloc-
ity. This establishes that the equation

L I (13.48) x
z
FIGURE 13.20 A body rotating about the
is of general validity. z axis.