Javier Junquera
Momento angular
In Chapter 9, we began by developing the mathematical form of linear momentum
and then proceeded to show how this new quantity was valuable in problem-solving.
We will follow a similar procedure for angular momentum.
Consider a particle of mass mlocated at the vector position rand moving with linear
momentum pas in Figure 11.4. In describing linear motion, we found that the net force
on the particle equals the time rate of change of its linear momentum, F!dp/dt
(see Eq. 9.3). Let us take the cross product of each side of Equation 9.3 with r, which
gives us the net torque on the particle on the left side of the equation:
Now let us add to the right-hand side the term , which is zero because dr/dt !
vand vand pare parallel. Thus,
We recognize the right-hand side of this equation as the derivative of r!p(see Equa-
tion 11.6). Therefore,
(11.9)
This looks very similar in form to Equation 9.3, F!dp/dt. This suggests that the
combination r!pshould play the same role in rotational motion that pplays in trans-
lational motion. We call this combination the angular momentum of the particle:
!
!
"
! d(r!p)
dt
!
"
! r!dp
dt "dr
dt
!p
d r
dt
!p
r!
!
F !
!
"
! r!dp
dt
!
340 CHAPTER 11 • Angular Momentum
This allows us to write Equation 11.9 as
(11.11)
which is the rotational analog of Newton’s second law, F!dp/dt. Note that torque
causes the angular momentum Lto change just as force causes linear momentum pto
change. Equation 11.11 states that the torque acting on a particle is equal to the
time rate of change of the particle’s angular momentum.
Note that Equation 11.11 is valid only if
"
and Lare measured about the same ori-
gin. (Of course, the same origin must be used in calculating all of the torques.) Fur-
thermore, the expression is valid for any origin fixed in an inertial frame.
The SI unit of angular momentum is kg ·m2/s. Note also that both the magnitude
and the direction of Ldepend on the choice of origin. Following the right-hand rule,
we see that the direction of Lis perpendicular to the plane formed by rand p. In
Figure 11.4, rand pare in the xy plane, and so Lpoints in the zdirection. Because
p!mv, the magnitude of Lis
L!mvr sin
#
(11.12)
where
#
is the angle between rand p. It follows that Lis zero when ris parallel to
p(
#
!0 or 180°). In other words, when the linear velocity of the particle is along a
line that passes through the origin, the particle has zero angular momentum with
respect to the origin. On the other hand, if ris perpendicular to p(
#
!90°), then
L!mvr.At that instant, the particle moves exactly as if it were on the rim of a wheel
rotating about the origin in a plane defined by rand p.
!
!
!
"
! dL
dt
The instantaneous angular momentum Lof a particle relative to the origin Ois
defined by the cross product of the particle’s instantaneous position vector rand its
instantaneous linear momentum p:
L"r!p(11.10)
Active Figure 11.3 As the skater
passes the pole, she grabs hold of
it. This causes her to swing around
the pole rapidly in a circular path.
Active Figure 11.4 The angular
momentum Lof a particle of mass
mand linear momentum plocated
at the vector position ris a vector
given by L!r!p. The value of L
depends on the origin about which
it is measured and is a vector per-
pendicular to both rand p.
At the Active Figures link
at http://www.pse6.com, you
can change the speed of the
skater and her distance to the
pole and watch her spin when
she grabs the pole.
At the Active Figures link
at http://www.pse6.com, you
can change the position vector
r and the momentum vector p
to see the effect on the angular
momentum vector.
O
z
L = r × p
rmp
φ
y
x
Angular momentum of a particle
