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Finally, integrating over z , we obtain
22
ρπRh
h
C
=
=
.
(14.28)
z
12
M
4
Thus, the center of mass of the cone is given by
C
=
0, 0,
h
4
.
14.2.2 Angular Momentum and Torque
Recall that the linear momentum p of a particle having mass m moving at a ve-
locity v is given by
p . Just as angular velocity is the rotational analog of
linear velocity, there exists a quantity called angular momentum that serves as
the rotational analog of linear momentum.
Suppose that a particle of mass m is rotating about some axis with an angular
velocity of
=
m
()
()
ω and that the position of the particle is given by the function
r
t
.
()
The angular momentum
L
t
of the particle is defined to be
()
()
()
L
t
r
t
p ,
t
(14.29)
()
()
where
t
=
mt
v
is the linear momentum of the particle.
p
Differentiating both sides of Equation (14.29) gives us
()
()
()
()
.
()
L
t
=× +×
r
t
p r
t
t
p
t
(14.30)
()
()
r and
()
()
Since
r
t
=
v
t
, the vectors
t
p
t
point in the same direction, so the
()
()
cross product
r
t
×
p
t
is zero. Thus,
()
()
()
()
.
()
L
t
=× =×
r
t
p r
t
t
mt
v
(14.31)
v is equal to the net force
()
()
The vector
mt
F
t
acting on the particle, so we can
write
()
()
()
L
t
r
t
F
t
.
(14.32)
()
The quantity on the right side of Equation (14.32) is called the torque
τ
being applied to the particle:
()
()
()
τ r
t
t
F .
t
(14.33)
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