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magnitude of v in Eq. 2-57 is v ω= r P sinφ , and since r = r P sinφ , Fig.
2-19,a, then v ω= r which agrees with Eq. 2-56. As a special case, the
position vector r can be chosen for r . Here r lies in the plane of
P
motion and again the velocity of point P is
×
v = ω r 2-58
Acceleration. The acceleration of P can be expressed in terms of
its normal and tangential components. Since a = dv / dt and
t
a = v 2 / ρ , where ρ = , v ω= r and α = dω / dt we have
r
n
a = α r 2-59
t
a = ω 2 r 2-60
n
The tangential component of acceleration, Figs. 2-20,a and 2-
20,b, represents the time rate of change in the velocity’s magnitude. If
the speed of P is increasing, then a acts in the same direction as v; if
t
the speed is decreasing, a acts in the opposite direction of v; and
t
finally, if the speed is constant, a is zero.
t
The normal component of acceleration represents the time rate
of change in the velocity’s direction. The direction of a is always
n
toward O, the center of the circular path, Figs. 2-20,a and 2-20,b.
Like the velocity, the acceleration of point P can be expressed in
terms of the vector cross product. Taking the time derivative of Eq. 2-
57 we have
dv dω dr
a = = ×r + ω × P 2-61
dt dt P dt
Recalling that α = dω / dt , and using Eq. 2-57
(dr P / dt = v = ω r ) yields
×
P
a =×α r P + ω × ( ×ω r P ) 2-62
From the definition of the cross product, the first term on the
right has a magnitude a = α r P sinφ α= r and by the right-hand rule,
t
α ×r is in the direction of a Fig. 2-20,a. Likewise, the second term
P
t
has a magnitude a = ω 2 r P sinφ ω= 2 r , and applying the right-hand
n
×
rule twice, first to determine the result v P = ω r then ω r it can
×
P
P
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