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0593_C03_fm Page 67 Monday, May 6, 2002 2:03 PM

Kinematics of a Particle 67

Hence, by differentiating, the velocity of P becomes:

V = d p dt = d( n r ) dt
P
r
= r d n dt = n r × n d dt) (3.7.3)
r( θ
r z
rd dt)
= ( θ n
z
where the fourth equality is obtained from Eq. (3.5.7). By differentiating again, we find
the acceleration of P to be:
d r d dt) ]
a = d V dt = ( [ θ n θ dt
P
P
[
d dt d )
= rd r dθ 2 +( θ n dt ]
2
θ
[ d dt) 2 z ] (3.7.4)
= rd r dθ 2 +( θ n × n θ
2
[ 2 2 n ) d dt) 2 r]
= r d ( θ dt θ −( θ n
where again we have used Eq. (3.5.7) to obtain dn /dt.
θ
P
Observe in Eq. (3.7.3) that V is tangent to C as expected. Alternatively, observe in Eq.
P
(3.7.4) that a has two components: one tangent to C and the other normal to C. The
tangential component of a arises from a change in the speed of P along C. The normal
P
component of a arises from a change in the direction of P as it moves along C.
P
To see these results more clearly let ω, v, and α be defined as:
θ
ω = d dt, v = rd θ dt = r ω, and α = d ω dt (3.7.5)

P
In terms of v, ω, and α, V and a are then:
P
V = v n = rω n θ (3.7.6)
P
θ
and

a = rα n − rω 2 n = ( dv dt) n −( v r) n r (3.7.7)
2
P
θ
θ
r
These expressions will be useful to us throughout the text. We consider a generalization
of these concepts in the next section.
Example 3.7.1: Movement of a Particle on the Rim of a Rotor
Consider a rotor R with radius r rotating about its axis with an angular speed and an
angular acceleration as shown in Figure 3.7.2. Let P be a particle on the rim of R which
is located by the angular coordinate θ as shown. If n and n are radial and tangential unit
θ
r
vectors, as shown, express the velocity and acceleration of P in terms of n and n .
θ
r
Solution: The angular speed is simply the time derivative of θ. Hence, we have:
˙
ω = θ and α = ω ˙ = θ ˙ (3.7.8)