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0593_C03_fm Page 63 Monday, May 6, 2002 2:03 PM

Kinematics of a Particle 63

It can be shown that the velocity of P relative to Q in R may be expressed as:

R P Q = ωω ×
/
V r (3.4.10)
Assuming this to be correct, and if the velocity of Q in R is also known as 5n – 8n + 3n 3
2
1
ft/sec, find the velocity of P relative to Q in R and the absolute velocity of P in R.
Solution: From Eqs. (3.4.8) to (3.4.10), the relative velocity is:
R P Q n )
/
V =− ( 2 n + 3 n − 4 n ) ×(2 n − 3 n + 7
1 2 3 1 2 3
(3.4.11)
= 33 n + 6 n ft sec
1 2
Then, from Eq. (3.4.6), the velocity of P is:
P
V = (5 n − 8 n + 3 n ) +(33 n + 6 n )
1 2 3 1 2
(3.4.12)
= 38 n + 2 n + 3 n
1 2 3
Observe that even though the distance between P and Q is constant, the relative velocity
of P and Q in R is not zero. Observe further, however, that relative to an observer in B
the relative velocity is zero. That is,

B P Q
/
V = 0 (3.4.13)

3.5 Differentiation of Rotating Unit Vectors

In Section 3.2, we observed that we can calculate the derivative of a vector in a reference
frame R by first expressing the vector in terms of unit vectors fixed in R and then by
differentiating the components. In this procedure, the vector derivative is expressed in
terms of scalar derivatives. Although in principle this procedure always works, it may
not always be the most convenient way to obtain the derivative. Consider, for example,
the velocity of a point P moving along a fixed curve C, as discussed in Section 3.3. The
P
velocity vector V is tangent to C as in Figure 3.5.1. Then, if n is a unit vector parallel to
P
P
V , we may express V as:
V = V n = v n (3.5.1)
P
P
P
where v is defined as V to simplify the notation; v is often called the speed of P.
Suppose now that we want to calculate the acceleration of P. By differentiating in Eq.
(3.5.1), we have:
a = ( dv dt) n v d dt (3.5.2)
+
P
R
n
The first term involves a scalar derivative and presents no difficulty. The second term
involves the derivative of a unit vector n that is not fixed in R. As with any other vector,