Page 105 - Introduction to Continuum Mechanics

90 Kinematics of a Continuum

i.e.,

(b)We can use either Eq. (iii) or Eq. (v) to find the acceleration.
Using Eq. (iii) and Eq. (3.4.5b), we obtain

i.e.,

Or, using Eq. (v) and Eqs. (3.4.8), we obtain

i.e.,

We note that (xiei+X2*2) = re r so that (vi) and (vii) are the same. We also note that in this
example, even though at every spatial point there is no change of velocity with time, for every
material point, there is a rate of change of velocity due to a change of direction at every point
as it moves along a circular path giving rise to a centripetal acceleration.

Example 3.4.2
Given the velocity field

(a) Find the acceleration field and (b) find the pathlinejc, = */ (X,f)
Solution, (a) With