The physical interpretation applied to the acceleration vector is as follows. Ob-
serve that the vectors
ˆer= cos θeˆ 1 + sin θˆe 2 and ˆeθ=−sin θˆe 1 + cos θˆe 2
are unit vectors and that these vectors are perpendicular to one another since they
satisfy the dot product relation eˆr·ˆeθ= 0. The vectors ˆer and ˆeθ represent unit
vectors in polar coordinates and are illustrated in the figure 6-13. The acceleration
vector can then be expressed in the form
a =−rω^2 ˆer+rα ˆeθ=a r+a t
where a r=−rω^2 ˆer is called the radial component of the acceleration or centripetal
acceleration and a t =rα ˆeθ is called the tangential component of the acceleration.
These components have the magnitudes
|a r|=rω^2 and |at|=rα
Note that if ωis a constant, then α= 0 and consequently the tangential component of
the acceleration will always be zero leaving only the radial component of acceleration.
Example 6-21. Transverse and Radial Components of Velocity
Consider the motion of a particle which is
described in polar coordinates by an equation
of the form r =f(θ), where θ is measured in
radians. Select a point Pwith coordinates (r, θ)
on the curve and construct the radius vector r
from the origin to the point P. Construct the
tangent to the curve at the point P and define
the angle ψbetween the radius vector and the
tangent. Label a fixed point on the curve, say
the fixed point x 0 where the curve intersects the x-axis. Let sdenote the arc length
along the curve measured from x 0 to the point P. The velocity of the particle Pas
it moves along the curve is given by the change in distance with respect to time t
and can be written v=
ds
dt.
