where the last simplification was obtained using the vector identity given by equation
(6.32) and the result ω·r = 0. The above results are derived under the assumption
that the angular velocity ω=dθdt was a constant.
Figure 6-13. Particle moving in circular motion.
In contrast, let us examine what happens if the angular velocity is not a constant.
The position vector to a particle undergoing circular motion is given by
r =rcos θˆe 1 +rsin θˆe 2
where θ=θ(t)is the angular displacement as a function of time. The velocity of the
particle is given by
v =dr
dt
=−rsin θdθ
dt
ˆe 1 +rcos θdθ
dt
ˆe 2
Let dθ
dt
=ω(t)denote the angular speed which is a function of time tand express the
velocity as
v =−rω (t) sin θˆe 1 +rω (t) cosθˆe 2
The acceleration is obtained by taking the derivative of the velocity to obtain
a =dvdt =−r
[
ωcos θdθdt +dωdt sin θ
]
ˆe 1 +r
[
−ω(t) sin θdθdt +dωdt cos θ
]
ˆe 2
a =−rω^2 cos θˆe 1 −rα sin θˆe 1 −rω^2 sin θˆe 2 +rα cos θ eˆ 2
where α=α(t) =
dω
dt is the angular acceleration. The acceleration vector can be
broken up into two components by writing
a =−rω^2 [cos θˆe 1 + sin θˆe 2 ] + rα [−sin θˆe 1 + cos θˆe 2 ]
