Equating like components produces the result that
r
dθ
ds = sin ψ and
dr
ds = cos ψ
The derivative of the position vector r =reˆrwith respect to time ttakes on the form
dr
dt
=rdˆer
dt
+dr
dt
ˆer=rdˆer
dθ
dθ
dt
+dr
dt
ˆer=rdθ
dt
ˆeθ+dr
dt
ˆer=vrˆer+vθeˆθ=v
where
vr=dr
dt
=vcos ψ is the radial component of the velocity
vθ=rdθ
dt
=vsin ψ is the transverse component of the velocity
ˆer= cosθˆe 1 + sin θˆe 2 is a unit vector in the radial direction
ˆeθ=−sin θˆe 1 + cos θˆe 2 is a unit vector in the transverse direction
dr
dt =v =vcos ψ
ˆer+vsin ψeˆθ is alternative form for the velocity vector
Note also that if dθdt =ωis the angular velocity, then one can write vθ=rω.
Example 6-22. Angular Momentum
Recall that a moment causes a rotational motion. Let us investigate what hap-
pens when Newton’s second law is applied to rotational motion. The angular mo-
mentum of a particle is defined as the moment of the linear momentum. Let H denote
the angular momentum; mv , the linear momentum; and r, the position vector of the
particle, then by definition the moment of the linear momentum is expressed
H =r ×(mv ) = r ×
(
mdr
dt
)
. (6 .58)
Differentiating this relation produces
dH
dt
=r ×
(
m
d^2 r
dt^2
)
+
dr
dt
×
(
m
dr
dt
)
.
Observe that the second cross product term is zero because the vectors are parallel.
Also note that by using Newton’s second law, involving a constant mass, one can
write
F =ma =mdv
dt =m
d^2 r
dt^2.