Begin2.DVI

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

d
r

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

d
r

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; m
v , the linear momentum; and
r, the position vector of the

particle, then by definition the moment of the linear momentum is expressed

H
 =
r ×(m
v ) = 
r ×

(

md
r

dt

)

. (6 .58)

Differentiating this relation produces

dH


dt

=
r ×

(

m

d^2 
r

dt^2

)

+

d
r

dt

×

(

m

d
r

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
 =m
a =md
v

dt =m

d^2 
r

dt^2.