since ˆerchanges with time, but ˆez=ˆe 3 remains constant. From equation (9.27) one

can calculate the derivatives

dˆer

dt =−sin θ

dθ

dt ˆe^1 + cos θ

dθ

dt ˆe^2 =

dθ

dt ˆeθ

dˆeθ

dt

=−cos θdθ

dt

ˆe 1 −sin θdθ

dt

ˆe 2 =−dθ

dt

ˆeθ

dˆez

dt

=0

(9 .30)

as these derivatives will be useful in simplifying any derivatives with respect to time

of vectors in cylindrical coordinates. The equations (9.30) allows one to obtain the

result

v =

d
r

dt =

dr

dt ˆer+r

dθ

dt ˆeθ+

dz

dt eˆz (9 .31)

which can also be represented in the form

v = ̇rˆer+rθ ̇ˆeθ+ ̇zeˆz

where the dot notation is used to represent time differentiation. Here vr= ̇ris the

radial component of the velocity ,vθ=rθ ̇is the azimuthal component of velocity and

vz= ̇zis the vertical component of the velocity.

The acceleration in cylindrical coordinates is obtained by differentiating the

velocity. Differentiate the equation (9.31) with respect to time tand show

a =d
v

dt

=d

(^2)
r
dt^2
=d
dt
[
dr
dt
ˆer+rdθ
dt
ˆeθ+dz
dt
ˆez
]
=dr
dt
dˆer
dt
+d
(^2) r
dt^2
eˆr+rdθ
dt
dˆeθ
dt
+rd
(^2) θ
dt^2
ˆeθ+dr
dt
dθ
dt
ˆeθ+d
(^2) z
dt^2
ˆez
=dr
dt
dθ
dt
eˆθ+d
(^2) r
dt^2
ˆer−r
(
dθ
dt
) 2
ˆer+rd
(^2) θ
dt^2
eˆθ+dr
dt
dθ
dt
ˆeθ+d
(^2) z
dt^2
ˆez

(
d^2 r
dt^2
−r
(
dθ
dt
) 2 )
ˆer+
(
rd
(^2) θ
dt^2

• 2 dr
  dt
  dθ
  dt
  )
  ˆeθ+d
  (^2) z
  dt^2
  ˆez
  
  a =( ̈r−r(θ ̇)^2 )ˆer+ (rθ ̈+ 2 ̇rθ ̇)ˆeθ+ ̈zˆez
  (9 .32)

where ̇ = dtd and ̈= d

2

dt^2 are shorthand notations for the first and second derivatives

with respect to time t. In calculating the derivatives in equation (9.32) make note

that the results from equation (9.30) have been employed.