Taylor Classical Mechanics - Problem 1.
47 Page 1 of 5
Problem 1.47
Let the position of a point P in three dimensions be given by the vector r = (x, y, z) in
rectangular (or Cartesian) coordinates. The same position can be specified by cylindrical polar
coordinates, ρ, ϕ, z, which are defined as follows: Let P ′ denote the projection of P onto the xy
plane; that is, P ′ has Cartesian coordinates (x, y, 0). Then ρ and ϕ are defined as the
two-dimensional polar coordinates of P ′ in the xy plane, while z is the third Cartesian coordinate,
unchanged. (a) Make a sketch to illustrate the three cylindrical coordinates. Give expressions for
ρ, ϕ, z in terms of the Cartesian coordinates x, y, z. Explain in words what ρ is (“ρ is the distance
of P from ”). There are many variants in notation. For instance, some people use r
instead of ρ. Explain why this use of r is unfortunate. (b) Describe the three unit vectors ρ̂, ϕ̂, ẑ
and write the expansion of the position vector r in terms of these unit vectors. (c) Differentiate
your last answer twice to find the cylindrical components of the acceleration a = r̈ of the particle.
To do this, you will need to know the time derivatives of ρ̂ and ϕ̂. You could get these from the
corresponding two-dimensional results (1.42) and (1.46), or you could derive them directly as in
Problem 1.48.
Solution
Part (a)
Below is a sketch that illustrates the three cylindrical coordinates (ρ, ϕ, z).
The relationships between the cylindrical and rectangular coordinates are derived in Problem 1.42.
p
x2 + y 2 = ρ2 → ρ = x2 + y 2
y
tan−1 if x and y are positive (Quadrant I)
x
−1 y
π + tan if x is negative and y is positive (Quadrant II)
y x
tan ϕ = ⇒ ϕ= .
x −1 y
π + tan if x and y are negative (Quadrant III)
x
−1 y
tan if x is positive and y is negative (Quadrant IV)
x
z=z → z=z
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Taylor Classical Mechanics - Problem 1.47 Page 2 of 5
ρ is the distance of (x, y, z) from the z-axis—the perpendicular distance, that is. ϕ is the angle
measured counterclockwise from the x-axis in the xy-plane. z is the vertical height from the
xy-plane. The problem with using r for ρ (as in Problem 1.42) is that r is commonly used in
physics texts to represent the distance from the origin to (x, y, z).
p
r = |r| = x2 + y 2 + z 2
p p
In calculus texts it’s the other way around: ρ = x2 + y 2 + z 2 and r = x2 + y 2 . One can tell
from context what meaning r has.
Part (b)
The unit vectors in cylindrical coordinates are illustrated below.
ρ̂ points radially outward from the z-axis; ϕ̂ is perpendicular to both ρ̂ and ẑ, pointing in the
direction of increasing ϕ; and ẑ points in the direction of the z-axis.
ρ
ρ̂ =
|ρ|
x x̂ + y ŷ + 0 ẑ
=p
x2 + y 2 + 0 2
x y
=p x̂ + p ŷ + 0 ẑ
x2 + y 2 x2 + y 2
x y
= x̂ + ŷ + 0 ẑ
ρ ρ
= cos ϕ x̂ + sin ϕ ŷ + 0 ẑ
ϕ̂ = ẑ × ρ̂
= ẑ × (cos ϕ x̂ + sin ϕ ŷ + 0 ẑ)
= cos ϕ (ẑ × x̂) + sin ϕ (ẑ × ŷ) + 0 (ẑ × ẑ)
= cos ϕ (ŷ) + sin ϕ (−x̂) + 0 (0)
= − sin ϕ x̂ + cos ϕ ŷ + 0 ẑ
ẑ = ẑ
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Taylor Classical Mechanics - Problem 1.47 Page 3 of 5
In terms of these cylindrical unit vectors,
r = x x̂ + y ŷ + z ẑ
!
p x y
= x2 + y 2 p x̂ + p ŷ + 0 ẑ + z ẑ
x2 + y 2 x2 + y 2
= ρ ρ̂ + z ẑ.
Part (c)
The aim here is to differentiate r with respect to t twice in order to obtain r̈ = d2 r/dt2 . Find the
first derivative.
dr
ṙ =
dt
d
= (ρ ρ̂ + z ẑ)
dt
d d
= (ρ ρ̂) + (z ẑ)
dt dt
dρ dρ̂ dz dẑ
= ρ̂ + ρ + ẑ + z
dt dt dt dt
dρ d dz dẑ
= ρ̂ + ρ (cos ϕ x̂ + sin ϕ ŷ + 0 ẑ) + ẑ + z
dt dt dt dt
dρ d d dz dẑ
= ρ̂ + ρ (cos ϕ x̂) + (sin ϕ ŷ) + ẑ + z
dt dt dt dt dt
dρ d dx̂ d dŷ dz dẑ
= ρ̂ + ρ (cos ϕ) x̂ + cos ϕ + (sin ϕ) ŷ + sin ϕ + ẑ + z
dt dt dt dt dt dt dt
dρ dϕ dx̂ dϕ dŷ dz dẑ
= ρ̂ + ρ − sin ϕ · x̂ + cos ϕ + cos ϕ · ŷ + sin ϕ + ẑ + z
dt dt dt
|{z} dt dt
|{z} dt dt
|{z}
=0 =0 =0
The derivative of any Cartesian unit vector with respect to time is zero.
dρ dϕ dϕ dz
ṙ = ρ̂ + ρ − sin ϕ · x̂ + cos ϕ · ŷ + ẑ
dt dt dt dt
dρ dϕ dz
= ρ̂ + ρ (− sin ϕ x̂ + cos ϕ ŷ) + ẑ
dt dt dt
dρ dϕ dz
= ρ̂ + ρ ϕ̂ + ẑ
dt dt dt
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Taylor Classical Mechanics - Problem 1.47 Page 4 of 5
Find the second derivative.
dṙ
r̈ =
dt
d dρ dϕ dz
= ρ̂ + ρ ϕ̂ + ẑ
dt dt dt dt
d dρ d dϕ d dz
= ρ̂ + ρ ϕ̂ + ẑ
dt dt dt dt dt dt
d dρ dρ dρ̂ dρ dϕ d dϕ dϕ dϕ̂ d dz dz dẑ
= ρ̂ + + ϕ̂ + ρ ϕ̂ + ρ + ẑ +
dt dt dt dt dt dt dt dt dt dt dt dt dt dt
d2 ρ dρ dρ̂ dρ dϕ d2 ϕ dϕ dϕ̂ d2 z dz dẑ
= 2
ρ̂ + + ϕ̂ + ρ 2
ϕ̂ + ρ + 2 ẑ +
dt dt dt dt dt dt dt dt dt dt |{z}
dt
=0
d2 ρ dρ d dρ dϕ d2 ϕ dϕ d d2 z
= ρ̂ + (cos ϕ x̂ + sin ϕ ŷ + 0 ẑ) + ϕ̂ + ρ ϕ̂ + ρ (− sin ϕ x̂ + cos ϕ ŷ + 0 ẑ) + ẑ
dt2 dt dt dt dt dt2 dt dt dt2
d2 ρ d2 ϕ d2 z
dρ d d dρ dϕ dϕ d d
= 2 ρ̂ + (cos ϕ x̂) + (sin ϕ ŷ) + ϕ̂ + ρ 2 ϕ̂ + ρ (− sin ϕ x̂) + (cos ϕ ŷ) + 2 ẑ
dt dt dt dt dt dt dt dt dt dt dt
d2 ρ
dρ d dx̂ d dŷ dρ dϕ
= 2 ρ̂ + (cos ϕ) x̂ + cos ϕ + (sin ϕ) ŷ + sin ϕ + ϕ̂
dt dt dt dt dt dt dt dt
d2 ϕ d2 z
dϕ d dx̂ d dŷ
+ ρ 2 ϕ̂ + ρ (− sin ϕ) x̂ − sin ϕ + (cos ϕ) ŷ + cos ϕ + 2 ẑ
dt dt dt dt dt dt dt
=0 =0
z}|{ z}|{
d2 ρ
dρ dϕ dx̂ dϕ dŷ dρ dϕ
= 2 ρ̂ + − sin ϕ · x̂ + cos ϕ + cos ϕ · ŷ + sin ϕ + ϕ̂
dt dt dt dt dt dt dt dt
d2 ϕ d2 z
dϕ dϕ dx̂ dϕ dŷ
+ ρ 2 ϕ̂ + ρ − cos ϕ · x̂ − sin ϕ + − sin ϕ · ŷ + cos ϕ + 2 ẑ
dt dt dt dt
|{z} dt dt
|{z} dt
=0 =0
The derivative of any Cartesian unit vector with respect to time is zero.
d2 ρ
dρ dϕ dϕ dρ dϕ
r̈ = 2 ρ̂ + − sin ϕ · x̂ + cos ϕ · ŷ + ϕ̂
dt dt dt dt dt dt
d2 ϕ d2 z
dϕ dϕ dϕ
+ ρ 2 ϕ̂ + ρ − cos ϕ · x̂ + − sin ϕ · ŷ + 2 ẑ
dt dt dt dt dt
d2 ρ dρ dϕ dρ dϕ
= 2
ρ̂ + (− sin ϕ x̂ + cos ϕ ŷ) + ϕ̂
dt dt dt dt dt
2
d2 ϕ dϕ d2 z
+ ρ 2 ϕ̂ − ρ (cos ϕ x̂ + sin ϕ ŷ) + 2 ẑ
dt dt dt
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Taylor Classical Mechanics - Problem 1.47 Page 5 of 5
The vectors in parentheses are unit vectors in cylindrical coordinates.
2
d2 ρ dρ dϕ dρ dϕ d2 ϕ dϕ d2 z
r̈ = 2 ρ̂ + ϕ̂ + ϕ̂ + ρ 2 ϕ̂ − ρ ρ̂ + 2 ẑ
dt dt dt dt dt dt dt dt
" 2 #
d2 ρ
2
d2 z
dϕ d ϕ dρ dϕ
= − ρ ρ̂ + ρ + 2 ϕ̂ + ẑ
dt2 dt dt2 dt dt dt2
Therefore, since a = r̈, the components of acceleration in cylindrical coordinates are
2ρ
2
d dϕ
aρ = −ρ
dt 2 dt
d2 ϕ
dρ dϕ .
aϕ = ρ 2 + 2
dt dt dt
2
a = d z
z
dt2
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