Cylindrical and Spherical Particle Kinematics

Cartesian coordinates are often the simplest coordinates for mechanics because the basis vectors are fixed. Many three-dimensional problems, however, possess axial or spherical geometry. In those cases cylindrical or spherical coordinates can describe the motion much more naturally.

The complication is that the curvilinear basis vectors generally move with the particle. Therefore velocity and acceleration are not obtained by differentiating only the scalar coordinates. One must also differentiate the basis vectors themselves.

This article derives the complete particle velocity and acceleration formulas for cylindrical coordinates

(1)

and spherical coordinates

(2)

using the convention that 𝜃 is the polar angle measured from the positive z axis and ϕ is the azimuth measured in the xy plane.

1 Why curvilinear basis vectors matter

In a Cartesian frame,

(3)

and the unit vectors satisfy

(4)

In cylindrical and spherical coordinates, some unit vectors point in directions defined by the particle’s current angular coordinates. When those angles change, the basis rotates. Thus a derivative such as

(5)

must be evaluated with the product rule:

(6)

The second term is the source of the angular pieces that distinguish curvilinear kinematics from Cartesian kinematics.

2 Cylindrical coordinates

Cylindrical coordinates are related to Cartesian coordinates by

(7)

(8)

and

(9)

Here ρ is the perpendicular distance from the z axis, ϕ is the azimuthal angle in the xy plane, and z is the ordinary vertical Cartesian coordinate.

The cylindrical unit vectors can be written in the Cartesian basis as

(10)

(11)

and

(12)

The position vector is

(13)

3 Cylindrical basis-vector derivatives

Differentiate e_ρ with respect to ϕ:

(14)

Likewise,

(15)

Since the basis vectors depend on time only through ϕ(t),

(16)

(17)

and

(18)

4 Cylindrical velocity

Differentiate

(19)

Using the product rule,

(20)

Substitute

(21)

to obtain

(22)

Thus the physical velocity components are

(23)

Notice that dϕ∕dt itself is an angular rate, not a linear speed. The azimuthal linear speed is ρ(dϕ∕dt).

5 Cylindrical acceleration

Differentiate the cylindrical velocity:

(24)

The derivative of the first term is

(25)

The derivative of the azimuthal term is

(26)

Using

(27)

gives

(28)

Combining terms,

(29)

The radial component is

(30)

and the azimuthal component is

(31)

The term −ρ(dϕ∕dt)² is the inward centripetal contribution. The term 2(dρ∕dt)(dϕ∕dt) appears because the radial distance and the rotating basis change simultaneously.

6 Polar-coordinate motion as a cylindrical special case

If the motion remains in a plane of constant z, then

(32)

and cylindrical kinematics reduces directly to planar polar kinematics:

(33)

(34)

This is the direct three-dimensional extension of M01-10.

7 Spherical coordinates and convention

Spherical-coordinate notation is not universal, so the angular convention must be stated explicitly. In this article,

• r is radial distance from the origin;
• 𝜃 is the polar angle measured downward from the positive z axis;
• ϕ is the azimuth measured in the xy plane from the positive x axis.

The Cartesian relations are

(35)

(36)

and

(37)

The spherical basis vectors are

(38)

(39)

and

(40)

The position vector is especially simple:

(41)

8 Spherical basis-vector derivatives

Because the spherical basis depends on both 𝜃 and ϕ, its derivatives are richer than in cylindrical coordinates.

The needed partial derivatives are

(42)

(43)

(44)

(45)

(46)

and

(47)

Applying the chain rule gives the time derivatives

(48)

(49)

and

(50)

9 Spherical velocity

Differentiate

(51)

Then

(52)

Substitute the derivative of e_r:

(53)

Thus the physical velocity components are

(54)

The factor r sin 𝜃 is the perpendicular distance from the particle to the z axis, so the azimuthal speed is exactly the circular-motion form ρ(dϕ∕dt) with

(55)

10 Spherical acceleration

Start from

(56)

Differentiate each term with the product rule and then substitute the three basis-vector derivatives. Collecting coefficients of e_r, e_𝜃, and e_ϕ gives

(57)

where

(58)

(59)

and

(60)

11 Interpreting the spherical terms

The radial acceleration contains three contributions:

(61)

The first is ordinary radial acceleration. The other two are inward curvature terms associated with angular motion.

The polar component

(62)

contains polar angular acceleration, radial-polar coupling, and a geometric contribution from azimuthal motion.

The azimuthal component

(63)

contains azimuthal angular acceleration plus coupling terms from changing radial distance and changing polar angle.

These are kinematic terms produced by coordinate geometry; they do not by themselves represent new physical forces.

12 Scale factors

The physical differential displacement in cylindrical coordinates is

(64)

Therefore the cylindrical scale factors are

(65)

In spherical coordinates,

(66)

so

(67)

The velocity formulas can be read directly from these physical line elements.

13 Coordinate singularities

Curvilinear coordinates may become singular even when physical space is perfectly regular.

In cylindrical coordinates, ϕ is undefined on the z axis where

(68)

In spherical coordinates, all angles are undefined at

(69)

and the azimuth ϕ is undefined on the polar axis where

(70)

These are coordinate singularities, not physical singularities. One should avoid interpreting divergent coordinate rates near such locations without first checking the actual Cartesian motion.

14 Worked example 1: uniform helical motion in cylindrical coordinates

A particle follows

(71)

Then

(72)

The velocity is

(73)

so

(74)

Its speed is

(75)

Because d²ρ∕dt² = d²ϕ∕dt² = d²z∕dt² = 0,

(76)

so

(77)

The vertical motion does not contribute to acceleration because dz∕dt is constant.

15 Worked example 2: expanding spiral with vertical motion

A particle has

(78)

with SI units understood. Find v and a at t = 1 s.

At t = 1,

(79)

(80)

and

(81)

Therefore

(82)

For the acceleration,

(83)

and

(84)

Hence

(85)

The nonzero azimuthal acceleration occurs even though d²ϕ∕dt² = 0 because ρ is increasing while the basis rotates.

16 Worked example 3: constant-r, constant-𝜃 spherical motion

A particle moves with

(86)

Thus

(87)

and all second coordinate derivatives vanish.

The velocity is purely azimuthal:

(88)

Therefore

(89)

The acceleration components are

(90)

(91)

and

(92)

Thus

(93)

Its magnitude is

(94)

This is exactly the centripetal acceleration for circular motion about the z axis. The circle has cylindrical radius

(95)

so

(96)

17 Practice problems

1. In cylindrical coordinates, a particle has ρ = 3.0 m, dρ∕dt = 1.0 m/s, dϕ∕dt = 2.0 rad/s, and dz∕dt = −0.50 m/s. Find its velocity vector in the cylindrical basis and its speed.
2. A particle moves on a cylinder of fixed radius ρ = 4.0 m with ϕ = 0.50t² and z = 2.0t. Find v and a at t = 2.0 s.
3. Show that for ρ = R = constant, z = constant, and dϕ∕dt = ω = constant, the cylindrical acceleration reduces to a = −ω²Re _ρ.
4. A particle has ρ = t², ϕ = t, and z = 0. Evaluate a _ρ and a_ϕ at t = 2 s.
5. In spherical coordinates, r = 4.0 m, 𝜃 = 45^∘, dr∕dt = 1.0 m/s, d𝜃∕dt = 0.20 rad/s, and dϕ∕dt = 0.50 rad/s. Find v_r, v_𝜃, and v_ϕ.
6. A particle moves radially outward with r = 2+t², while 𝜃 and ϕ remain constant. Find its velocity and acceleration in the spherical basis.
7. A particle moves with r = 3.0 m, 𝜃 = 90^∘, and ϕ = 2.0t. Use the spherical formulas to find its speed and acceleration, and show that the result matches uniform circular motion in the xy plane.
8. Starting from the Cartesian expression for e_r, explicitly verify ∂e_r∕∂𝜃 = e_𝜃 and ∂e_r∕∂ϕ = sin 𝜃 e_ϕ.

18 Answer check

1. v = 1.0e_ρ + 6.0e_ϕ − 0.50e_z m/s; |v| = 6.10 m/s.
2. At t = 2 s, dϕ∕dt = 2.0 rad/s and d²ϕ∕dt² = 1.0 rad/s². Thus v = 8.0e _ϕ + 2.0e_z m/s and a = −16.0e_ρ + 4.0e_ϕ m/s².
3. Direct substitution gives a_ρ = −Rω², a _ϕ = 0, and a_z = 0.
4. At t = 2: ρ = 4, dρ∕dt = 4, d²ρ∕dt² = 2, dϕ∕dt = 1, d²ϕ∕dt² = 0. Hence a _ρ = −2 m/s² and a_ϕ = 8 m/s².
5. v_r = 1.0 m/s, v_𝜃 = 0.80 m/s, v_ϕ = 1.41 m/s.
6. v = 2te_r and a = 2e_r.
7. v = r(dϕ∕dt) = 6.0 m/s and a = −r(dϕ∕dt)²e _r = −12.0e_r m/s²; at 𝜃 = 90^∘, e _r lies in the xy plane and points radially outward.
8. Differentiation gives the stated basis identities exactly.

19 Summary

Cylindrical and spherical coordinates simplify problems with axial or radial geometry, but their basis vectors move with the particle. The central cylindrical results are

(97)

(98)

For the stated spherical convention,

(99)

with acceleration components

(100)

(101)

and

(102)

The safest derivation strategy is always the same: write the position vector, determine how the local basis vectors change, and then differentiate carefully with the product and chain rules.

References