Polar Coordinate Particle Kinematics

Cartesian coordinates are often the simplest language for particle motion, but many mechanical systems have a natural center or axis. Orbital motion, particles moving on disks, sliders on rotating arms, central force motion, and many planar mechanisms are described more directly by Plane polar coordinates.

A particle in the plane is located by two coordinates,

r = r(t), 𝜃 = 𝜃(t),

(1)

where r is the distance from the origin and 𝜃 is the angular coordinate. The important difference from Cartesian coordinates is that the polar unit vectors rotate as 𝜃 changes. That moving basis is responsible for the extra terms in polar coordinate velocity and acceleration.

The main results developed in this article are

|------------˙---| -v-=-r˙er +-r𝜃e𝜃-|

(2)

and

|-----------2------------------| a-=--(¨r −-r𝜃˙)er-+-(r¨𝜃-+-2˙r𝜃˙)e𝜃.

(3)

These equations are kinematic identities. They follow from geometry and differentiation; no force law has yet been used.

1 The polar basis

In a fixed Cartesian basis e_x,e_y, define the radial and transverse unit vectors by

|----------------------| |er = cos𝜃 ex + sin 𝜃ey | -----------------------

(4)

and

|------------------------| e 𝜃 = − sin 𝜃ex + cos 𝜃ey.| --------------------------

(5)

The vector e_r points outward from the origin through the particle. The vector e_𝜃 is perpendicular to e_r and points in the direction of increasing 𝜃.

Figure 1. Plane polar coordinates and the local basis. The radial and transverse unit vectors are attached to the particle and change direction as the angular coordinate changes.

The particle position vector has the particularly simple form

|--------| -r =-rer.-

(6)

The simplicity of the position formula hides an important complication: e_r is generally time dependent.

2 Why the unit vectors change

Differentiate the Cartesian expression for e_r:

der ----= − sin 𝜃 ˙𝜃ex + cos𝜃 ˙𝜃ey. dt

(7)

The quantity in parentheses is exactly e_𝜃, so

|----------| -˙er-=-˙𝜃-e𝜃.|

(8)

Similarly,

de-𝜃= − cos𝜃 ˙𝜃e − sin𝜃 ˙𝜃e , dt x y

(9)

which gives

|------------| -˙e𝜃-=-−-˙𝜃er.-|

(10)

The same result has a geometric interpretation. During a small angular change Δ𝜃, the unit radial vector rotates through the same angle. Its change is approximately transverse and has magnitude Δ𝜃:

Δe ≃ e Δ 𝜃. r 𝜃

(11)

Dividing by Δt and taking the limit gives the radial basis derivative stated above.

Figure 2. Geometric origin of the polar basis derivative. A small angular change rotates the radial unit vector toward the transverse direction; dividing that change by time produces the basis rotation term used in the velocity and acceleration derivations.

This basis rotation is a coordinate effect, not evidence that the physical reference frame itself is rotating. Polar coordinates may be used perfectly well inside an inertial Cartesian frame.

3 Differential displacement

Because

r = rer,

(12)

a differential change is

dr = er dr + r der.

(13)

For a small angular change,

der = e𝜃 d 𝜃,

(14)

|--------------------| |dr = dr er + rd𝜃 e𝜃.| ---------------------

(15)

The two orthogonal displacement components therefore have magnitudes dr and r d𝜃. The corresponding line element is

|------------------| | 2 2 2 2 | -ds-=--dr-+--r-d𝜃-.-

(16)

This already anticipates the two terms that appear in the velocity.

4 Velocity in polar coordinates

Differentiate the position vector directly:

d v = --(rer). dt

(17)

By the product rule,

v = ˙rer + r˙er.

(18)

Using the radial basis derivative derived above,

|----------------| |v = ˙r e + r𝜃˙e .| --------r------𝜃--

(19)

Thus the radial and transverse velocity components are

|--------------------| | ˙ | -vr-=-˙r,----v-𝜃 =-r-𝜃.

(20)

The speed is

|----∘-----------| | 2 2 ˙2| -v-=---r˙-+-r-𝜃-.-

(21)

Figure 3. Polar velocity decomposes into a radial component and a transverse component, aligned with the local polar basis directions.

The factor r multiplying the angular rate is essential. Angular speed has units of rad/s, while the tangential velocity component must have units of m/s.

5 Acceleration in polar coordinates

Differentiate the velocity:

( ) a = d- ˙re + r ˙𝜃e . dt r 𝜃

(22)

Differentiate the first term:

d -- (r˙er ) = ¨rer + r˙𝜃˙e 𝜃. dt

(23)

Differentiate the second term:

d --(r˙𝜃e𝜃) = (˙r˙𝜃 + r¨𝜃)e𝜃 + r˙𝜃˙e𝜃. dt

(24)

Since

e˙𝜃 = − 𝜃˙er,

(25)

we obtain

-d (r ˙𝜃e𝜃) = (r˙˙𝜃 + r𝜃¨)e 𝜃 − r𝜃˙2er. dt

(26)

Collect the radial and transverse parts:

|-----------2------------------| a-=--(¨r −-r𝜃˙)er-+-(r¨𝜃-+-2˙r𝜃˙)e𝜃.

(27)

Therefore

|------------| -ar =-¨r −-r˙𝜃2-

(28)

and

|--------------| |a = r¨𝜃 + 2˙r˙𝜃.| --𝜃-------------

(29)

Figure 4. Radial and transverse acceleration components. Each component contains terms caused by changes in coordinate magnitudes and by rotation of the local polar basis.

6 Physical meaning of the acceleration terms

The four pieces have distinct kinematic meanings.

The term

¨rer

(30)

is the direct radial acceleration caused by changing radial speed.

The term

− r˙𝜃2e r

(31)

is inward and remains even when the radius and angular speed are constant. It is the familiar centripetal acceleration.

The term

r ¨𝜃e𝜃

(32)

is the transverse acceleration associated with angular acceleration.

Finally,

2˙r𝜃˙e 𝜃

(33)

appears when radial motion and angular motion occur simultaneously. It results from differentiating both the transverse speed and the rotating basis. It is sometimes described as a Coriolis like kinematic term. In the present derivation, however, we are simply using polar coordinates in an inertial frame; no fictitious force has been introduced.

7 Important special cases

Pure radial motion

If 𝜃 is constant,

˙𝜃 = ¨𝜃 = 0,

(34)

v = ˙rer, a = ¨rer.

(35)

Circular motion

If r = R is constant,

˙r = ¨r = 0,

(36)

and therefore

v = R ˙𝜃e𝜃,

(37)

a = − R𝜃˙2er + R ¨𝜃e𝜃.

(38)

For uniform circular motion, the angular speed is the constant ω and the angular acceleration is zero, giving

|----------------------| | 2 v2 | |a = − Rω er = − --er.| ------------------R-----

(39)

This recovers the result derived geometrically in M01-08.

Radial sliding on a uniformly rotating arm

If the angular speed ω is constant while r changes,

a = (¨r − rω2 )er + 2 ˙rωe𝜃.

(40)

The transverse term is nonzero even though the angular speed is constant.

8 Worked example 1: a spiral trajectory

A particle moves according to

1 1 r(t) = 2 + -t2 mm, 𝜃(t) = -t2 mrad. 2 4

(41)

Find the velocity and acceleration at t = 2 s.

The required derivatives are

˙r = t, ¨r = 1,

(42)

1 1 𝜃˙= 2t, 𝜃¨= 2.

(43)

At t = 2 s,

r = 4.0 mm, ˙r = 2.0 mm ∕s, ˙𝜃 = 1.0 mrad ∕s.

(44)

Hence

v = 2er + 4e 𝜃 mm ∕s,

(45)

and

|-----√------------------| |v | = 20 = 4.47 mm ∕s.| --------------------------

(46)

The acceleration components are

ar = 1 − (4)(1)2 = − 3.0 mm ∕s2,

(47)

a 𝜃 = (4)(0.5) + 2(2)(1) = 6.0 mm ∕s2.

(48)

Therefore

|------------------------| |a = − 3er + 6e𝜃 mm ∕s2.| -------------------------

(49)

9 Worked example 2: slider on a rotating arm

A bead slides outward on a radial arm according to

r (t) = 1.0 + 0.20t mm,

(50)

while the arm rotates at constant angular speed

˙ 𝜃 = 2.0 mrad ∕s.

(51)

Find v and a at t = 3 s.

At that instant,

r = 1.6 mm, ˙r = 0.20 mm ∕s, ¨r = 0,

(52)

with 𝜃 = 0. Thus

|---------------------------| v = 0.20er + 3.20e 𝜃 mm ∕s. | ----------------------------

(53)

For the acceleration,

ar = 0 − (1.6)(2)2 = − 6.4 mm ∕s2,

(54)

and

a = 0 + 2(0.20 )(2) = 0.80 mm ∕s2. 𝜃

(55)

Therefore

|----------------------------| | 2 | -a =-−-6.4er-+-0.80e𝜃-mm--∕s-.-

(56)

The transverse acceleration is present even though the angular speed is constant because the radial distance is changing.

10 Worked example 3: uniform circular motion as a special case

A particle moves on a circle of radius

R = 5.0 mm

(57)

with

𝜃(t) = 0.80t.

(58)

Then

˙ ¨ r˙= r¨= 0, 𝜃 = 0.80 mrad ∕s, 𝜃 = 0.

(59)

The velocity is

|-----------------| v-=--4.0e-𝜃 mm-∕s.--

(60)

The acceleration is

|------------------2-| a-=--− 3.20er-mm-∕s-.-

(61)

The period is

T = -2π- = 7.85 ms. 0.80

(62)

Thus the general polar coordinate formula reduces exactly to uniform circular motion when the radius and angular speed are constant.

11 Connection to angular momentum

Polar kinematics also prepares the geometry used in central force mechanics. Since

r = re r

(63)

and

v = r˙er + r𝜃˙e 𝜃,

(64)

the specific angular momentum is

2 r × v = r ˙𝜃ez.

(65)

For a particle of mass m,

L = mr2 ˙𝜃e . z

(66)

The conservation or evolution of this quantity is a dynamical question, but the r² times angular rate structure is already visible from kinematics.

12 Practice problems

Starting from e_r = cos 𝜃e_x + sin 𝜃e_y, differentiate with respect to time and show that the derivative of e_r points in the transverse direction with magnitude equal to the angular rate.
A particle has r = 3.0 m, radial speed 2.0 m/s, and angular speed 4.0 rad/s. Find v_r, v_𝜃, and the speed.
At one instant a particle has r = 3.0 m, radial speed 2.0 m/s, radial acceleration contribution −1.0 m/s², angular speed 4.0 rad/s, and angular acceleration 0.50 rad/s². Find a_r and a_𝜃.
A particle moves with 𝜃 constant and r(t) = 1 + 3t − 0.5t² m. Find its velocity and acceleration at t = 2 s.
A particle moves on a circle of radius 2.5 m with constant angular speed 3.0 rad/s. Find its speed and acceleration vector in polar components.
A bead moves on an arm rotating at constant ω = 5.0 rad/s. At an instant r = 0.80 m, the radial speed is −0.30 m/s, and the second time derivative of r is zero. Find a_r and a_𝜃.
For r(t) = 2t m and 𝜃(t) = 0.50t² rad, find v and a at t = 1 s.
Show directly from ds² = dr²+r²d𝜃² that dividing by dt² produces the polar coordinate speed formula derived in the text.

13 Answer check

Differentiate term by term; the resulting direction is e_𝜃, multiplied by the angular rate.
v_r = 2.0 m/s, v_𝜃 = 12.0 m/s, v = = 12.17 m/s.
a_r = −49.0 m/s²; a_𝜃 = 17.5 m/s².
At t = 2 s the radial speed is 1.0 m/s and the radial acceleration is −1.0 m/s², so v = 1.0e_r m/s and a = −1.0e_r m/s².
v = 7.5 m/s; a = −22.5e_r m/s².
a_r = −20.0 m/s²; a_𝜃 = −3.0 m/s².
At t = 1 s: r = 2 m, radial speed 2 m/s, radial second derivative zero, angular speed 1 rad/s, and angular acceleration 1 rad/s²; v = 2e_r +2e_𝜃 m/s and a = −2e_r +6e_𝜃 m/s².
Divide the line element relation by dt² and take the positive square root.

14 Summary

For plane polar coordinates,

r = rer,

(67)

with rotating basis relations

e˙r = 𝜃˙e 𝜃, ˙e𝜃 = − ˙𝜃er.

(68)

These lead to

|--------------| v = ˙re + r˙𝜃e | -------r------𝜃-

(69)

and

|------------------------------| | ˙2 ¨ ˙ | a-=--(¨r −-r𝜃-)er-+-(r𝜃-+-2˙r𝜃)e𝜃.

(70)

The additional terms compared with Cartesian formulas arise because the local polar basis changes direction with time. This is the essential idea that generalizes to cylindrical, spherical, and other curvilinear coordinate systems.

References

[1] J. R. Taylor, Classical Mechanics, University Science Books, 2005.

[2] J. B. Marion and S. T. Thornton, Classical Dynamics of Particles and Systems, 5th ed., Brooks/Cole, 2004.

[3] H. Goldstein, C. Poole, and J. Safko, Classical Mechanics, 3rd ed., Addison-Wesley, 2002.

[4] PhysicsLibrary, M00-05, Coordinate Systems for Mechanics.

[5] PhysicsLibrary, M01-08, Uniform Circular Motion.

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