GRE Physics Companion: Polar Coordinate Particle Kinematics

This companion is designed for rapid review after M01-10. The fastest approach is usually to identify which polar quantities are zero before substituting numbers.

1 Fast triage

Memorize the structure, not isolated terms:

|--------------| v-=--˙rer +-r˙𝜃e𝜃-

(1)

and

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

(2)

Then eliminate terms using the motion description. For a fixed radius, the first and second time derivatives of r are zero. For constant angular speed, the angular acceleration is zero. For pure radial motion, the angular rate is zero.

Figure 1. GRE speed triage for polar kinematics. Substitute only after identifying which radial and angular derivatives vanish.

2 Common traps

The polar unit vectors are not fixed Cartesian vectors. Even with constant r, velocity is generally nonzero because e_r changes direction.

The transverse quantity equal to radius times angular rate is a linear speed; angular rate alone is not.

The radial centripetal term points inward.

The mixed transverse term can survive even when the angular speed is constant.

Figure 2. Common polar kinematics traps: forgetting the moving basis, omitting the factor of radius in transverse speed, losing the inward sign, and dropping the mixed radial angular acceleration term.

3 Worked GRE example 1: fixed radius

A particle moves at fixed radius r = 3 m with angular speed 4 rad/s and angular acceleration 2 rad/s². Find the polar acceleration components.

Because r is fixed,

˙r = ¨r = 0.

(3)

Therefore

a = − r˙𝜃2 = − 3(4)2 = − 48 mm ∕s2, r

(4)

and

a = r𝜃¨= 3(2 ) = 6 mm ∕s2. 𝜃

(5)

Hence

|------------------------| a-=--− 48er-+-6e-𝜃 mm-∕s2.-

(6)

4 Worked GRE example 2: radial sliding at constant angular speed

At an instant, r = 2 m, the radial speed is 1.5 m/s, the second time derivative of r is zero, and the angular speed is constant at ω = 3 rad/s. Find the transverse acceleration.

Since 𝜃 = 0,

a 𝜃 = 2 ˙r˙𝜃 = 2(1.5 )(3) = 9 mm ∕s2.

(7)

Thus

|--------------| | 2 | a-𝜃 =-9-mm-∕s-.-

(8)

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

5 GRE speed questions

A particle moves on a circle of radius R with constant angular speed ω. Its radial acceleration is (A) 0 (B) +Rω² (C) −Rω² (D) −2Rω².
At an instant, r = 2 m, the radial speed is 1 m/s, the angular speed is 3 rad/s, and both the radial second derivative and angular acceleration are zero. The transverse acceleration is (A) 0 (B) 3 (C) 6 (D) 18 m/s².

6 Answers and rationales

C. Fixed radius uniform circular motion gives a_r = −Rω².
C. The mixed transverse term gives a_𝜃 = 2(1)(3) = 6 m/s².

References

[1] PhysicsLibrary, M01-10, Polar Coordinate Particle Kinematics.

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

"GRE Physics Companion: Polar Coordinate Particle Kinematics" is owned by bloftin.
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