Uniform Circular Motion

Uniform circular motion is motion along a circular path at constant speed. The word uniform refers to the constancy of the speed, not to the velocity vector. Because the direction of the velocity changes continuously, the object is accelerated even though its speed does not change.

For a circle of radius R, the velocity is tangent to the path while the acceleration points toward the center. This inward acceleration is called the centripetal acceleration.

1 Circular position and angular position

Let a particle move on a circle of radius R centered at the origin. Its position can be described by the angular coordinate 𝜃 measured from the positive x axis:

(1)

The magnitude of the position vector is constant,

(2)

but its direction changes as the particle moves.

For uniform circular motion, the angular speed is constant:

(3)

Therefore

(4)

Angles used in the relations below are measured in radians. One full revolution corresponds to

(5)

2 Period, frequency, and angular speed

The period T is the time required for one revolution. The frequency f is the number of revolutions per unit time:

(6)

During one period the angular displacement is 2π, so

(7)

Thus

(8)

The SI unit of ω is radians per second. Since the radian is dimensionless, ω has dimensions of inverse time.

3 Arc length and tangential speed

For a circular arc, the distance traveled along the path is

(9)

Differentiating with respect to time gives

(10)

Because ds∕dt is the speed v,

(11)

The speed is constant when both R and ω are constant. The velocity vector itself is not constant because its direction rotates with the particle.

4 Geometric derivation of centripetal acceleration

Consider two nearby points on the circular path separated by a small angular displacement Δ𝜃. The velocity vectors at the two points have the same magnitude v but different directions.

For a sufficiently small angle, the velocity triangle is geometrically similar to the position triangle. Therefore

(12)

The arc traveled during the same interval is

(13)

Since the speed is v,

(14)

Hence

(15)

Substitution into the velocity change relation gives

(16)

Dividing by Δt and taking the limit as Δt → 0 gives

(17)

The change in velocity points inward in the small angle limit, so the acceleration vector is directed toward the center of the circle.

5 Equivalent forms of centripetal acceleration

Using

(18)

in

(19)

we obtain

(20)

Since

(21)

we may also write

(22)

In terms of frequency,

(23)

These forms describe the same acceleration and differ only in which quantities are most convenient to use.

6 Vector derivation

A second derivation follows directly from differentiating the position vector. For constant angular speed,

(24)

Thus

(25)

Differentiating gives

(26)

Its magnitude is

(27)

Differentiating again gives

(28)

Therefore

(29)

The minus sign shows directly that the acceleration points opposite the outward position vector and therefore toward the center.

7 Velocity and acceleration are perpendicular

For uniform circular motion,

(30)

The acceleration is therefore perpendicular to the velocity at every instant. A perpendicular acceleration changes the direction of the velocity without changing its magnitude. This is why the speed remains constant.

Equivalently, the rate of change of v² is

(31)

Hence v², and therefore the speed v, remains constant.

8 Centripetal does not name a new force

Uniform circular motion is a kinematic description. The inward acceleration must be produced by the net force in the dynamics problem, but centripetal force is not a separate additional force to place on a Free-body diagram.

Newton’s second law later gives the required inward net force magnitude as

(32)

Depending on the physical situation, this inward net force may be supplied by Tension, gravity, friction, a normal force, an electromagnetic force, or a combination of forces. The present article focuses on the kinematics of the acceleration itself.

9 Worked example 1: speed, period, and centripetal acceleration

A particle moves in a circle of radius

(33)

with constant speed

(34)

The angular speed is

(35)

The period is

(36)

The frequency is

(37)

The centripetal acceleration is

(38)

Thus

(39)

10 Worked example 2: wheel speed from revolutions per minute

A point on the rim of a wheel of radius

(40)

rotates at 120 revolutions per minute.

First convert to revolutions per second:

(41)

Therefore

(42)

The tangential speed is

(43)

The centripetal acceleration is

(44)

11 Worked example 3: position, velocity, and acceleration vectors

A particle moves counterclockwise in a circle of radius R = 2.0 mm with angular speed ω = 3.0 mrad∕s. Let 𝜃₀ = 0 at t = 0. Find r, v, and a at

(45)

The angular position is

(46)

Therefore

(47)

Using the vector velocity formula,

(48)

we obtain

(49)

Similarly,

(50)

so

(51)

The velocity is tangent to the circle and the acceleration points inward, exactly as required.

12 Worked example 4: two points on the same rigidly rotating disk

Two marked points lie at radii

(52)

on a rigid disk rotating uniformly with angular speed

(53)

Because the disk is rigid, both points have the same angular speed. Their linear speeds are

(54)

and

(55)

Thus

(56)

Their centripetal accelerations are

(57)

and

(58)

Hence

(59)

At fixed angular speed, both v and a_c are proportional to radius. The familiar quadratic dependence a_c ∝ v² applies when the radius is held fixed instead.

13 Beyond uniform circular motion

If the speed changes while the object remains on a circular path, the acceleration generally has two components. The inward normal component is still

(60)

while the tangential component is

(61)

Uniform circular motion is the special case

(62)

The full treatment of moving polar coordinate basis vectors appears later in the mechanics sequence.

14 Practice problems

1. A wheel rotates uniformly at 30 revolutions per minute. Find its frequency, period, and angular speed.
2. A particle moves in a circle of radius 2.5 m with period 4.0 s. Find its speed.
3. A CAR follows a circular path of radius 50 m at constant speed 15 m/s. Find the centripetal acceleration.
4. A particle has centripetal acceleration 20 mm∕s² at radius 0.80 m. Find its angular speed.
5. Two points on the same rigid disk are at radii R and 3R. Compare their angular speeds, linear speeds, and centripetal accelerations.
6. A particle moves counterclockwise with R = 1.5 m and ω = 2.0 rad/s. At 𝜃 = 0, write the directions and magnitudes of v and a.
7. A point travels through three complete revolutions on a circle of radius 0.40 m. What distance does it travel?
8. A rotor spins uniformly at 5.0 Hz for 12 s. How many revolutions occur, and what angular displacement is swept out in radians?
9. A satellite is idealized as moving uniformly on a circular path of radius 7.0 × 10⁶ m with period 6000 s. Find its speed and centripetal acceleration.
10. A particle moves at constant speed around a circle. Explain why its acceleration can be nonzero even though dv∕dt = 0 for the scalar speed.

15 Answer check

1. f = 0.500 mHz, T = 2.00 ms, ω = π mrad∕s ≃ 3.14 mrad∕s.
2. v = 2πR∕T = 3.93 mm∕s.
3. a_c = v²∕R = 4.50 mm∕s².
4. ω = = 5.00 mrad∕s.
5. Same ω; outer point has 3 times the speed and 3 times the centripetal acceleration.
6. v = 3.0 mm∕s in +y; a_c = 6.0 mm∕s² in −x.
7. s = 3(2πR) = 2.4π mm ≃ 7.54 mm.
8. 60 revolutions; Δ𝜃 = 120π mrad.
9. v = 2πR∕T ≃ 7.33 × 10³ mm∕s; a _c ≃ 7.67 mm∕s².
10. The velocity vector changes direction even when its magnitude is constant; acceleration is dv∕dt, not merely dv∕dt.

16 Summary

Uniform circular motion has constant speed but continuously changing velocity direction. Its central kinematic relations are

(63)

(64)

and

(65)

The velocity is tangent to the circle, the acceleration is inward, and the two vectors are perpendicular. The inward acceleration arises because the velocity direction changes, not because the speed changes.

References