Uniform Circular Motion

Uniform circular motion (UCM) is a motion whose trajectory is a circle and whose speed magnitude is constant. It is nevertheless an accelerated motion: the acceleration vector, directed toward the center, reflects the constant change in the direction of the velocity.

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How to measure it in class

FizziQ’s gyroscope measures the angular velocity ω and the accelerometer measures the centripetal acceleration: you can therefore experimentally verify the relation a = ω²R.

Steps:

Firmly attach the smartphone flat on a rotating platform (turntable, salad spinner, office chair) at a known distance R from the axis, measured at the center of the phone.
Select the gyroscope and linear acceleration simultaneously in FizziQ, then start recording.
Set it rotating and let it spin at the most constant rate possible for about ten seconds.
Read off the plateau the angular velocity ω (in rad/s) given by the gyroscope around the vertical axis, and the acceleration measured in the plane of rotation.
Compare the measured acceleration with ω²R: the two values should agree to within a few percent.
Repeat at two or three different rotation speeds and plot a as a function of ω²: the points should line up on a straight line of slope R.

Scientific activities on this topic

Several experiments easily carried out with a smartphone or tablet make it possible to measure the centripetal acceleration experienced by an object following a uniform circular trajectory:

Could an astronaut survive if placed in a salad spinner?
Verify the relation between rotation speed and centripetal acceleration by simultaneously recording accelerometer and gyroscope data
Measuring the centripetal acceleration of a pendulum at its lowest point
Athletics: hammer release speed during training
Centrifuge: a = ω²r - verify that the acceleration is proportional to the square of the angular velocity.
Centripetal acceleration: a = ω²R - directly measure the acceleration directed toward the center.
Orbital period of the Moon - determine the revolution period by observation.
Geostationary orbit: Meteosat - relate the 24 h period and the altitude of a weather satellite.

Learn more

Uniform does not mean without acceleration

In uniform circular motion, the velocity vector is tangent to the circle of the trajectory, and therefore perpendicular to the radius throughout the motion. The magnitude of this vector is constant since the motion is uniform. However, it is not constant in direction. Yet acceleration measures the change in the velocity vector, not only in its magnitude: UCM is therefore an accelerated motion, with a permanently non-zero acceleration, even though the speed does not change in value. This is the most frequent error in this chapter. A motion with truly zero acceleration is a uniform linear motion, and it alone.

Centripetal acceleration

The acceleration vector points constantly toward the center of the circle, and its value is:

a = v*v/R

v is the speed (m.s-1)

R is the radius of the circular trajectory (m)

a is the acceleration (m.s-2)

Angular velocity and period

The angular velocity is a measure of the rotation speed of the moving object and is represented by the Greek letter ω (omega). It is defined as the change in angle θ per unit of time t, that is ω = θ/t. In UCM, the angular velocity is constant.

The centripetal acceleration is directed toward the center of the trajectory and has the value v²/R. According to Newton’s second law, it requires the existence of a net force also directed toward the center: tension of a string, friction force of the tires in a bend, gravitational attraction for a satellite. Without this force, the moving object would go off in a straight line along the tangent, in accordance with the principle of inertia.

The rotation period is the time needed for an object in circular motion to complete one full revolution. It is represented by T and is determined by the formula T = 2πr/v, where π is a mathematical constant (approximately equal to 3.14). The rotation frequency, written f, is the inverse of the rotation period, that is f = 1/T.

UCM as a model for orbits

Kepler’s laws describe the motion of the planets around the Sun. The first states that orbits are ellipses with the Sun at one focus: so it is not exactly a uniform circular motion. The second, the law of areas, indicates that the segment connecting the planet to the Sun sweeps equal areas in equal times, which implies that the planet moves faster when it is closer to the Sun. The third relates the revolution period to the semi-major axis of the orbit. UCM is the limiting case where the ellipse becomes a circle: it is an excellent approximation for the Earth, whose orbital eccentricity is only 0.017, and an exact model for artificial satellites placed in circular orbit.

Orders of magnitude

Ferris wheel: ω ≈ 0.03 rad/s, a ≈ 0.03 m/s². Merry-go-round: a ≈ 1 m/s². Salad spinner: ω ≈ 60 rad/s, R = 10 cm, a ≈ 360 m/s² (37 g). Washing machine drum during spin: up to 400 g. Laboratory centrifuge: several thousand g. Satellite in low orbit: v ≈ 7,700 m/s, period 90 min. Geostationary satellite: altitude 35,786 km, period 24 h. Earth around the Sun: v ≈ 29,800 m/s, a ≈ 0.006 m/s².

Formula

The centripetal acceleration, directed toward the center of the circle:

a = v² / R = ω²·R

The linear speed follows from the angular velocity:

v = ω·R

The rotation period and frequency:

T = 2π·R / v = 2π / ω and f = 1 / T

The centripetal force needed to maintain the motion, according to Newton’s second law:

F = m·v² / R = m·ω²·R

where:

a: centripetal acceleration (m/s²)
v: magnitude of the velocity, constant (m/s)
R: radius of the circular trajectory (m)
ω: angular velocity (rad/s)
T: rotation period (s)
f: rotation frequency (Hz)
m: mass of the moving object (kg)
F: value of the net force, directed toward the center (N)

Application examples

A car taking a bend of 50 m radius at 20 m/s (72 km/h) experiences a centripetal acceleration of 20²/50 = 8.0 m/s², that is 0.8 g: it is the friction of the tires that provides it.
A satellite in low orbit at 400 km altitude travels its orbit at 7,700 m/s and completes one revolution in about 90 minutes; its centripetal acceleration, 8.7 m/s², is exactly the acceleration of gravity at that altitude.
A salad spinner of 10 cm radius rotating at 600 revolutions per minute (ω = 63 rad/s) imposes 63² × 0.10 ≈ 400 m/s², about 40 g: the water is expelled through the holes.
A geostationary satellite must have a period of 24 h to remain above the same point on the equator, which fixes its altitude at 35,786 km.
A hammer thrower spins the 7.26 kg implement at the end of a 1.2 m wire: at 25 m/s, the tension of the wire is 7.26 × 25²/1.2 ≈ 3,800 N.
On a merry-go-round, a child 3 m from the axis rotating at 0.6 rad/s experiences 0.6² × 3 ≈ 1.1 m/s², a clearly perceptible value.

FAQ

Q: Is there an acceleration in uniform circular motion? A: Yes, always, and this is the key point. The word “uniform” only concerns the magnitude of the velocity. The direction of the velocity vector changes constantly, so the velocity vector varies, so the acceleration is non-zero. It equals v²/R and points toward the center.

Q: Does the centrifugal force exist? A: Not in an inertial reference frame. The only real force is directed toward the center. The “centrifugal force” is an inertial force, which only appears if you choose to work in the rotating, non-inertial reference frame. The passenger of a car in a bend is pressed against the door because their body tends to continue in a straight line.

Q: What happens if the centripetal force suddenly disappears? A: The moving object goes off in a straight line, along the tangent to the circle at the point where it was, and not radially outward. This is exactly the principle of inertia. This is what happens when a string breaks or a car loses its grip.

Q: What is the difference between the speed v and the angular velocity ω? A: v is a linear speed, in m/s, tangent to the trajectory. ω is a rotation rate, in rad/s, identical for all points of the same rotating solid. They are related by v = ωR: on a merry-go-round, everyone has the same ω, but the outer seats have a larger speed v.

Q: Is the motion of a point on a bicycle wheel uniform circular? A: It depends on the reference frame. In the frame’s reference frame, yes, if the bicycle rolls at constant speed. In the ground reference frame, this point traces a cycloid, a curve that is not a circle. It is a good example to show that you cannot characterize a motion without having specified the reference frame.

Centripetal Acceleration - Period - Frequency - Inertial Reference Frame - Uniform Linear Motion - Kinematics