B. Dynamics · Dynamics: Forces and Motion in a Plane

Uniform Circular Motion: Period, Frequency and Centripetal Acceleration

B3.3 B2.1 A1.12About 70 minutes

Learning goals

I can describe uniform circular motion using radius, period, frequency and speed, and convert between rpm, hertz and seconds.
I can explain why an object moving in a circle at constant speed is accelerating toward the centre.
I can derive a = v²/r using similar triangles, and then a = 4π²r/T² and a = 4π²rf².
I can calculate centripetal acceleration in everyday and astronomical situations.

Words to know

uniform circular motion
Motion in a circle at a constant speed.

period (T)
The time for one complete revolution, in seconds.

frequency (f)
The number of revolutions per second, in hertz (Hz). f = 1 ÷ T. Revolutions per minute (rpm) ÷ 60 gives hertz.

tangential velocity
The instantaneous velocity of an object in circular motion. It is tangent to the circle, perpendicular to the radius.

centripetal acceleration (ac)
The acceleration of an object in uniform circular motion. It points toward the centre and has magnitude v²/r. "Centripetal" means "centre-seeking".

Spinning salad, spinning planet

A salad spinner whirls lettuce around at a steady rate and flings the water off. Earth spins once a day, carrying you around a huge circle. The Moon circles Earth about once a month. In all three cases the speed is constant, but the direction changes every instant, so the velocity changes. In Vectors in Two Dimensions you saw that a car turning a corner at constant speed has an acceleration pointing into the turn. In this lesson you will find exactly how big that acceleration is for motion in a circle, and derive the equations yourself.

Describing circular motion

Period T: the time for one revolution (s).
Frequency f: revolutions per second (Hz). f = 1 ÷ T. To convert rpm to Hz, divide by 60.
Speed: in one period the object travels one circumference, 2πr, so v = 2πr ÷ T = 2πrf.

The velocity at every instant is tangent to the circle. If the string breaks or the track ends, the object leaves along that tangent line, not outward along the radius.

In uniform circular motion, the velocity is always tangent to the circle and the acceleration always points toward the centre. The two are perpendicular, so the acceleration changes only the direction of the velocity, never its size.

Deriving a = v²/r

Consider an object moving at constant speed v around a circle of radius r. In a short time Δt it moves from point 1 to point 2, travelling an arc Δs = vΔt, while the radius turns through a small angle Δθ.

Position triangle: the two radii (each r) and the chord between points 1 and 2 form an isosceles triangle with apex angle Δθ. For a small angle, the chord is almost equal to the arc, Δs.
Velocity triangle: v₁ and v₂ are each perpendicular to their radius, so the angle between them is also Δθ. Drawn tail to tail, v₁, v₂ and Δv = v₂ − v₁ form an isosceles triangle with sides v, v and |Δv|.
The two triangles have the same apex angle, so they are similar: |Δv| ÷ v = Δs ÷ r.
So |Δv| = vΔs ÷ r = v(vΔt) ÷ r. Divide by Δt: a = |Δv| ÷ Δt = v² ÷ r.
As Δt shrinks to zero, Δv becomes perpendicular to v, pointing along the radius toward the centre.

Substituting v = 2πr ÷ T gives a = (2πr ÷ T)² ÷ r, so a = 4π²r ÷ T², and with T = 1 ÷ f, a = 4π²rf².

Left: in time Δt the object moves through angle Δθ. Right: the velocity vectors, drawn tail to tail, make a triangle with the same angle Δθ. The two isosceles triangles are similar, which gives |Δv| ÷ v = Δs ÷ r. (The angle is exaggerated to make it visible.)

The three forms

ac = v² ÷ r = 4π²r ÷ T² = 4π²rf², always directed toward the centre. For a fixed speed, a tighter circle (smaller r) means a larger acceleration. For a fixed period (everything on one spinning disc), a larger r means a larger acceleration.

Worked example

Example 1: The salad spinner

The basket of a salad spinner has a radius of 12 cm and spins at 300 rpm. Find (a) the frequency and period, (b) the speed of the lettuce at the edge, and (c) its centripetal acceleration, in m/s² and as a multiple of g.

(a) f = 300 ÷ 60 = 5.0 Hz; T = 1 ÷ 5.0 = 0.20 s.
(b) v = 2πrf = 2π(0.12)(5.0) = 3.77 m/s.
(c) a = v² ÷ r = 3.77² ÷ 0.12 = 118 m/s². Check with a = 4π²rf² = 4π²(0.12)(25) = 118 m/s² ✓.
118 ÷ 9.8 = 12, so a ≈ 12g.

(a) 5.0 Hz, 0.20 s (b) 3.8 m/s (c) about 1.2 × 10² m/s², or 12g. Water drops cling to lettuce with a force much smaller than 12 times their weight, so they fly off through the holes, along tangent lines.

Worked example

Example 2: Riding the spinning Earth

Earth's equatorial radius is 6.38 × 10⁶ m, and it rotates once in about 24 h. Find the speed and centripetal acceleration of a person standing on the equator.

T = 24 × 3600 = 86 400 s.
v = 2πr ÷ T = 2π(6.38 × 10⁶) ÷ 86 400 = 464 m/s.
a = v² ÷ r = 464² ÷ (6.38 × 10⁶) = 0.034 m/s².

About 464 m/s (roughly 1670 km/h) but only 0.034 m/s², about 0.3% of g. That is why the ground can be treated as an inertial frame for most everyday problems, as you saw in Inertial and Non-inertial Frames of Reference.

Worked example

Example 3: Changing one variable

A car takes a curve of radius 50 m at 20 m/s. Find its centripetal acceleration. What happens to it if the car doubles its speed? If the curve's radius is doubled at the original speed?

a = v² ÷ r = 20² ÷ 50 = 8.0 m/s².
a ∝ v²: doubling v multiplies a by 4, giving 32 m/s².
a ∝ 1 ÷ r: doubling r halves a, giving 4.0 m/s².

8.0 m/s²; 32 m/s² at double speed (four times as much); 4.0 m/s² on the wider curve. This is why speed matters so much on curves.

Just for fun

Why did the physics student get dizzy studying circular motion?

The questions kept going around in circles, and every answer pointed straight back to the centre.

Hands-on activity

Measure a spinning turntable

You need: something that spins steadily (a lazy Susan, a record player, a bicycle wheel with the bike turned upside down, or a desk chair), masking tape, a ruler or measuring tape, and a stopwatch.

Put a tape mark near the edge and another halfway to the centre. Measure each radius.
Spin it as steadily as you can. Count the revolutions in 20 s (start counting at zero). Find f and T. Repeat 3 times.
Calculate v and ac for each mark. Which mark has the larger speed? The larger acceleration? Does that match a = 4π²rf²?
Tangent test (with a grown-up handling the scissors): cut a gap of about one quarter out of the rim of a paper plate. Lay the plate right side up on the table and roll a marble around the inside of the rim. Predict where the marble will go when it reaches the gap, then watch. Sketch its path. (Keep marbles away from young children.)

Your turn

A record turns at 45 rpm. Find its frequency and period.

A child sits 4.0 m from the centre of a carousel that turns once every 8.0 s. Find the child's speed and centripetal acceleration.

A car rounds a curve of radius 100 m at 25 m/s. Find its centripetal acceleration.

By what factor does the centripetal acceleration change if (a) the speed triples at the same radius, (b) the period halves at the same radius?

A ball on a string moves counterclockwise in a horizontal circle. At the instant it is due north of the centre, in which directions do its velocity and acceleration point?

The Moon orbits Earth at a radius of 3.84 × 10⁸ m with a period of 27.3 days. Find its centripetal acceleration.

More practice: warm-up

What is the frequency of an object with a period of 0.25 s?

Which way does centripetal acceleration point?

Convert 120 rpm to hertz.

In uniform circular motion, what stays constant: speed, velocity, or both?

Write the speed of an object in terms of r and T.

More practice: on your own

A Ferris wheel of radius 20 m turns once every 60 s. Find a rider's speed and centripetal acceleration.

The International Space Station orbits at a radius of about 6.78 × 10⁶ m with a period of about 92 min. Find its speed and centripetal acceleration.

A rotating space station of radius 1.0 km is to give a centripetal acceleration of 9.8 m/s² at its rim (artificial gravity). Find the period and the rate in rpm.

Two coins sit on a turntable, one at 5.0 cm and one at 10 cm from the centre. Compare their periods, speeds and centripetal accelerations.

The tip of a fan blade is 0.25 m from the axis and the fan turns at 1200 rpm. Find the tip's speed and acceleration.

Challenge

Starting from a = v² ÷ r, derive a = 4π²rf².

An object moves at 10 m/s around a circle, turning through 60° in 0.50 s. In Vectors in Two Dimensions you found its average acceleration over that interval was 20 m/s². Find the radius and the instantaneous centripetal acceleration, and explain why they differ.

A laboratory centrifuge must produce 1000g at a radius of 10 cm. Find the required speed and rate in rpm.

Check your understanding

Quiz

An object moves in a circle at constant speed. Its acceleration is:

A wheel turns at 600 rpm. Its period is:

If the speed of an object in a circle of fixed radius doubles, its centripetal acceleration:

In the derivation of a = v²/r, the key step uses:

A stone whirled on a string is released when it is moving north. Seen from above, it then moves:

A point 0.50 m from the axis of a wheel with period 2.0 s has a centripetal acceleration of about:

For parents and teachers

Big idea: An object moving in a circle at constant speed is accelerating toward the centre, with a = v²/r = 4π²r/T² = 4π²rf². Students derive these equations from similar triangles (B3.3) and use the terms period, frequency and centripetal (B2.1).

How to help: Ask your teen to explain, without equations, why something moving at a steady speed can be accelerating. Then ask what would happen if the "pull to the centre" suddenly stopped (it goes off along the tangent).

Common difficulties:

Believing the acceleration points outward (the outward feeling is the fictitious force from the previous lesson).
Forgetting to convert rpm to Hz, or minutes to seconds.
Thinking objects released from a circle fly straight outward.
Using diameter instead of radius.

Safety: a grown-up should cut the paper plate; keep marbles away from young children.