PHYSICS · LESSON 08 OF 9

Circular motion

An object moving in a circle at constant speed is accelerating towards the centre, a = v²/r. Some real force (tension, friction, gravity, normal force) must supply the net inward force mv²/r.

What this lesson explains

Cars rounding bends, centrifuges, rotating machinery, satellites and roller coaster loops all move on curved paths. Designing a safe road bend (friction and banking), a centrifuge (separation force) or a rotating shaft (stresses) needs the relationship between speed, radius and inward force.

Before you begin

Radians

An angle in radians is arc length divided by radius: θ = s/r. One revolution = 2π rad = 360°. rpm (revolutions per minute) × 2π/60 gives rad/s.

Newton’s second law along a chosen axis

Take the positive axis pointing towards the centre: ΣF_{toward centre} = m·v² / r.

The idea, made visible

Velocity is a vector. In uniform circular motion the speed is constant, but the direction of the velocity changes continuously (it is always tangent to the circle). A change of velocity is an acceleration, so the object is accelerating even though its speed does not change.

The acceleration points towards the centre (centripetal means “centre-seeking”) and has magnitude a_c = v² / r. Faster motion or a tighter circle means more acceleration: doubling the speed quadruples it.

The velocity (blue) is tangent to the path; the acceleration (red) points to the centre. Both change direction continuously while their sizes stay constant.

By Newton’s second law, a net force towards the centre is required: ΣF_in = mv² / r. This is not a new kind of force. It is the net result of real forces: the tension in a string whirling a ball, friction between tyres and road, gravity on a satellite, the normal force from a banked track or loop. On a free-body diagram you draw the real forces only, never a separate “centripetal force”.

If the inward force disappears (a string breaks), the object does not fly outward along the radius; it continues in a straight line along the tangent, as the first law says. The outward “push” you feel in a turning car is your body’s inertia, not a real outward force.

Circular motion is often described with angles: the angular speed ω = Δθ/Δt in rad/s, with v = ωr, and the period T (time for one revolution), with v = 2πr / T and ω = 2π / T.

What provides the inward force?
situation	force(s) pointing to the centre	equation
ball on a string (horizontal circle, ignoring gravity sag)	tension T	T = mv²/r
car on a flat bend	static friction f_s	f_s = mv²/r ≤ μ_smg
car on a banked bend (no friction)	horizontal part of normal force	N sin θ = mv²/r, N cos θ = mg
top of a vertical loop	weight + normal force (both down)	mg + N = mv²/r
satellite in circular orbit	gravity	GMm/r² = mv²/r

Key terms

Uniform circular motion: Motion on a circle at constant speed.
Centripetal acceleration: a_c = v² / r = ω²r, directed towards the centre.
Period T and frequency f: T is the time for one revolution; f = 1/T revolutions per second (Hz).
Angular speed ω: ω = Δθ / Δt = 2π / T (rad/s); v = ωr.
Banked curve: A road tilted at angle θ so that the horizontal component of the normal force helps provide the inward force.

The formulas and what they mean

Symbols, meanings and units
Symbol	Meaning	Unit
r	radius of the circular path	m
v	speed	m/s
a_c	centripetal acceleration	m/s²
T	period	s
ω	angular speed	rad/s
μ_s	coefficient of static friction	—

Centripetal acceleration

a_c = v² / r = ω²r

Conditions and limits: Uniform circular motion; directed towards the centre. For non-uniform circular motion this is the radial part of the acceleration.

Net inward force

ΣF_in = mv² / r

Conditions and limits: Sum of the components of the real forces along the radius, towards the centre positive.

Speed, period, angular speed

v = 2πr / T = ωr; ω = 2π / T = 2πf

Conditions and limits: Constant speed.

Maximum speed on a flat curve

v_max = √μ_sgr

Conditions and limits: Level road, friction alone provides the inward force; the tyres do not slip (static friction).

Ideal banking speed

tan θ = v² / rg

Conditions and limits: Banked curve with no friction needed at this speed.

Why it works: Where a = v²/r comes from

In a short time Δt the object moves through a small angle Δθ. Its velocity keeps size v but turns through the same angle Δθ.

The change in velocity has magnitude |Δv| ≈ vΔθ.
For a small angle, the arc swept by the tip of the velocity arrow is v × Δθ.
The angle turned is Δθ = vΔt / r.
Arc length travelled (vΔt) divided by radius.
a = ∣Δv∣ / Δt = vΔθ / Δt = v·v / r = v² / r.
Divide by Δt.
Δv points towards the centre (perpendicular to v).
As Δt → 0 the change is at right angles to the velocity, i.e. inward.

Check units: (m/s)²/m = m/s².

A first worked example

Circular-motion problems

Identify the circle: its centre, its radius and the plane it lies in.
Draw the free-body diagram with real forces only.
Take one axis pointing towards the centre (and usually one vertical).
Write ΣF_{toward centre} = mv²/r and, if needed, ΣF_vertical = 0.
Solve for the unknown; for friction on a flat curve use f_s ≤ μ_sN to find the limit.

Centripetal acceleration

Problem. An object moves at 6 m/s around a circle of radius 3 m. Find its acceleration.

a_c = v² / r = 36 / 3 = 12 m/s².
Directed towards the centre.

Result: 12 m/s² towards the centre.

What it means: More than g (9.8 m/s²), even though the speed is modest, because the circle is tight.

A different case

Net inward force

Problem. A 2 kg object travels in a circle of radius 4 m at 4 m/s. Find the net inward force.

F = mv² / r = 2 × 16 / 4 = 8 N.
Newton’s second law towards the centre.

Result: 8 N towards the centre.

What it means: This is the net force; it could be supplied by tension, friction or any other real force.

More worked cases

Each case below uses a different skill. Every step and result is shown.

Maximum speed on a flat bend

Problem. A car rounds a flat bend of radius 50 m. μ_s between tyres and road is 0.80. What is the maximum speed without skidding?

Inward force: static friction f_s ≤ μ_sN = μ_smg.
Flat road: N = mg.
At the limit: μ_smg = mv² / r ⇒ v = √μ_sgr.
Mass cancels.
v = √0.80 × 9.8 × 50 = √392 = 19.8 m/s.
Substitute.

Result: About 19.8 m/s (≈ 71 km/h).

What it means: On a wet road with μₛ = 0.40 the limit falls to 14 m/s (about 50 km/h).

Banked curve

Problem. At what angle should a curve of radius 100 m be banked so that cars at 20 m/s need no friction?

Vertical: N cos θ = mg. Horizontal (inward): N sin θ = mv² / r.
Normal force is perpendicular to the tilted road.
Divide: tan θ = v² / rg = 400 / 100 × 9.8 = 0.408.
N and m cancel.
θ = arctan 0.408 = 22.2°.
Solve for the angle.

Result: About 22°.

What it means: Cars faster than 20 m/s need friction pointing down the slope; slower cars need it pointing up the slope.

Top of a vertical loop

Problem. A roller coaster car goes over the inside top of a loop of radius 10 m. What is the minimum speed at the top for the car to stay on the track?

At the top both weight and normal force point down (towards the centre): mg + N = mv² / r.
Inside of the loop.
The minimum speed is when the track just stops pushing: N = 0, so mg = mv² / r.
Contact is about to be lost.
v_min = √gr = √9.8 × 10 = 9.90 m/s.
Solve.

Result: About 9.9 m/s.

What it means: Slower than this, gravity alone would pull the car inward more than needed and it would leave the circular path.

Period and angular speed

Problem. A centrifuge spins at 3000 rpm with the sample 0.10 m from the axis. Find ω, v and the acceleration as a multiple of g.

ω = 3000 × 2π / 60 = 314 rad/s.
Convert rpm to rad/s.
v = ωr = 314 × 0.10 = 31.4 m/s.
Linear speed.
a = ω²r = 314² × 0.10 = 9.87 × 10³ m/s² ≈ 1000g.
Centripetal acceleration.

Result: ω ≈ 314 rad/s, v ≈ 31.4 m/s, a ≈ 1000g.

What it means: That large effective gravity is what separates particles quickly.

Common misunderstandings

Misunderstanding: “Constant speed means no acceleration.”
Correct idea: The direction changes, so the velocity changes: a = v²/r inward.
Misunderstanding: Adding a separate “centripetal force” to the free-body diagram.
Correct idea: mv²/r is the required net inward force, supplied by real forces.
Misunderstanding: “There is an outward centrifugal force.”
Correct idea: In an inertial frame there is none; the outward feeling is inertia. If released, objects move along the tangent.
Misunderstanding: Using the diameter instead of the radius.
Correct idea: r is measured from the centre of the circle.
Misunderstanding: Using rpm directly as ω.
Correct idea: Convert: ω (rad/s) = rpm × 2π/60.

Going further (optional): Angular kinematics with constant angular acceleration (check whether your outline includes it)

A wheel speeding up or slowing down has an angular acceleration α = Δω/Δt (rad/s²). If α is constant, the rotational equations mirror the linear ones: ω = ω₀ + αt and θ = ω₀t + 1 / 2αt².

Example: a wheel starts at 2 rad/s and has α = 3 rad/s² for 4 s. Then ω = 2 + 3(4) = 14 rad/s and θ = 2(4) + 1 / 2(3)(16) = 32 rad (about 5.1 turns). A point 0.50 m from the axle then moves at v = rω = 0.50 × 14 = 7 m/s. Every point on the wheel shares ω, but points farther out move faster.

Keep in mind

a_c = v² / r = ω²rCentripetal acceleration
ΣF_in = mv² / rNet inward force
v = 2πr / T = ωr; ω = 2π / T = 2πfSpeed, period, angular speed
v_max = √μ_sgrMaximum speed on a flat curve
tan θ = v² / rgIdeal banking speed

Scope of this lesson

Uniform circular motion and simple vertical circles. Angular kinematics is an optional section below; torque is an optional section of Newton laws and force diagrams. Rotational dynamics (moment of inertia) and gravitation beyond the circular-orbit equation are not covered.

Next: Laboratory graphs and evidence. Plot measured data, draw the best straight line, and read physics from its slope and intercept — with units and a realistic uncertainty.

Original study text. Sources and credits.

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