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Circular
Motion

Master the physics of objects moving in circles: from centripetal acceleration to orbital dynamics and everything in between.

AP Physics 1

What Is Circular Motion?

When an object travels along a circular path at any speed, it undergoes circular motion. Even if the speed is constant, the direction is always changing: that means the velocity is always changing, and the object is always accelerating.

Core Vocabulary

Period \(T\)
Time for one complete revolution, measured in seconds.
Frequency \(f\)
Revolutions per second: \(f = \frac{1}{T}\), measured in Hz.
Radius \(r\)
Distance from the center of the circle to the object.

Uniform vs Non-Uniform

Uniform Circular Motion
Constant speed, changing direction. Acceleration points radially inward (centripetal only).
Non-Uniform Circular Motion
Speed changes along the path. Acceleration has both a centripetal and a tangential component.

Centripetal Acceleration

\(r\)
\(v\)
\(a_c\)
 center
$$a_c = \frac{v^2}{r}$$

Centripetal Force

By Newton's second law, if there is a centripetal acceleration, there must be a net force directed toward the center. This net inward force is called the centripetal force:

Key Equations

1
Speed: \(\displaystyle v = \frac{2\pi r}{T}\)
2
Centripetal acceleration: \(\displaystyle a_c = \frac{v^2}{r} = \frac{4\pi^2 r}{T^2}\)
3
Centripetal force: \(\displaystyle F_c = \frac{mv^2}{r} = \frac{4\pi^2 m r}{T^2}\)
4
Period-frequency link: \(\displaystyle T = \frac{1}{f}\)

Centripetal vs Centrifugal

Centripetal Centrifugal
Direction Toward center Away from center
Type Real net force Fictitious (pseudo) force
Frame Inertial (lab) frame Non-inertial (rotating) frame
AP Physics 1 ✓ Always use ✕ Avoid

How to Solve Problems

1
Draw a free-body diagram. Identify every real force on the object (gravity, tension, normal, friction).
2
Choose a coordinate system. Set the positive radial axis toward the center of the circle.
3
Write Newton's second law radially: \(\sum F_{\text{radial}} = \frac{mv^2}{r}\). Forces toward center are positive.
4
Solve for the unknown. Substitute known values and solve algebraically before plugging in numbers.

Example 1: Car on a Curve

A 1200 kg car rounds a flat curve of radius \(r = 50\) m at speed \(v = 20\) m/s. What friction force is required?

\(f\)
\(v\)
 r = 50 m

Example 2: Vertical Loop (Top)

A roller-coaster car of mass \(m = 500\) kg enters a vertical loop of radius \(r = 10\) m. At the top of the loop, its speed is \(v = 12\) m/s. Find the normal force on the car.

\(mg\)
\(N\)
 center r = 10 m

Example 3: Minimum Speed at the Top

What is the minimum speed at the top of a vertical loop (radius \(r\)) so the car stays on the track?

Sources of Centripetal Force

Tension
Ball on a string swung in a circle.
Friction
Car turning on a flat road.
Gravity
Moon orbiting Earth, planets orbiting the Sun.
Normal Force
Rider on a loop-the-loop or banked curve.

Circular-motion laboratory geometry and force measurement

A clamped glass tube has one continuous string through it, a metal mass hanging below and a rubber stopper to the right below the tube exit. The exposed string slopes downward toward the stopper.In a steady conical orbit, tension has an upward component that balances weight and an inward component that produces centripetal acceleration. The horizontal radius is r = ℓ sin θ.

Tension must also support the stopper

Tension must also support the stopper. In a steady conical orbit, tension has an upward component that balances weight and an inward component that produces centripetal acceleration. The horizontal radius is r = ℓ sin θ.In a steady conical orbit, tension has an upward component that balances weight and an inward component that produces centripetal acceleration. The horizontal radius is r = ℓ sin θ.

Time the orbit to find acceleration

Circular orbit with horizontal radius r and equations for period, speed and radial acceleration from timed revolutions.Time N complete revolutions over Δt to find period P = Δt/N. For steady circular motion, use the horizontal orbit radius r to calculate v = 2πr/P and inward acceleration 4π²r/P².Open full diagram

Banked Curves

\(mg\)
\(N\)
\(N\sin\theta\)
\(\theta\)

Additional worked example

The following is a separate example from the long-form lesson. Its values may differ from the example already on this slide.

A car of mass 1200 kg travels around a banked circular curve with radius 80 m. The banking angle is 20° and the road surface is frictionless. Determine the speed at which the car can negotiate the curve without sliding up or down the bank.

Show answer and explanation

Step 1 — Draw Free-Body Diagram & Identify Forces: Only two forces act on the car: the gravitational force mg directed downward and the normal force N directed perpendicular to the banked surface. Because there is no friction, the only force with a horizontal component is the normal force. Step 2 — Choose Coordinate System & Decompose: Use a radial-vertical coordinate system. The radial direction (horizontal, toward center) is positive inward. Decompose N into components: N sin θ (radial, inward) and N cos θ (vertical, upward). Gravity has no radial component. Step 3 — Apply Newton's Second Law (vertical): In the vertical direction, the car is in equilibrium (no vertical acceleration): N cos θ − mg = 0, so N = mg / cos θ. N = mg / cos θ Step 4 — Apply Newton's Second Law (radial): The net inward force equals the centripetal force: N sin θ = mv²/r. Substitute N from Step 3: (mg / cos θ) × sin θ = mv²/r, which simplifies to mg tan θ = mv²/r. mg tan θ = mv²/r Step 5 — Solve for Speed: Cancel m from both sides: g tan θ = v²/r, so v = √(rg tan θ). Substituting values: v = √(80 × 9.8 × tan 20°) = √(80 × 9.8 × 0.364) = √(285.4) ≈ 16.9 m/s. Converting to more intuitive units, this is roughly 38 mph or 61 km/h. v ≈ 16.9 m/s Step 6 — Check the Result: Note that mass canceled entirely—the design speed depends only on r, g, and θ. This makes physical sense: a heavier car needs more centripetal force but also has a larger normal force component, and these scale identically with mass. The result ~17 m/s for a 20° bank is reasonable for a highway exit ramp.

On a frictionless 20° bank of radius 80 m, the normal force has upward component N cos θ = mg and inward component N sin θ = mv²/r. Dashed arrows are components of N, not additional forces. The ideal speed is 16.9 m/s.
On a frictionless 20° bank of radius 80 m, the normal force has upward component N cos θ = mg and inward component N sin θ = mv²/r. Dashed arrows are components of N, not additional forces. The ideal speed is 16.9 m/s.

Common Mistakes

✕ Adding a "centrifugal force"
In an inertial frame, there is no outward force. Only draw real forces on the FBD.
✕ Setting \(F_c\) as a separate force
Centripetal force is the net inward force, not an additional force. Don't add it alongside tension, gravity, etc.
✕ Wrong direction at the bottom
At the bottom of a loop, normal force points up (toward center) and gravity points down (away). They don't add: \(N - mg = \frac{mv^2}{r}\).

Summary Table

Quantity Formula Direction
Speed \(\frac{2\pi r}{T}\) Tangent to circle
Centripetal acceleration \(\frac{v^2}{r}\) Toward center
Centripetal force \(\frac{mv^2}{r}\) Toward center
Min speed (top of loop) \(\sqrt{gr}\) Tangent
Banked curve angle \(\tan\theta = \frac{v^2}{rg}\) Surface tilt

Question 1 of 5

An object moves in a circle at constant speed. Which statement is true about its acceleration?

Question 2 of 5

A 0.5 kg ball on a string moves in a horizontal circle of radius 2 m with a speed of 4 m/s. What is the tension in the string?

Question 3 of 5

At the top of a vertical loop, a roller-coaster car barely maintains contact with the track. Which equation gives the minimum speed?

Question 4 of 5

A car rounds a flat curve. If the speed doubles while the radius stays the same, the required friction force:

Question 5 of 5

At the bottom of a vertical loop, a 600 kg car moves at 15 m/s through a loop of radius 20 m. What is the normal force on the car? (Use \(g = 10 \text{ m/s}^2\))