# Circular motion and centripetal force

Canonical: https://duckyhelper.com/learn/physics/circular-motion/
Updated: 2026-10-01

An object moving in a circle at a constant speed is still accelerating, because its direction keeps changing. This centripetal acceleration points toward the center of the circle and equals \(a_c = v^2/r\). By Newton's second law it needs a net inward force, \(F_c = mv^2/r\). That force always comes from something real, such as tension in a string, friction on a tire, gravity or a normal force.

## Key ideas

$$
a_c = \frac{v^2}{r} \qquad F_c = m\frac{v^2}{r} \qquad v = \frac{2\pi r}{T}
$$

- \(a_c\) is the centripetal acceleration, in m/s², pointing to the center.
- \(F_c\) is the net inward force, in newtons (N).
- \(v\) is the speed (m/s), \(r\) is the radius of the circle (m) and \(m\) is mass (kg).
- \(T\) is the period, the time for one full circle, in seconds (s). Frequency is \(f = 1/T\).

"Centripetal force" is not a new kind of force. It is a job title: whatever real force or combination of forces points toward the center is doing it.

**What provides the centripetal force**

| Situation | Inward force |
| --- | --- |
| Ball whirled on a string | Tension in the string |
| Car turning on a flat road | Static friction between tires and road |
| Moon orbiting Earth | Gravity |
| Rider at the top of a loop | Normal force from the track plus weight |

## Worked examples

**Example 1: a car on a curve**

Problem: A 1200 kg car goes around a flat curve of radius 50.0 m at 15.0 m/s. Find its centripetal acceleration and the friction force needed.

1. Acceleration toward the center.

   $$
   a_c = \frac{v^2}{r} = \frac{(15.0\ \text{m/s})^2}{50.0\ \text{m}} = 4.50\ \text{m/s}^2
   $$
2. Friction is the only inward force, so it must equal \(ma_c\).

   $$
   F_c = (1200\ \text{kg})(4.50\ \text{m/s}^2) = 5400\ \text{N}
   $$

Answer: 4.50 m/s² toward the center, needing \(5.40 \times 10^3\) N of friction

**Example 2: a ball on a string**

Problem: A 0.20 kg ball is whirled in a horizontal circle of radius 0.80 m, making one turn every 0.50 s. Ignoring gravity's small effect on the string, what is the tension?

1. Speed from the period.

   $$
   v = \frac{2\pi r}{T} = \frac{2\pi(0.80\ \text{m})}{0.50\ \text{s}} = 10.05\ \text{m/s}
   $$
2. Tension provides the centripetal force.

   $$
   T_{\text{string}} = m\frac{v^2}{r} = (0.20\ \text{kg})\frac{(10.05\ \text{m/s})^2}{0.80\ \text{m}} = 25.3\ \text{N}
   $$

Answer: 25 N

**Example 3: top speed on a flat curve**

Problem: Tires and road have \(\mu_s = 0.60\). What is the fastest a car can take a flat curve of radius 40. m without skidding?

1. At the limit, maximum static friction supplies the centripetal force. On flat ground \(F_N = mg\).

   $$
   \mu_s mg = m\frac{v^2}{r}
   $$
2. Mass cancels. Solve for \(v\).

   $$
   v = \sqrt{\mu_s g r} = \sqrt{(0.60)(9.8\ \text{m/s}^2)(40.\ \text{m})} = 15.3\ \text{m/s}
   $$

Answer: 15 m/s (about 34 mph)

**Example 4: the top of a loop**

Problem: A roller coaster goes through a vertical loop of radius 10.0 m. What is the slowest it can go at the top and still stay on the track?

1. At the top, weight and the track's normal force both point down, toward the center. The slowest speed is when the normal force just reaches zero, so weight alone provides the centripetal force.

   $$
   mg = m\frac{v^2}{r}
   $$
2. Solve.

   $$
   v = \sqrt{gr} = \sqrt{(9.8\ \text{m/s}^2)(10.0\ \text{m})} = 9.90\ \text{m/s}
   $$

Answer: 9.90 m/s

## Common mistakes and how to fix them

- **Drawing a centripetal force arrow on the free body diagram.** Fix: draw only real forces (tension, friction, gravity, normal). Then set their inward sum equal to \(mv^2/r\).
- **Thinking there is an outward force.** The outward feeling in a turning car is your body's inertia. No force pushes you out. Fix: in the ground's frame, the net force points in.
- **Using the diameter instead of the radius.** Fix: \(r\) is half the diameter.
- **Forgetting to square the speed.** Doubling the speed takes four times the force.

**Practice problems**

1. Which way does the acceleration point for an object in uniform circular motion?
   A. Toward the center
   B. Away from the center
   C. Along the direction of motion
   D. There is no acceleration

   Answer: Toward the center. The speed is constant but the direction changes, and that change always points toward the center.

2. A toy car moves at 6.0 m/s around a circular track of radius 3.0 m. What is its centripetal acceleration?

   Answer: 12 m/s². \(a_c = v^2/r = 6.0^2 \div 3.0 = 36 \div 3.0 = 12\) m/s².

3. If a car takes the same curve at twice the speed, how does the friction force it needs change?
   A. It doubles
   B. It becomes 4 times as large
   C. It halves
   D. It stays the same

   Answer: It becomes 4 times as large. \(F_c = mv^2/r\), and \(2^2 = 4\). This is why speeding on curves is dangerous.

4. A ball whirled on a string moves in a circle. If the string breaks, which way does the ball go?
   A. Straight out from the center
   B. Straight along the tangent to the circle
   C. Toward the center
   D. It keeps circling

   Answer: Straight along the tangent to the circle. With no inward force, Newton's first law takes over: the ball keeps the velocity it had, which points along the tangent.

5. A child sits 2.0 m from the center of a merry-go-round that turns once every 4.0 s. How fast is the child moving?

   Answer: 3.1 m/s. \(v = 2\pi r / T = 2\pi(2.0) \div 4.0 = 3.14\) m/s, which is 3.1 m/s.

## Frequently asked questions

### What is the difference between centripetal and centrifugal force?

Centripetal means center-seeking, the real net inward force. Centrifugal force is the outward push you seem to feel inside a turning car. Physicists call it a fictitious force: it appears only when you describe motion from inside the turning car. High school and AP problems use the ground's view, with no centrifugal force.

### Why do banked curves help?

On a banked curve, the road tilts toward the center, so part of the normal force points inward. That helps friction, or even replaces it, in turning the car. Race tracks and highway ramps are banked so cars can turn faster and more safely.

### Does centripetal force do work?

No. In uniform circular motion, the inward force is always at 90° to the motion, so it does zero work and the speed stays the same. It only changes the direction, see [work and energy](https://duckyhelper.com/learn/physics/work-and-energy/).

## Sources

- [OpenStax College Physics 2e, 6.2 Centripetal Acceleration](https://openstax.org/books/college-physics-2e/pages/6-2-centripetal-acceleration), accessed 2026-10-01
- [OpenStax College Physics 2e, 6.3 Centripetal Force](https://openstax.org/books/college-physics-2e/pages/6-3-centripetal-force), accessed 2026-10-01

## Related

- [Newton's laws of motion](https://duckyhelper.com/learn/physics/newtons-laws-of-motion/)
- [How to draw free body diagrams](https://duckyhelper.com/learn/physics/free-body-diagrams/)
- [Friction: static and kinetic](https://duckyhelper.com/learn/physics/friction/)
- [Physics study guides](https://duckyhelper.com/learn/physics/)

## Try asking Ducky

- "Check my free body diagram for a car on a curve. Did I add a fake centripetal force?"
- "Why does a bucket of water not spill at the top of the swing? Explain with my numbers."
- "Walk me through the loop problem, but let me set up the forces first."

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