Centripetal Force and Acceleration, Explained

Centripetal force is the net inward force that keeps an object moving in a circle. Its magnitude is mv^2/r and it always points toward the center. It is not a new force: tension, gravity, friction, or a normal force plays the role. Centrifugal force does not exist in an inertial frame.

AP Physics: Unit 2 (topics 2.9 Circular Motion). Circular motion is Topic 2.9 in Unit 2 (Force and Translational Dynamics) of AP Physics 1, a unit weighted at 18 to 23 percent of the multiple-choice section. AP Physics C: Mechanics treats the same material with calculus in its own Unit 2.

Centripetal force is a job, not a new force

Centripetal force is the net force pointing toward the center of a circle that keeps an object moving along that circle. It is a role that real forces play, not a separate force of nature. When a car rounds a flat curve, friction from the road points toward the center of the turn and acts as the centripetal force. When a ball swings on a string, tension does the job. For the Moon orbiting Earth, gravity does it.

That is why you never add an arrow labeled centripetal force to a correct free-body diagram. You draw the real forces (tension, gravity, friction, normal force), then notice that their sum points toward the center. Newton's second law still runs the show: the net force equals mass times acceleration, and for circular motion that acceleration points inward.

What is the formula for centripetal force?

An object moving in a circle of radius $r$ at constant speed $v$ accelerates with magnitude

a_c = \frac{v^2}{r}

directed toward the center of the circle. The speed is constant, but the velocity is not: its direction changes at every instant, and a changing velocity is an acceleration. Multiply by mass and you get the net inward force the motion requires:

F_{net} = ma_c = \frac{mv^2}{r}

Two things to notice. Doubling the speed quadruples the required force, because $v$ is squared. Tightening the turn (smaller $r$ ) also raises the required force. If a problem gives the period $T$ instead of the speed, convert with $v = 2\pi r/T$ , since the object covers one circumference $2\pi r$ each period. The AP Physics 1 equation sheet prints $a_c = v^2/r$ ; you supply the mass.

How to find centripetal force in any problem

Every circular motion problem follows the same recipe.

Draw a free-body diagram showing only real forces: gravity, normal force, tension, friction. No centripetal arrow, no centrifugal arrow.
Find the center of the circle and take the direction toward it as positive.
Write Newton's second law along the radial direction: forces pointing toward the center are positive, forces pointing away are negative.
Set that net force equal to $mv^2/r$ .
Solve for the unknown, then sanity-check units and size.

The equation $F_{net} = mv^2/r$ is not a new law. It is Newton's second law with the acceleration replaced by $v^2/r$ . If you want to check your algebra on any of these setups, plug the numbers into the centripetal force calculator.

Centripetal vs centrifugal: only one is real

The word centrifugal causes more confusion than anything else in this topic, so let's settle it. In an inertial reference frame (the ground frame AP problems use), there is no outward centrifugal force on an object moving in a circle. None. The feeling of being thrown outward in a turning car is your inertia: your body keeps moving in a straight line (Newton's first law) while the car curves underneath you. The door then pushes you inward, and your brain misreads that inward push as evidence of an outward force.

Here is the quick test. If a real outward force balanced the inward one, the net force would be zero, and zero net force means straight-line motion at constant velocity, not a circle. The object curves precisely because the inward force is unbalanced. Centrifugal force only shows up as a bookkeeping trick for observers inside a rotating (non-inertial) frame; in an inertial frame it simply does not exist.

Which real force does the job

On the exam, the phrase centripetal force is shorthand for whichever real force, or combination of real forces, points toward the center. Match the situation to its source:

Situation	What supplies the inward force
Car on a flat curve	Static friction from the road
Ball on a string, horizontal circle	Horizontal component of tension
Moon or satellite in orbit	Gravity
Rider against the wall of a spinning ride	Normal force from the wall
Bottom of a vertical loop	Normal force (or tension) minus gravity
Top of a vertical loop	Normal force (or tension) plus gravity

The last two rows matter most. At the bottom of a loop, inward is up, so $F_N - mg = mv^2/r$ and the normal force exceeds the weight (you feel heavy). At the top, inward is down, so $F_N + mg = mv^2/r$ and the normal force can drop all the way to zero at the minimum speed.

Mistakes that cost points on the AP exam

Circular motion is Topic 2.9 in Unit 2 of AP Physics 1, and Unit 2 carries 18 to 23 percent of the multiple-choice section, so these errors are expensive.

Adding a centripetal force arrow to a free-body diagram. Graders read that as an extra, nonexistent force. Label real forces only.
Adding a centrifugal arrow so the diagram balances. The forces should not balance; an unbalanced inward net force is exactly what circular motion requires.
Using kinetic friction for a car rounding a curve. A tire rolling without skidding does not slide against the road, so static friction applies, and it obeys the inequality $|F_f| \leq |\mu F_N|$ .
Assuming constant speed means zero acceleration. Speed can stay fixed while the direction of the velocity changes, and changing velocity means acceleration: magnitude $v^2/r$ , pointed at the center.

Friction provides the centripetal force: car on a flat curve

A 1200 kg car rounds a flat curve of radius 50.0 m at a constant speed of 15.0 m/s. (a) What net force keeps the car moving in the circle? (b) What minimum coefficient of static friction between the tires and the road makes this possible?

Identify the force doing the job. On a flat curve, the only horizontal force on the car is static friction from the road, and it points toward the center of the turn. The free-body diagram shows weight $mg$ down, normal force $F_N$ up, and friction toward the center. No extra centripetal arrow.
Find the required net inward force: $F_{net} = \frac{mv^2}{r} = \frac{(1200\ \text{kg})(15.0\ \text{m/s})^2}{50.0\ \text{m}} = \frac{(1200)(225)}{50.0}\ \text{N} = 5400\ \text{N}$
Use the vertical direction to get the normal force. The car does not accelerate vertically, so $F_N = mg = (1200\ \text{kg})(9.8\ \text{m/s}^2) = 11760\ \text{N}$
Static friction must supply the full 5400 N, and friction obeys $|F_f| \leq |\mu F_N|$ . So $\mu_s \geq \frac{5400\ \text{N}}{11760\ \text{N}} = 0.459$

The net (centripetal) force is 5400 N toward the center of the curve, and the minimum coefficient of static friction is 0.46. If the road is icier than that, the required inward force exceeds what friction can supply and the car slides toward the outside of the curve.

Tension at the bottom of a vertical circle

A 0.500 kg ball on a 0.800 m string swings in a vertical circle. At the lowest point its speed is 4.00 m/s. What is the tension in the string at that point?

Set up the radial equation. At the lowest point, tension $T$ points up (toward the center) and gravity $mg$ points down (away from the center). Taking toward the center as positive: $T - mg = \frac{mv^2}{r}$
Compute the required net inward force: $\frac{mv^2}{r} = \frac{(0.500\ \text{kg})(4.00\ \text{m/s})^2}{0.800\ \text{m}} = \frac{(0.500)(16.0)}{0.800}\ \text{N} = 10.0\ \text{N}$
Compute the weight: $mg = (0.500\ \text{kg})(9.8\ \text{m/s}^2) = 4.90\ \text{N}$
Solve for the tension: $T = \frac{mv^2}{r} + mg = 10.0\ \text{N} + 4.90\ \text{N} = 14.9\ \text{N}$

The tension is 14.9 N, about three times the ball's 4.9 N weight. Tension peaks at the bottom of a vertical circle because it must support the weight and supply the inward net force at the same time.

Minimum speed at the top of a loop

A roller coaster car travels around a vertical loop of radius 6.0 m. What is the minimum speed the car can have at the top of the loop without leaving the track?

At the top of the loop, both gravity and the normal force point downward, which is toward the center, so they add: $F_N + mg = \frac{mv^2}{r}$
The minimum speed occurs when the track barely pushes on the car. Set $F_N = 0$ : $mg = \frac{mv_{min}^2}{r}$
The mass cancels from both sides. Solve for the speed: $v_{min} = \sqrt{gr} = \sqrt{(9.8\ \text{m/s}^2)(6.0\ \text{m})} = \sqrt{58.8\ \text{m}^2/\text{s}^2} = 7.7\ \text{m/s}$

The minimum speed at the top is 7.7 m/s. Any slower and gravity alone would exceed the required inward force $mv^2/r$ , so the track cannot push outward to fix it; the car leaves the rails and briefly becomes a projectile.

Frequently asked questions

Is centripetal force a real force?

A net inward force really does act on anything moving in a circle, but centripetal force is not a separate kind of force. The term names a role. Tension, gravity, friction, or a normal force (or a combination of them) supplies the inward net force equal to mv^2/r. You never add it as an extra force on a free-body diagram.

Is centrifugal force real?

Not in an inertial reference frame, which is the frame AP problems use. The outward push you feel in a turning car is your inertia: your body tends to keep moving in a straight line while the car curves under you. The only real horizontal force on you is the inward push from the seat or door.

Do I draw centripetal force on a free-body diagram?

No. Draw only real forces: gravity, normal force, tension, friction, and so on. The centripetal force is the net result of those arrows, not an additional arrow. The guide on how to draw a free-body diagram walks through the full process.

What is the formula for centripetal force?

Centripetal acceleration is a = v^2/r, pointed at the center of the circle, so the required net force is F = mv^2/r. If you know the period T instead of the speed, first convert with v = 2πr/T. The centripetal force calculator handles both versions.

Why is there an acceleration if the speed is constant?

Acceleration is the rate of change of velocity, and velocity includes direction. In uniform circular motion the direction of the velocity changes at every instant, so the object accelerates even though its speed never changes. That acceleration has magnitude v^2/r and points toward the center.