Centripetal vs Centrifugal Force: Which Is Real?

Centripetal is a direction, not a force: it means pointing toward the center, and real forces like friction, tension or gravity do the pointing. Centrifugal is not a force at all in AP Physics 1, because no second object exerts it. Every arrow on a free-body diagram must be exerted by something.

AP Physics: Unit 2 (topics 2.2 Forces and Free-Body Diagrams, 2.9 Circular Motion). This comparison sits across two topics of AP Physics 1 Unit 2, Force and Translational Dynamics, weighted at 18 to 23 percent of the multiple-choice section over about 22 to 27 class periods. The word centrifugal does not appear in the AP Physics 1, AP Physics 2, AP Physics C: Mechanics or AP Physics C: Electricity and Magnetism Course and Exam Descriptions (all Effective Fall 2024, V.1), and neither does the word fictitious. The phrase centripetal force does not appear in the AP Physics 1 CED either: every occurrence of centripetal in that document is followed by acceleration. EK 2.9.A.1 defines centripetal acceleration as the component of an object's acceleration directed toward the center of the object's circular path, with a_c = v^2/r, and EK 2.9.A.2 says that acceleration can result from a single force, more than one force, or components of forces. EK 2.9.A.2.ii and 2.9.A.2.iii use the phrase the net force producing centripetal acceleration. The rule that decides what belongs on a diagram is EK 2.2.A.1.i, a force exerted on an object is always due to the interaction of that object with another object, together with EK 2.2.A.1.ii, an object cannot exert a net force on itself, EK 2.2.B.2, and the Topic 2.2 boundary statement on drawing forces rather than components. EK 2.6.C.4, the equivalence principle, is the only essential knowledge statement in the AP Physics 1 CED that mentions a noninertial reference frame. Suggested skills are 1.A, 2.B, 2.C and 3.C for Topic 2.2, and 1.B, 2.A, 2.D, 3.A and 3.C for Topic 2.9.

Every arrow needs an object exerting it

Before any talk of frames of reference, apply the rule the AP Physics 1 CED actually gives you for deciding what goes on a diagram.

EK 2.2.A.1.i: a force exerted on an object or system is always due to the interaction of that object with another object or system.

That single sentence settles the question. Every force you draw has to be nameable in the form "the force exerted on this object by that object." Run the test on a car rounding a flat curve:

Gravitational force: exerted on the car by the Earth. Passes.
Normal force: exerted on the car by the road surface. Passes.
Friction: exerted on the car by the road surface. Passes.
An outward force: exerted on the car by what? There is nothing on the outside of the curve touching the car, and no field pulling it outward. Fails.

So there is no outward arrow. Not because outward forces are forbidden by decree, but because no object is available to exert one, and EK 2.2.A.1.i requires one. EK 2.2.B.2 says the same thing from the diagram's side: the free-body diagram of an object or system shows each of the forces exerted on the object by the environment. An arrow with no exerter is not a force that got left out; it is not a force.

EK 2.2.A.1.ii closes the last escape route: an object or system cannot exert a net force on itself. The passenger's sense of being flung outward cannot be drawn as a force the passenger exerts on the passenger.

What about the inward arrow, then? Also not a separate arrow, but for a different reason. Friction is already drawn. Adding a second inward arrow labeled centripetal would count the same interaction twice. The Topic 2.2 boundary statement is explicit about what belongs on the diagram: AP Physics 1 only expects students to depict the forces exerted on objects, not the force components on free-body diagrams, and individual forces must be drawn as individual straight arrows, originating on the dot and pointing in the direction of the force.

The procedure for turning that diagram into an answer is in centripetal force and acceleration. This page is about why one of these two words earns a place in your vocabulary and the other does not.

Neither word names a force in the CED

This is the fact that reframes the whole comparison, and it is checkable.

The word "centrifugal" does not appear anywhere in the AP Physics 1 Course and Exam Description. Nor in the AP Physics 2, AP Physics C: Mechanics, or AP Physics C: Electricity and Magnetism CEDs. Neither does the word "fictitious." You will not meet either term in a required course content statement, a boundary statement, or an exam convention.

More surprising: the phrase "centripetal force" does not appear in the AP Physics 1 CED either. Every occurrence of the word centripetal in that document is followed by the word acceleration. What the CED defines is centripetal acceleration, at EK 2.9.A.1: centripetal acceleration is the component of an object's acceleration directed toward the center of the object's circular path. When the CED needs to talk about the forces responsible, it writes a phrase instead of a noun. EK 2.9.A.2.ii speaks of components of the static friction force and the normal force contributing to the net force producing centripetal acceleration of an object traveling in a circle on a banked surface, and EK 2.9.A.2.iii speaks of a component of tension contributing to the net force producing centripetal acceleration experienced by a conical pendulum.

Read those phrasings and the CED's model is unmistakable. Centripetal is an adjective describing a direction, and it modifies acceleration. The forces are the ordinary ones you already know how to name, and their vector sum happens to point that way.

EK 2.9.A.2 states the general case: centripetal acceleration can result from a single force, more than one force, or components of forces exerted on an object in circular motion. One force, several forces, or parts of forces. Nothing in that list is a force called centripetal.

So the honest comparison is not between two forces, one real and one not. It is between a legitimate direction word and a word the course does not use.

Centripetal vs centrifugal, side by side

Question you are asking	Centripetal	Centrifugal
What the word names	A direction, toward the center of the circular path	Nothing in AP Physics 1
Appears in the AP Physics 1 CED	Yes, always as "centripetal acceleration"	No, not once
Appears in any of the four 2026 CEDs	Yes, in AP Physics 1 and AP Physics C: Mechanics	No
Which object exerts it	Whichever real object does: road, string, Earth, wall	None, which is why it is not a force
Goes on a free-body diagram	No, and neither does an inward arrow labeled centripetal; you draw the real forces	No
Direction	Toward the center, EK 2.9.A.1.ii	Would be away from the center
Magnitude of the acceleration	$a_c = \frac{v^2}{r}$	Not applicable
Required net force	$\sum F_{\text{inward}} = \frac{mv^2}{r}$ , and it is unbalanced	Adding one would balance the forces and forbid the circle
On the AP Physics 1 equation sheet	$a_c = \frac{v^2}{r}$ , an acceleration, with no force version printed	Nothing
What the sensation actually is	The real inward push you feel from a seat, door or wall	Your inertia, that is, your body continuing in a straight line
CED essential knowledge	2.9.A.1, 2.9.A.1.i, 2.9.A.1.ii, 2.9.A.2	None

Two rows in that table are the page.

The equation-sheet row. The AP Physics 1 sheet prints $a_c = \frac{v^2}{r}$ and stops there. There is no $F_c = \frac{mv^2}{r}$ printed anywhere on it. You reach the force by combining that acceleration with Newton's second law from EK 2.5.A.2, $\vec{a}_{\text{sys}} = \frac{\sum \vec{F}}{m_{\text{sys}}}$ , which is the same law you use everywhere else. The sheet is telling you which of the two quantities the course treats as primary.

The required net force row. This is the contradiction that finishes the argument, and the next section works it out.

Adding an outward force forbids the motion you are looking at

Suppose you insist on an outward force equal in magnitude to the inward one, so the diagram looks tidy and balanced. Follow that assumption through the CED's own statements and it destroys itself in three steps.

Balanced is a defined term. EK 2.4.A.2: translational equilibrium is a configuration of forces such that the net force exerted on a system is zero, $\sum_i \vec{F}_i = 0$ .
Newton's first law then applies. EK 2.4.A.3: if the net force exerted on a system is zero, the velocity of that system will remain constant. Constant velocity means constant magnitude and constant direction.
A circle has a changing direction of velocity at every instant. So a balanced-force diagram predicts a straight line at constant speed, which is not what you are looking at.

Run it the other way and you get the correct statement. EK 2.5.A.1: unbalanced forces are a configuration of forces such that the net force exerted on a system is not equal to zero. Circular motion needs unbalanced forces. The net force must be nonzero and it must point inward, because EK 2.9.A.1.ii says the centripetal acceleration is directed toward the center, and EK 2.5.A.2 says the acceleration is in the same direction as the net force.

So the outward arrow is not merely unnecessary. It is contradicted by the observed motion. Draw it and your algebra predicts the car leaving the curve in a straight line, which is exactly the motion the road is preventing.

Worked example three below puts a number on how fast that wrong prediction fails: for a rider on a ride of radius $3.5$ m turning once every $2.4$ s, the balanced-force model has the rider $0.45$ m outside the wall after $0.20$ s.

There is a version of the outward push that is real, and it is worth naming so you do not confuse it with the fake one. Newton's third law. EK 2.3.A.1 says the third law describes the interaction of two objects in terms of the paired forces that each exerts on the other, $\vec{F}_{\text{A on B}} = -\vec{F}_{\text{B on A}}$ . The wall pushes the rider inward, so the rider pushes the wall outward with an equal magnitude. That outward force is real, it has both objects named, and it acts on the wall. It never appears on the rider's free-body diagram, because a free-body diagram shows only the forces exerted on the object it is drawn for.

What the feeling actually is

The sensation is real. The explanation people reach for is the one that fails the exerter test.

Sitting in a car that turns left, you feel pressed against the right-hand door. What is happening, in the CED's terms:

Your body was moving in a straight line, and EK 2.4.A.3 says it keeps doing so while no net force acts on it.
The car changes direction underneath you.
The door arrives at your body and pushes it inward, toward the center of the turn. That normal force is what you feel, and EK 2.7.A.2.ii defines it: the perpendicular component of the force exerted on an object by the surface with which it is in contact, directed away from the surface.

You feel an inward push and interpret it as evidence of an outward pull. The pressure on your right side is genuine; the outward force is the inference, and it is the wrong one.

The CED has a related idea that people sometimes reach for here, and it is worth separating carefully because it is about something else. EK 2.6.C.4 states the equivalence principle: an observer in a noninertial reference frame is unable to distinguish between an object's apparent weight and the gravitational force exerted on the object by a gravitational field. That is the only essential knowledge statement in the AP Physics 1 CED that mentions a noninertial reference frame, and it is a statement about gravity and apparent weight, not a licence to add an outward arrow in a turn. Apparent weight itself is defined at EK 2.6.C.1 as the magnitude of the normal force exerted on the system, which is again the inward push, measured.

The distinction between the real force and the sensation is the reason this comparison is worth a page. The physics of the situation is settled by asking which object pushes, and the answer is always the object you are touching.

The frames question, and why the exam never asks it

There is a legitimate rotating-frame calculation in which an outward term appears in the equations. It is not on the AP Physics 1 syllabus, and the CED closes the door in four separate places rather than one.

The exam convention. The AP Physics 1 Table of Information prints a short list of conventions used on the exam, and the first line reads: the frame of reference of any problem is assumed to be inertial unless otherwise stated.
The Topic 1.4 boundary statement. Unless otherwise stated, the frame of reference of any problem may be assumed to be inertial.
The learning objective itself. LO 1.4.B asks you to describe the motion of objects as measured by observers in different inertial reference frames. The word inertial is in the objective, not smuggled in later.
The appendix. In its list of what students may assume unless otherwise stated, the CED includes that frames of reference are inertial.

What an inertial frame is, per EK 2.4.A.5: one from which an observer would verify Newton's first law of motion. A rotating frame is not one, which is precisely why Newton's first law appears to fail in it and why an extra term has to be inserted to make the bookkeeping balance.

Two more limits worth knowing. EK 1.4.B.2.ii says the acceleration of any object is the same as measured from all inertial reference frames, so switching between the legitimate frames never changes an acceleration and therefore never changes a force. And the Topic 1.4 boundary statement adds that adding or subtracting vectors to find relative velocities is restricted to motion along one dimension for AP Physics 1, which rules out the two-dimensional frame transformations a rotating-frame treatment would need.

So the correct exam answer is not "centrifugal force exists in a rotating frame." It is that the course works in inertial frames, and in an inertial frame there is no outward force, because no object exerts one. If a question ever supplied a rotating frame explicitly, it would have to say so, and the CED gives you no equation for that case.

Where the confusion costs a mark

Each of these is a scoring event on a free-response or multiple-choice item.

Drawing an outward arrow on a free-body diagram. A grader reads it as a claim that a fifth object is pushing the car outward. EK 2.2.A.1.i requires an exerting object and there is none.
Drawing an inward arrow labeled centripetal alongside the real forces. This double counts, because the real force is already drawn and its inward component is the thing you were about to call centripetal. The Topic 2.2 boundary statement says to depict the forces, not the components.
Drawing a force and its two components as three arrows on the same diagram. Same boundary statement, same fix: draw each force once, then resolve it in your algebra.
Writing that the forces balance in circular motion. EK 2.4.A.3 makes balanced forces mean constant velocity, which rules out a circle. Say the net force is unbalanced and directed toward the center.
Writing "centripetal force" as though it were a separate force of nature in a justification. Name the real force instead: static friction from the road, the horizontal component of the tension, the gravitational force from the planet, the normal force from the wall. That is what EK 2.9.A.2 asks for.
Claiming an object in circular motion at constant speed has zero acceleration. EK 2.9.A.1 defines centripetal acceleration as a component of the acceleration directed toward the center, and $a_c = \frac{v^2}{r}$ is nonzero whenever the object is moving.
Using kinetic friction for a car on a curve that is not skidding. Tyres rolling without sliding are a static friction situation, governed by $\lvert \vec{F}_f \rvert \leq \lvert \mu \vec{F}_N \rvert$ ; the split is covered in static vs kinetic friction.
Calling the outward push on the wall a centrifugal force on the rider. It is a real third law partner, and it acts on the wall. Getting the object right is the whole of the mark.
Explaining the sensation as an outward force in a written answer. Explain it as inertia plus the inward normal force you feel from the surface you are touching, and reference EK 2.4.A.3.

What the CED asks, and how the exam frames it

This comparison sits across two topics of AP Physics 1 Unit 2, Force and Translational Dynamics, weighted at 18 to 23 percent of the multiple-choice section and estimated at about 22 to 27 class periods.

Topic 2.9, Circular Motion, has two learning objectives. 2.9.A asks you to describe the motion of an object traveling in a circular path, and carries EK 2.9.A.1 through 2.9.A.5, including the three worked sources of centripetal acceleration: the top of a vertical loop with its derived result $v = \sqrt{gr}$ , the banked surface, and the conical pendulum. 2.9.B asks you to describe circular orbits using Kepler's third law, with the derived relation $T^2 = \frac{4\pi^2}{GM}R^3$ . Two boundary statements apply: AP Physics 1 only expects students to quantitatively analyze banked curves in which no friction is required to maintain uniform circular motion, with friction on a banked curve limited to qualitative descriptions, and AP Physics 1 does not expect students to know Kepler's first or second laws of planetary motion. Suggested skills for Topic 2.9 are 1.B, 2.A, 2.D, 3.A and 3.C.

Topic 2.2, Forces and Free-Body Diagrams, supplies the rule that decides what you may draw. LO 2.2.A asks you to describe a force as an interaction between two objects or systems, which is where EK 2.2.A.1.i and 2.2.A.1.ii live, and LO 2.2.B asks you to describe the forces exerted on an object or system using a free-body diagram, which carries EK 2.2.B.1 through 2.2.B.4 and the boundary statement quoted above. Suggested skills for Topic 2.2 are 1.A, 2.B, 2.C and 3.C.

One CED sample instructional activity for Topic 2.9 is a direct rehearsal of this page: it asks a teacher to describe something a driver could be doing, such as turning the steering wheel to the right while pressing the brake, and have students walk out the motion while holding out one arm representing the velocity vector and the other arm representing the acceleration vector. Two arms, and neither of them is an outward force.

For the routine rather than the distinction, centripetal force and acceleration has the five-step method and the table of which real force does the job in each geometry, and the centripetal force calculator will check a $\frac{mv^2}{r}$ . How to draw a free-body diagram has the drawing rules, and the free-body diagram builder lets you test whether a diagram you drew produces the motion you expected. The CED framing is on Topic 2.9 and Topic 2.2, and the circular motion and gravitation practice set has problems to work.

A car on a flat curve: name every force, then count them

A $1400$ kg car rounds a flat, unbanked curve of radius $55$ m at a steady $18$ m/s. List every force exerted on the car and name the object exerting each one. Find the centripetal acceleration, the net inward force required, and the smallest coefficient of static friction that makes the turn possible. Take the direction toward the center of the curve as positive. Use $g = 9.8\ \text{m/s}^2$ .

Run EK 2.2.A.1.i on every candidate force before drawing anything. Gravitational force, exerted on the car by the Earth, $13720$ N downward, since $mg = (1400)(9.8)$ . Normal force, exerted on the car by the road surface, vertically upward. Static friction, exerted on the car by the road surface, horizontally. That is the complete list, and there are three arrows on the diagram.
Ask the fourth question explicitly, because it is the one this page exists for. An outward force would have to be exerted on the car by some object on the outside of the curve. Nothing is there. No object, no force, no arrow.
Vertical direction first, to get the normal force. The car does not accelerate vertically, so $F_N - mg = 0$ and $F_N = 13720\ \text{N}$ . Note that this is the one situation in which $F_N = mg$ holds, because the surface is horizontal and there is no vertical acceleration.
Centripetal acceleration from EK 2.9.A.1.i: $a_c = \frac{v^2}{r} = \frac{(18)^2}{55} = \frac{324}{55} = 5.89\ \text{m/s}^2$ , directed toward the center of the curve.
Net inward force from Newton's second law, EK 2.5.A.2: $\sum F = ma_c = (1400)(5.89) = 8250\ \text{N}$ inward, to three significant figures. Carrying the unrounded acceleration gives $8247$ N.
Which force supplies it? Gravity is vertical and the normal force is vertical, so neither has a horizontal component. Only static friction is available, so friction alone must supply the whole $8250$ N, and it must point toward the center of the curve. This is EK 2.9.A.2's "single force" case.
Smallest coefficient of static friction. Static friction obeys $\lvert \vec{F}_f \rvert \leq \lvert \mu_s \vec{F}_N \rvert$ , so $8247 \leq \mu_s (13720)$ , giving $\mu_s \geq \frac{8247}{13720} = 0.601$ .
Notice that the mass cancelled: $\mu_s \geq \frac{ma_c}{mg} = \frac{a_c}{g} = \frac{5.89}{9.8} = 0.601$ . Every vehicle takes this curve at this speed or none does, whatever it weighs.
Finally, check what an outward arrow would have done to this calculation. Add an $8250$ N outward force and the horizontal net force becomes zero. EK 2.4.A.3 then says the velocity stays constant, so the car travels in a straight line at $18$ m/s and never rounds the curve. The tidy balanced diagram predicts the wrong motion.

Three forces, each with a named exerter: gravity from the Earth ( $13720$ N down), the normal force from the road ( $13720$ N up), and static friction from the road ( $8250$ N toward the center). $a_c = 5.89\ \text{m/s}^2$ , the required net inward force is $8250$ N, and $\mu_s \geq 0.601$ . There is no fourth force, because no fourth object is in contact with the car on the outside of the curve.

A conical pendulum: two forces, one inward resultant, no third arrow

A $0.25$ kg ball on the end of a $0.80$ m string swings in a horizontal circle with the string making a steady $25^\circ$ with the vertical. Find the radius of the circle, the tension in the string, the net inward force, the ball's speed, its centripetal acceleration, and the period. Take the direction toward the center as positive horizontally and upward as positive vertically. Use $g = 9.8\ \text{m/s}^2$ , $\sin 25^\circ = 0.4226$ , $\cos 25^\circ = 0.9063$ .

Forces on the ball, with exerters named: the gravitational force from the Earth, $mg = (0.25)(9.8) = 2.45\ \text{N}$ downward, and the tension from the string, along the string toward the pivot. Two arrows. That is the whole diagram, and EK 2.9.A.2.iii describes exactly this case as a component of tension contributing to the net force producing centripetal acceleration.
Geometry. The ball moves in a horizontal circle whose radius is the horizontal distance from the ball to the vertical axis: $r = L\sin 25^\circ = (0.80)(0.4226) = 0.338\ \text{m}$ .
Vertical direction. The ball stays at the same height, so its vertical acceleration is zero and the vertical component of the tension must cancel gravity: $T\cos 25^\circ - mg = 0$ , so $T = \frac{2.45}{0.9063} = 2.70\ \text{N}$ .
Horizontal direction. The only horizontal force is the horizontal component of the tension, and it points toward the axis: $\sum F = T\sin 25^\circ = (2.703)(0.4226) = 1.142\ \text{N}$ inward. That $1.142$ N is not a new force. It is a component of the one tension arrow already drawn.
Speed, from $\sum F = \frac{mv^2}{r}$ : $v = \sqrt{\frac{(1.142)(0.338)}{0.25}} = \sqrt{1.545} = 1.24\ \text{m/s}$ . Cross-check with the symbolic result, which follows from dividing the two direction equations: $v = \sqrt{gr\tan 25^\circ} = \sqrt{(9.8)(0.338)(0.4663)} = \sqrt{1.545} = 1.24\ \text{m/s}$ . The tension cancelled out, which is why the mass does not appear.
Centripetal acceleration: $a_c = \frac{v^2}{r} = \frac{1.545}{0.338} = 4.57\ \text{m/s}^2$ , or equivalently $g\tan 25^\circ = (9.8)(0.4663) = 4.57\ \text{m/s}^2$ , directed horizontally toward the axis.
Period, from EK 2.9.A.5.iii: $T_{\text{period}} = \frac{2\pi r}{v} = \frac{2\pi(0.338)}{1.243} = 1.71\ \text{s}$ .
Now count the arrows against the numbers. Two forces produced three distinct magnitudes: $2.45$ N of gravity, $2.70$ N of tension, and $1.14$ N of inward resultant. The $1.14$ N has no exerter of its own, because it is not a force; it is what the two real forces add up to. Anyone drawing it as a third arrow has drawn the answer instead of the physics.
One sanity check on direction. The tension is larger than the weight, $2.70$ N against $2.45$ N, which it has to be: the string must cancel the full weight vertically using only part of its magnitude, leaving the rest to point inward.

$r = 0.338$ m, $T = 2.70$ N, net inward force $1.14$ N, $v = 1.24$ m/s, $a_c = 4.57\ \text{m/s}^2$ , period $1.71$ s. Two forces act on the ball, gravity from the Earth and tension from the string. The inward $1.14$ N is the horizontal component of the tension, not a separate force, and there is no outward force at any point in the calculation.

How fast the balanced-force model fails

A $60$ kg rider stands with their back against the inside wall of a cylindrical ride of radius $3.5$ m, which completes one revolution every $2.4$ s. Find the rider's speed, centripetal acceleration, and the horizontal force the wall exerts on them. Then assume an outward force of equal magnitude also acts, and find where that assumption places the rider $0.20$ s later. Take inward as positive.

Speed from the period, using $T = \frac{2\pi r}{v}$ from EK 2.9.A.5.iii rearranged: $v = \frac{2\pi r}{T} = \frac{2\pi(3.5)}{2.4} = \frac{21.99}{2.4} = 9.16\ \text{m/s}$ .
Centripetal acceleration: $a_c = \frac{v^2}{r} = \frac{(9.163)^2}{3.5} = \frac{83.96}{3.5} = 24.0\ \text{m/s}^2$ , which is about $2.4$ times $g$ . Directed horizontally toward the axis of the cylinder.
Horizontal forces on the rider: the normal force from the wall, pointing away from the wall surface and therefore inward, by EK 2.7.A.2.ii. Nothing else is horizontal. So $F_N = ma_c = (60)(23.99) = 1440\ \text{N}$ inward, to three significant figures.
The rider's weight, $mg = (60)(9.8) = 588\ \text{N}$ , is vertical and is balanced by the floor, so it plays no part in the horizontal equation. Note the wall's push is about $2.4$ times the rider's weight, which is why the sensation is so strong.
Now the assumption under test. Suppose an outward force of $1440$ N also acts. The horizontal net force is then $1440 - 1440 = 0$ , so the forces are balanced in the sense of EK 2.4.A.2.
EK 2.4.A.3 then applies without exception: with zero net force the velocity remains constant, in magnitude and direction. The rider's velocity at that instant is $9.16$ m/s tangential, so the prediction is a straight line along the tangent at $9.16$ m/s.
Follow that straight line for $0.20$ s. The rider covers $(9.163)(0.20) = 1.833\ \text{m}$ along the tangent, so their distance from the axis becomes $\sqrt{(3.5)^2 + (1.833)^2} = \sqrt{12.25 + 3.36} = \sqrt{15.61} = 3.95\ \text{m}$ .
Compare with the wall, which sits at $3.5$ m from the axis. The balanced-force prediction has the rider $0.45$ m outside the wall after one fifth of a second. In the same $0.20$ s the ride has turned through $\omega t = \frac{2\pi}{2.4}(0.20) = 0.524$ rad, that is $30^\circ$ , so the real rider is still against the wall $30^\circ$ further round.
That is the refutation in numbers. The outward force does not just fail to help; assuming it predicts the rider leaving through a solid wall within a fifth of a second. The forces on an object in circular motion are unbalanced, and they must be, by EK 2.5.A.1.

$v = 9.16$ m/s, $a_c = 24.0\ \text{m/s}^2$ , about $2.4g$ , and the wall pushes inward with $1440$ N. Assuming a matching outward force balances the horizontal forces, which by Newton's first law predicts straight-line motion: after $0.20$ s that puts the rider $3.95$ m from the axis, $0.45$ m outside a wall at $3.5$ m. The real rider has instead swept $30^\circ$ round and is still in contact.

Frequently asked questions

What is the difference between centripetal and centrifugal force?

Centripetal means directed toward the center of a circular path, and it describes the direction that real forces point when they hold something in a circle: static friction from a road, tension in a string, gravity on a satellite, a normal force from a wall. Centrifugal would mean directed away from the center, and no such force acts on the object, because no object is positioned to exert it. The AP Physics 1 CED requires at essential knowledge 2.2.A.1.i that every force be due to the interaction of the object with another object, and there is no second object on the outside of a turn pushing outward.

Is centrifugal force real?

There is no outward force acting on an object moving in a circle, so nothing outward belongs on its free-body diagram. The word centrifugal does not appear anywhere in the 2026 AP Physics 1 Course and Exam Description, nor in the AP Physics 2 or either AP Physics C description, and neither does the word fictitious. What is real is the inward push you feel from the seat, door or wall you are touching, and your own inertia, which is your body continuing in a straight line while the vehicle curves away underneath you. A rotating frame of reference does introduce an outward term into the equations, but AP Physics 1 assumes inertial frames unless a problem says otherwise.

Should you draw centripetal force on a free-body diagram?

No, and neither should you draw a centrifugal one. A free-body diagram shows the forces exerted on the object by the environment, which the AP Physics 1 CED states at essential knowledge 2.2.B.2, so each arrow must be a real interaction with a named second object: gravity from the Earth, the normal force from a surface, tension from a string, friction from a surface. Adding a separate inward arrow labeled centripetal double counts, because the inward direction is already supplied by one of those real forces or by a component of one. The Topic 2.2 boundary statement says AP Physics 1 expects students to depict the forces exerted on objects, not the force components.

Why do you feel pushed outward when a car turns?

Because your body was traveling in a straight line and the car changed direction underneath you. The AP Physics 1 CED states at essential knowledge 2.4.A.3 that a system with zero net force keeps a constant velocity, so until something pushes you, you keep going straight. What eventually pushes you is the door or the seat, and that push is directed inward, toward the center of the turn. So the pressure you feel on your outer side is a genuine inward force from a surface you are touching, and interpreting it as evidence of an outward force is the mistake. Nothing on the outside of the car is touching you.

Does the AP Physics 1 CED use the phrase centripetal force?

No. Every occurrence of the word centripetal in the AP Physics 1 Course and Exam Description is followed by the word acceleration. Essential knowledge 2.9.A.1 defines centripetal acceleration as the component of an object's acceleration directed toward the center of the object's circular path, and where the CED needs to refer to the forces responsible it uses a phrase instead, such as the net force producing centripetal acceleration, in essential knowledge 2.9.A.2.ii and 2.9.A.2.iii. Essential knowledge 2.9.A.2 adds that centripetal acceleration can result from a single force, more than one force, or components of forces. Centripetal is a direction word in this course, not the name of a force.

Is there a centripetal force formula on the AP Physics 1 equation sheet?

No. The AP Physics 1 equation sheet prints the centripetal acceleration, a c equals v squared over r, and nothing else about circular motion forces. There is no printed line of the form F equals m v squared over r. You get the force by combining that acceleration with Newton's second law, which the sheet prints as the acceleration of a system equalling the net force divided by the system's mass. That is the same law you use for every other dynamics problem, which is the point: circular motion introduces a new acceleration, not a new force.

Are the forces balanced on an object moving in a circle?

No, and this is the sharpest way to rule out an outward force. The AP Physics 1 CED defines balanced forces as translational equilibrium at essential knowledge 2.4.A.2, and essential knowledge 2.4.A.3 says that if the net force on a system is zero, the velocity of that system remains constant, in direction as well as magnitude. An object moving in a circle has a velocity whose direction changes at every instant, so its forces cannot be balanced. Essential knowledge 2.5.A.1 names the correct description: unbalanced forces, with a net force that is not zero, pointing toward the center of the circular path.