Centripetal Force & Acceleration Calculator

Centripetal acceleration is a_c = v^2/r and centripetal force is F_c = m v^2/r. Enter speed (or period), radius, and mass, and the calculator above returns both. F_c is not a separate force: it is supplied by real forces such as tension, friction, gravity, or the normal force.

Mass m (kg)Speed v (m/s)Radius r (m)

centripetal force (F_c)

64 N

Centripetal acceleration: 32 m/s^2. Both a_c and F_c point toward the center of the circle.

Steps

1.a_c = v^2 / r = (4 m/s)^2 / 0.5 m = 16 / 0.5 = 32 m/s^2
2.F_c = m v^2 / r = m a_c = 2 kg x 32 m/s^2 = 64 N

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% of the multiple-choice section. AP Physics C: Mechanics also covers circular motion within its Unit 2, with calculus-based methods.

What the calculator above computes

The calculator above computes centripetal acceleration with $a_c = v^2/r$ and centripetal force with $F_c = mv^2/r$ . Enter a speed and a radius to get $a_c$ ; add a mass to get $F_c$ . If you know the period instead of the speed, switch to period mode and the tool finds $v = 2\pi r/T$ first, then runs the same two equations.

A few unit reminders before you plug in numbers. Speed goes in meters per second, radius in meters, mass in kilograms, and period in seconds. The outputs come back in meters per second squared and newtons. If a problem hands you a diameter, halve it: the formulas want the radius. If it hands you rpm, convert to a period first ( $T$ in seconds equals 60 divided by the rpm value) before using period mode.

Where the formulas come from

$a_c = v^2/r$ is printed on the AP Physics 1 equation sheet. It says that an object moving in a circle of radius $r$ at constant speed $v$ accelerates toward the center, because its velocity direction changes even when its speed does not. The faster the object moves or the tighter the circle, the larger the acceleration, and the dependence on speed is quadratic: doubling $v$ quadruples $a_c$ .

The force equation is not a new law. It is Newton's second law pointed along the radius: $F_{net} = ma$ becomes $F_c = ma_c = mv^2/r$ when the acceleration is centripetal. That is why the calculator treats $F_c$ as mass times the acceleration it just found, and why you can check any circular motion answer with the net force calculator: the net inward force must equal $mv^2/r$ .

Using the period instead of speed

Plenty of problems give you the time for one revolution rather than the speed: a lab turntable, a carousel, a moon. In one period $T$ , the object covers one circumference, so

v = \frac{2\pi r}{T}

Substituting that into $a_c = v^2/r$ gives a direct formula, $a_c = 4\pi^2 r/T^2$ , and multiplying by mass gives $F_c = 4\pi^2 m r/T^2$ . The calculator above does this substitution for you in period mode.

Frequency works too, since $T = 1/f$ . A wheel spinning at 2.0 revolutions per second has $T = 0.50$ s. Watch the rpm trap: 120 rpm is 2.0 revolutions per second, so $T = 0.50$ s, not 120 s. Converting to seconds per revolution before you touch the formulas avoids nearly every period-mode error.

Centripetal force is supplied by real forces

Centripetal force never appears on a correct free-body diagram, because it is not its own interaction. It is a requirement: for anything to move in a circle, the real forces acting on it must add up to a net inward force of $mv^2/r$ . Something physical always plays that role.

Situation	What supplies $F_c$
Car on a flat curve	Static friction from the road
Ball on a string	Tension (its horizontal component)
Satellite in orbit	Gravity
Rider pressed against a spinning wall	Normal force

So when a problem asks for the tension or friction needed for circular motion, it is really asking you to set that real force equal to $mv^2/r$ . The centripetal force guide works through each of these cases, and how to draw a free-body diagram shows why the inward arrow is always a labeled real force.

Mistakes the calculator cannot catch

The tool runs the arithmetic; it cannot tell whether you set the problem up correctly. The errors that cost the most points:

Using diameter instead of radius. A loop 10 m across has $r = 5$ m, and since $a_c$ scales with $1/r$ , plugging in 10 m instead of 5 m halves your answer.
Adding a centripetal force arrow to a free-body diagram alongside tension or friction. Graders read that as a misunderstanding of Newton's second law.
Assuming constant speed means zero acceleration. In uniform circular motion the speed is constant but the velocity is not, so $a_c$ is never zero.
Mixing directions in vertical circles. At the top of a loop, gravity points toward the center and helps supply $F_c$ . At the bottom, gravity points away from the center, so the supporting force must exceed $mg$ .

Where this fits in AP Physics 1

Circular motion is Topic 2.9, the final topic of Unit 2 (Force and Translational Dynamics) in AP Physics 1. Unit 2 carries 18 to 23% of the multiple-choice section weight, tied with Unit 3 for the largest share of any unit, so circular motion questions and their free-body diagrams are reliable point sources. A four-function, scientific, or graphing calculator is allowed on both sections of the exam, so practicing clean numeric substitution like the worked examples below pays off directly.

Expect circular motion to mix with friction (flat curves), tension (strings and pendulums), and gravity (loops and orbits) rather than appear alone. The Unit 2 page collects the related guides, and the friction calculator helps with flat-curve setups where static friction is the inward force.

Car on a flat curve

A 1200 kg car rounds a flat curve of radius 50.0 m at a constant 15.0 m/s. Find the centripetal acceleration and the centripetal force, and identify the real force that supplies it.

Find the acceleration: $a_c = v^2/r = (15.0 \text{ m/s})^2/(50.0 \text{ m}) = (225 \text{ m}^2/\text{s}^2)/(50.0 \text{ m}) = 4.50 \text{ m/s}^2$ , directed toward the center of the curve.
Apply Newton's second law along the radius: $F_c = ma_c = (1200 \text{ kg})(4.50 \text{ m/s}^2) = 5400 \text{ N}$ .
Identify the real force. On a flat road the only horizontal force on the car is static friction from the pavement, so friction supplies all 5400 N. Plausibility check: the coefficient needed is $\mu = F_c/(mg) = (5400 \text{ N})/[(1200 \text{ kg})(9.8 \text{ m/s}^2)] = 5400/11760 = 0.46$ , well within the range of rubber on dry asphalt.

$a_c = 4.50 \text{ m/s}^2$ toward the center; $F_c = 5400 \text{ N}$ , supplied entirely by static friction.

Ball on a cord, starting from the period

A 0.150 kg ball on a cord travels in a horizontal circle of radius 0.800 m, completing one revolution every 0.600 s. Find the speed, the centripetal acceleration, and the net inward force.

Convert period to speed: $v = 2\pi r/T = 2\pi(0.800 \text{ m})/(0.600 \text{ s}) = (5.027 \text{ m})/(0.600 \text{ s}) = 8.38 \text{ m/s}$ . Keep the unrounded 8.378 m/s for the next step.
Compute the acceleration: $a_c = v^2/r = (8.378 \text{ m/s})^2/(0.800 \text{ m}) = (70.19 \text{ m}^2/\text{s}^2)/(0.800 \text{ m}) = 87.7 \text{ m/s}^2$ .
Cross-check with the direct period formula: $a_c = 4\pi^2 r/T^2 = 4\pi^2(0.800 \text{ m})/(0.600 \text{ s})^2 = (31.58 \text{ m})/(0.360 \text{ s}^2) = 87.7 \text{ m/s}^2$ . Same result.
Find the force: $F_c = ma_c = (0.150 \text{ kg})(87.7 \text{ m/s}^2) = 13.2 \text{ N}$ , supplied by the cord's tension.

$v = 8.38 \text{ m/s}$ , $a_c = 87.7 \text{ m/s}^2$ , and $F_c = 13.2 \text{ N}$ toward the center, supplied by tension in the cord.

Minimum speed at the top of a vertical circle

You swing a bucket of water in a vertical circle of radius 1.20 m. What minimum speed at the top of the circle keeps the water in the bucket?

At the top, both gravity and the normal force from the bucket bottom point downward, toward the center. Newton's second law along the radius gives $mg + F_N = mv^2/r$ .
At minimum speed the water is on the verge of leaving the bucket, so the contact force drops to $F_N = 0$ . Gravity alone supplies the centripetal force: $mg = mv^2/r$ .
Mass cancels, so solve for speed: $v = \sqrt{gr} = \sqrt{(9.8 \text{ m/s}^2)(1.20 \text{ m})} = \sqrt{11.76 \text{ m}^2/\text{s}^2} = 3.43 \text{ m/s}$ .

The minimum speed at the top is $3.43 \text{ m/s}$ . At exactly this speed gravity by itself is the centripetal force; any slower and gravity exceeds the required $mv^2/r$ , so the water leaves the circular path.

Frequently asked questions

Is centripetal force a real force?

No. Centripetal force is the name for the net inward force required for circular motion, equal to mv^2/r. It is always supplied by real forces: tension, friction, gravity, a normal force, or some combination. On a free-body diagram you draw those real forces, never a separate centripetal arrow.

Can I use the period instead of the speed?

Yes. In one period T the object covers one full circumference, so v = 2(pi)r/T. The calculator's period mode applies this conversion first, then computes acceleration and force as usual. If you are given rpm, convert to a period first: T in seconds is 60 divided by the rpm value.

Why is there an acceleration if the speed is constant?

Acceleration is any change in velocity, and velocity includes direction. In uniform circular motion the direction of motion changes continuously, so the object accelerates toward the center at v^2/r even though the speedometer reading never changes.

What units does the calculator expect?

SI units throughout: meters per second for speed, meters for radius, kilograms for mass, seconds for period. Outputs are m/s^2 for acceleration and newtons for force. Watch for problems that give a diameter (halve it to get the radius) or rpm (convert to seconds per revolution).

What about centrifugal force?

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 turns under you. In the inertial reference frames used in AP Physics 1, no outward centrifugal force acts on the object, and drawing one on a free-body diagram costs points.