Centripetal Force Formula: F = mv²/r, Unit and Examples

The centripetal force formula is $F_c = \frac{mv^2}{r}$ : multiply the mass by the square of its speed and divide by the radius of the circle it follows. In terms of angular velocity the same force is $F_c = m\omega^2 r$ . The force always points toward the centre of the circle, and it is measured in newtons.

Centripetal force is not a new kind of force. It is the name for whatever net force happens to point inward: tension in a string, friction under a car's tyres, gravity on a satellite, or the normal force of a wall. Newton's second law, $F = ma$ , combined with the centripetal acceleration $a_c = \frac{v^2}{r}$ , gives the formula directly.

Quick facts

Formula
$F_c = \frac{mv^2}{r}$
With ω
$F_c = m\omega^2 r$
SI unit
newton (N)
Dimensional formula
[M¹ L¹ T⁻²]
Direction
Toward the centre, at right angles to the velocity
Work done
Zero for uniform circular motion, since force is perpendicular to motion
Where used
Turns, loops, orbits, banked roads, spinning machines

What the letters in the centripetal force formula stand for

Inputs to F = mv²/r
Symbol	Stands for	Unit to use
$F_c$	Net inward force needed for the circle	N
$m$	Mass of the body	kg
$v$	Speed along the path (tangential speed)	m/s
$r$	Distance from the middle of the circle to the body	m
$\omega$	Angular velocity, with $v = \omega r$	rad/s

Notice the square on v. Double the speed round the same bend and the inward force needed goes up four times. Halve the radius at the same speed and the force doubles. These two facts explain why sharp, fast turns are the ones that skid. The angular velocity formula gives ω when a problem states revolutions instead of a speed.

Who supplies the inward force

Real sources of centripetal force
Situation	Force acting inward	Equation you write
Stone whirled on a string, horizontal circle	Tension	$T = \frac{mv^2}{r}$
Car on a flat bend	Static friction	$\mu_s mg \ge \frac{mv^2}{r}$
Satellite in a circular orbit	Gravity	$\frac{GMm}{r^2} = \frac{mv^2}{r}$
Electron in the Bohr model	Electric attraction	$\frac{ke^2}{r^2} = \frac{mv^2}{r}$
Car on a banked road, no friction	Horizontal part of the normal force	$N\sin\theta = \frac{mv^2}{r}$

Draw a free body diagram and list only the real forces. Then add up their components toward the centre and set that sum equal to mv²/r. Never draw an extra arrow labelled centripetal force alongside them; that counts the same force twice. Banked curves, the final row, are worked out in full on the banking of road formula, and the gravity row leads to the orbital velocity formula.

Worked problems

Example 1: a car on a curve

A 1200 kg car takes a flat curve of radius 50 m at 15 m/s. How much inward force does the road provide?

$F_c = \frac{1200 \times 15^2}{50} = \frac{1200 \times 225}{50} = 5400\ \text{N}$

Example 2: stone on a string, using rev/s

A 0.5 kg stone on a 0.8 m string makes 2 revolutions every second in a horizontal circle. Find the tension, ignoring gravity's tilt.

$\omega = 2\pi f = 2\pi \times 2 = 4\pi \approx 12.57\ \text{rad/s}$ .

$T = m\omega^2 r = 0.5 \times (12.57)^2 \times 0.8 \approx 63.2\ \text{N}$

Example 3: least friction coefficient, with km/h

A 1000 kg car rounds a flat bend of radius 80 m at 72 km/h. What is the smallest coefficient of static friction that stops it sliding? Take g = 9.8 m/s².

72 km/h is 20 m/s after dividing by 3.6.

$F_c = \frac{1000 \times 400}{80} = 5000\ \text{N}$ . Friction must supply this: $\mu_s \times 1000 \times 9.8 \ge 5000$ .

$\mu_s \ge \frac{5000}{9800} \approx 0.51$

The mass cancels, so the same bend needs μ of about 0.51 for any car at that speed.

Deriving F = mv²/r

A body moving round a circle at steady speed still changes direction, so its velocity changes. Over a short time the velocity vector turns through a small angle, and the change in velocity points toward the centre. Working through the similar triangles of position and velocity gives an acceleration of size $\frac{v^2}{r}$ . Newton's second law then fixes the force: $F = ma = \frac{mv^2}{r}$ . Substituting $v = \omega r$ gives $m\omega^2 r$ .

The force does no work because it is always perpendicular to the motion. That is why the speed stays the same even though a force acts all the time.

Centripetal and centrifugal compared

Seen from the ground, only the inward centripetal force exists. A passenger inside the turning car feels thrown outward; that is the centrifugal force, a pseudo force used only in the rotating frame. It has the same size, $\frac{mv^2}{r}$ , but points away from the centre. Pick one frame and stay in it. The circular motion section shows how both views lead to the same answers.

Traps to avoid

Leaving speed in km/h. A factor of 3.6 squared is about 13, so the error is large.
Using diameter for r. The radius is half the diameter.
Counting centripetal force as an extra force in the diagram.
For a vertical circle, forgetting that weight also has a component toward the centre. At the top of a loop, mg and the track force both point down, so $N + mg = \frac{mv^2}{r}$ .
Thinking the body flies outward when the string snaps. It moves off along the tangent, in a straight line.

Frequently asked questions

What is the formula of centripetal force?

It is F = mv² / r, or F = mω²r using angular velocity. It gives the net inward force in newtons needed to keep a mass m moving at speed v on a circle of radius r.

What is the formula for centripetal force in terms of period?

Put v = 2πr / T into mv² / r to get F = 4π²mr / T². This form is handy when a problem gives the time for one revolution.

Does centripetal force do work?

No. In uniform circular motion it stays perpendicular to the direction of motion, so it does zero work and the kinetic energy does not change.

Which direction does the force point?

Always toward the centre of the circle. Its direction keeps turning as the body moves, so it is not a constant vector even when its size is constant.

How is centripetal force different from centrifugal force?

Centripetal force is the real net inward force seen from an inertial frame. Centrifugal force is an outward pseudo force of the same size, used only by an observer who rotates with the body.