Centrifugal Force Formula: F = mω²r, Pseudo Force Explained

The centrifugal force formula is $F = m\omega^2 r$ , which is the same as $\frac{mv^2}{r}$ in terms of speed. It gives the outward push that an observer turning with a rotating system feels on a mass m at distance r from the axis. It has the same size as the centripetal force but the opposite direction.

The key word is observer. Someone standing on the ground watching a spinning merry-go-round sees no outward force at all, only an inward one. A rider on it feels pushed outward, and to make Newton's laws work in her turning frame she must add this extra term. That is why centrifugal force is called a pseudo force (or fictitious force): it is real in its effects for the rider, but no other body exerts it.

Quick facts

Formula
$F = m\omega^2 r = \frac{mv^2}{r}$
SI unit
newton (N)
Dimensional formula
[M¹ L¹ T⁻²]
Direction
Radially outward from the axis
Nature
Pseudo force, only in a rotating (non inertial) frame
Reaction force
None, since no body exerts it
Syllabus
Class 11 laws of motion, JEE

Every term of the centrifugal force formula

Symbols for F = mω²r
Symbol	Meaning	SI unit
$F$	Outward pseudo force in the rotating frame	N
$m$	Mass of the object	kg
$\omega$	Angular velocity of the frame	rad/s
$r$	Perpendicular distance from the rotation axis	m

Because ω is squared, a small rise in spin rate makes a big difference. A washing machine that goes from 600 to 1200 rpm puts four times the outward force on the wet clothes. Use $\frac{mv^2}{r}$ when the speed is given; $v = \omega r$ links the two forms, as on the angular velocity page.

Real force or pseudo force

Newton's laws hold in an inertial frame, one that is not accelerating. A rotating platform is accelerating toward its centre at every point, so it is not inertial. To keep using $F = ma$ inside it, you add a pseudo force of $-m\vec{a}_{frame}$ to every mass. For rotation that term points outward with size $m\omega^2 r$ : the centrifugal force.

In the ground frame you write: net inward real force = mv²/r. In the rotating frame you write: real forces + centrifugal force = 0. Both give the same tension, friction or normal force. Mixing the two in one diagram is the most common way to get a wrong answer.

A moving object in a rotating frame also feels a Coriolis force. School and JEE problems usually involve bodies at rest in the rotating frame, where only the centrifugal term matters.

Solved examples

Example 1: a rider on a merry-go-round

A 60 kg rider sits 2 m from the centre of a platform turning at 1.5 rad/s. What outward force does she feel?

$F = m\omega^2 r = 60 \times (1.5)^2 \times 2 = 60 \times 2.25 \times 2 = 270\ \text{N}$

Friction from the seat must provide 270 N inward to keep her in place.

Example 2: a sock in a spin dryer

A 50 g wet sock sits against the drum of radius 25 cm turning at 1200 rpm. Find the centrifugal force on it.

Convert: m = 0.05 kg, r = 0.25 m, $\omega = \frac{2\pi \times 1200}{60} = 40\pi \approx 125.7\ \text{rad/s}$ .

$F = 0.05 \times (125.7)^2 \times 0.25 \approx 197\ \text{N}$

That is about 400 times the sock's weight, which squeezes the water out through the holes.

Example 3: Earth's spin at the equator

Earth turns at ω = 7.27 × 10⁻⁵ rad/s and its radius is 6.37 × 10⁶ m. Find the centrifugal acceleration at the equator.

$\omega^2 R = (7.27 \times 10^{-5})^2 \times 6.37 \times 10^6 \approx 0.034\ \text{m/s}^2$

This small outward term is one reason the measured g is a little lower at the equator than at the poles.

Where you meet it

Cream separators and lab centrifuges, which drive denser parts outward.
The tilt a passenger feels in a turning bus, and the outward lean of liquid in a spinning bucket.
The variation of acceleration due to gravity with latitude, $g' = g - \omega^2 R\cos^2\phi$ .
Design of banked roads, when the problem is solved from the driver's point of view.

Centrifugal and centripetal force compared

Two names, two frames
Feature	Centripetal	Centrifugal
Frame	Inertial (ground)	Rotating
Direction	Toward centre	Away from centre
Real?	Yes, it is a net real force	No, a pseudo force
Size	$\frac{mv^2}{r}$	$\frac{mv^2}{r}$
Has a reaction pair	Yes	No

They are not an action reaction pair. Action and reaction act on two different bodies; these two are alternative descriptions of the same body in two frames. More circular motion results sit on the rotation and circles hub.

Frequently asked questions

What is the formula of centrifugal force?

Centrifugal force is F = mω²r, equal to mv² / r. It acts outward on a mass m at distance r from the axis of a frame turning at angular velocity ω.

Is centrifugal force a real force?

No. It is a pseudo force that appears only when you describe motion from a rotating frame. A ground observer explains the same motion with just the real inward pull.

What is the unit of centrifugal force?

It is measured in newtons, like any force, with dimensions [M¹ L¹ T⁻²].

Are centripetal and centrifugal forces action and reaction?

No. Newton's third law pairs act on different bodies, while these two refer to the same body seen from two different frames.

Why does water fly out of clothes in a spin dryer?

In the drum's frame, the water feels a large outward force mω²r. The drum wall holds the cloth in, but water can pass through the holes, so it leaves.