In short

Centripetal force is not a new kind of force — it is the name for whatever real force (tension, gravity, friction, normal force) acts toward the centre of a circular path. Any object moving in a circle of radius r at speed v needs a net inward force of F_c = \frac{mv^2}{r} = m\omega^2 r. If that force disappears, the object flies off along the tangent.

A car turns left on a Delhi flyover. You are sitting in the passenger seat, and your body presses against the right-side door. It feels like something is pushing you outward — but nothing is. What is actually happening is that the car is being pushed inward (by friction between the tyres and the road), and your body, which wants to keep going in a straight line, has to be dragged along. That inward push — the one that bends the car's path from a straight line into a curve — is what physics calls centripetal force.

The word comes from Latin: centrum (centre) + petere (to seek). Centre-seeking. And the single most important thing about it is this: centripetal force is not a new force. You will never draw it on a free body diagram as a separate arrow labelled "F_c." It is always some real, identifiable force — gravity, tension, friction, the normal force — that happens to point toward the centre of the circle. The label "centripetal" tells you the role the force is playing, not what kind of force it is.

The physics: Newton's second law in a circle

You already know from uniform circular motion that an object moving in a circle at constant speed is accelerating. The velocity vector changes direction every instant, and that change in direction requires an acceleration pointing toward the centre:

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

where v is the speed and r is the radius of the circle.

Now apply Newton's second law. If there is an acceleration, there must be a net force causing it. That net force must point in the same direction as the acceleration — toward the centre:

F_c = ma_c = \frac{mv^2}{r} \tag{1}

Why: Newton's second law says \vec{F}_{\text{net}} = m\vec{a}. Since \vec{a} points radially inward (centripetal acceleration), the net force must also point radially inward. The magnitude is m times the centripetal acceleration.

You can also write this using angular velocity \omega, since v = \omega r:

F_c = \frac{m(\omega r)^2}{r} = m\omega^2 r \tag{2}

Why: substitute v = \omega r into equation (1). The r in the numerator partially cancels the r in the denominator, leaving one power of r in the numerator.

Equations (1) and (2) are the same physics written two ways. Use whichever form matches the data you have — speed v or angular velocity \omega.

The critical point: Equation (1) does not create a new force. It tells you how much net inward force is needed. Your job, in every problem, is to look at the physical situation and identify which real force (or combination of forces) supplies that inward push.

Identifying the real force — the whole game

Here is the question that separates students who understand centripetal force from those who memorise it: what is providing the centripetal force in this situation?

Four examples showing which real force provides centripetal force Four panels: (1) stone on a string — tension provides centripetal force, (2) car on a flat road — friction provides it, (3) satellite in orbit — gravity provides it, (4) clothes in a washing machine drum — normal force from the drum wall provides it. Stone on a string T v Tension → centripetal force Car on a flat curve f v Friction → centripetal force Satellite in orbit Earth mg v Gravity → centripetal force Clothes in a drum N v Normal force → centripetal force
Four situations, four different forces — but in each case, the real force points toward the centre. Centripetal force is the role, not the actor.

The table below summarises the most common scenarios:

Situation What provides F_c Direction of F_c
Stone whirled on a string (horizontal) Tension in the string Along the string, toward your hand
Car turning on a flat road Static friction between tyres and road Toward the centre of the curve
Satellite orbiting Earth Gravitational pull of Earth Toward Earth's centre
Clothes in a washing machine drum Normal force from the drum wall Toward the drum's axis
Ball on the inside of a bowl Normal force from the bowl surface Toward the centre of curvature
Electron orbiting a nucleus Electrostatic (Coulomb) force Toward the nucleus

In every case, you can point at a real, physical contact or field force. There is no mysterious "centripetal force" that appears from nowhere. If someone asks "what provides the centripetal force?", they are asking you to name the real force and explain why it points inward.

Deriving F_c = mv^2/r from Newton's second law

The formula deserves a proper derivation, not just a statement. Start from Newton's second law applied to circular motion.

Assumptions: The object moves in a circle of constant radius r at constant speed v. The net force on the object is entirely radial (no tangential component — the speed is not changing).

Step 1. The object moves along a circular arc. At any instant, its velocity \vec{v} is tangent to the circle.

Why: by definition, velocity points along the direction of motion. For motion along a curve, that direction is always tangent to the curve at that point.

Step 2. In a small time interval \Delta t, the velocity vector rotates by a small angle \Delta\theta = \frac{v\,\Delta t}{r}.

Why: the object covers arc length v\,\Delta t, and the angle subtended by an arc of length s on a circle of radius r is \theta = s/r.

Step 3. The change in the velocity vector has magnitude:

|\Delta\vec{v}| = v\,\Delta\theta = v \cdot \frac{v\,\Delta t}{r} = \frac{v^2}{r}\,\Delta t

Why: for a small angle, the magnitude of the change in a vector of fixed length v that rotates by \Delta\theta is v\,\Delta\theta. This change points radially inward — toward the centre of the circle — because the velocity vector rotated but did not change in length.

Step 4. The acceleration is the rate of change of velocity:

a_c = \frac{|\Delta\vec{v}|}{\Delta t} = \frac{v^2}{r}

This acceleration points toward the centre. It is the centripetal acceleration.

Step 5. Apply Newton's second law. The net force equals mass times acceleration:

\boxed{F_c = ma_c = \frac{mv^2}{r}} \tag{1}

Why: Newton's second law — \vec{F}_{\text{net}} = m\vec{a} — gives both the magnitude and the direction. The net force must point inward, toward the centre, because that is where the acceleration points.

This is the centripetal force equation. It does not introduce a new force. It tells you the required magnitude of the net inward force for an object of mass m to follow a circular path of radius r at speed v.

What happens when the force is not enough

Tie a stone to a string and whirl it in a horizontal circle. Now imagine the string breaks. What happens?

The stone does not fly radially outward, away from the centre. It flies off along the tangent — in the direction it was moving at the instant the string snapped. This is Newton's first law: without a net force, the stone continues in a straight line along its instantaneous velocity.

Object flying off along the tangent when centripetal force disappears A dashed circle shows the original path. The string breaks at the top of the circle. The object continues along a straight tangent line to the right. A dotted radial line shows the direction outward from centre — the object does NOT go that way. centre actual path (tangent) NOT this way string (broken) direction of prior motion v (at break)
When the string breaks, the stone flies off along the tangent — not radially outward. The "outward push" you feel is your body resisting the inward acceleration, not a real outward force.

This is the answer to the common question "why do things fly outward in a circle?" They don't fly outward. They fly forward — along the tangent — because the inward force that was bending their path into a curve vanished. The sensation of being "pushed outward" in a turning car is your body's inertia resisting the inward acceleration. In a non-inertial (rotating) reference frame, you can introduce a fictitious centrifugal force to account for this — but in the ground frame, the only real force is the one pointing inward.

Think of it this way: when you spin a ball on a Diwali sparkler string and let go, the ball shoots off in a straight line. It does not spiral outward. It leaves along the tangent at the instant of release. This is centripetal force — or rather, its absence — in action.

How much centripetal force? Putting in numbers

To build intuition for the magnitudes involved, consider a few quick calculations.

A merry-go-round. A child of mass 30 kg sits on a merry-go-round of radius 2 m that completes one revolution every 4 seconds. The angular velocity is \omega = 2\pi/T = 2\pi/4 = \pi/2 rad/s. The centripetal force needed is:

F_c = m\omega^2 r = 30 \times \left(\frac{\pi}{2}\right)^2 \times 2 = 30 \times 2.47 \times 2 \approx 148 \text{ N}

That is about half the child's weight. The force comes from friction between the child and the seat (and their grip on the railing). If the merry-go-round spins faster, the required force increases as \omega^2 — and at some speed, friction alone cannot provide enough, and the child slides outward.

An ISRO satellite. A satellite in low Earth orbit (altitude ~400 km) orbits at about 7,670 m/s with r \approx 6{,}770 km. For a 500 kg satellite:

F_c = \frac{mv^2}{r} = \frac{500 \times (7670)^2}{6{,}770{,}000} \approx 4{,}340 \text{ N}

This entire centripetal force is provided by Earth's gravity. The satellite is in free fall — gravity is bending its path into a circle. No string, no friction, no engine thrust. Just gravity, doing the job of the centripetal force.

Worked examples

Example 1: Ball on a string — finding the tension

A ball of mass 500 g is attached to a string of length 1.2 m and swung in a horizontal circle at 3 m/s. Neglect gravity (imagine the circle is horizontal and the string is horizontal). Find the tension in the string.

Free body diagram: ball on a horizontal string A ball is shown at the right end of a horizontal string. The tension force T points left along the string toward the centre. The velocity v points upward (tangent to the circle). A dashed circle shows the circular path. The centre of the circle is marked at the left end of the string where the hand holds it. Coordinate axes show the radial direction (inward, positive) and the tangential direction. centre (hand) m FBD of ball T v +r (inward) +tangent r = 1.2 m v = 3 m/s m = 0.5 kg
Left: the ball moves in a horizontal circle, with the string providing the inward tension. Right: the free body diagram shows only one horizontal force — tension, pointing toward the centre. This tension is the centripetal force.

Step 1. Identify the forces on the ball.

The only horizontal force is the tension T in the string, directed radially inward (toward the centre). Gravity acts downward and the string's vertical component balances it — but since the problem says to neglect gravity (horizontal circle), the entire tension is the centripetal force.

Why: the free body diagram is the starting point of every force problem. For horizontal circular motion with gravity neglected, the only force acting inward is the string tension.

Step 2. Apply Newton's second law in the radial direction.

T = F_c = \frac{mv^2}{r}

Why: the net inward force equals the required centripetal force. Here, the tension is the only inward force, so T = mv^2/r directly.

Step 3. Substitute the values: m = 0.5 kg, v = 3 m/s, r = 1.2 m.

T = \frac{0.5 \times 3^2}{1.2} = \frac{0.5 \times 9}{1.2} = \frac{4.5}{1.2}
\boxed{T = 3.75 \text{ N}}

Why: plug in the numbers. The tension is 3.75 N — roughly the weight of a 380 g object. Not a huge force, but enough to keep a 500 g ball on a 1.2 m string moving at 3 m/s.

Step 4. Check: does the answer make physical sense?

If the ball moved faster — say 6 m/s instead of 3 m/s — the required tension would be \frac{0.5 \times 36}{1.2} = 15 N, which is four times as much. The force scales as v^2, so doubling the speed quadruples the tension. Spin a ball too fast, and the string snaps — it cannot supply the required centripetal force.

Result: The tension in the string is 3.75 N. The tension is the centripetal force — it is the real force that bends the ball's path from a straight line into a circle.

What this shows: You never write "F_c" as a separate force on the FBD. The tension does the job. The label "centripetal" just tells you that the tension is directed toward the centre and has magnitude mv^2/r.

Example 2: Car on a flat circular track — maximum safe speed

A car of mass 1,200 kg drives around a flat (unbanked) circular track of radius 50 m. The coefficient of static friction between the tyres and the road is \mu_s = 0.4. Find the maximum speed at which the car can take the turn without skidding.

Top-down free body diagram: car on a flat circular track Top-down view showing a car on a circular track. The friction force points inward (toward the centre of the curve) and the velocity points along the tangent. A side-view FBD shows weight downward, normal force upward, and friction inward. Top-down view centre car f v r = 50 m Side-view FBD mg N f (toward centre) Vertical: N = mg Radial: f = mv²/r
Left: top-down view showing friction pointing inward as the centripetal force. Right: side-view FBD — the normal force balances weight vertically, while friction acts horizontally toward the centre of the curve.

Step 1. Identify the forces on the car.

Three forces act on the car: weight mg downward, normal force N upward, and static friction f horizontally toward the centre of the curve. Vertically, the car does not accelerate, so N = mg. The only horizontal force is friction, directed inward — this is the centripetal force.

Why: on a flat road, there is no banking. The only force available to push the car inward is friction between the tyres and the tarmac. Gravity and the normal force are both vertical and cancel each other out.

Step 2. Write Newton's second law in the radial direction.

f = \frac{mv^2}{r}

The maximum static friction is f_{\max} = \mu_s N = \mu_s mg. The car skids when the required centripetal force exceeds this maximum.

Why: static friction can provide any force up to \mu_s N, after which the tyres lose grip and the car skids outward (along the tangent, not radially outward).

Step 3. Set the maximum friction equal to the required centripetal force and solve for v_{\max}.

\mu_s mg = \frac{mv_{\max}^2}{r}

The mass m cancels:

\mu_s g = \frac{v_{\max}^2}{r}
v_{\max}^2 = \mu_s g r
v_{\max} = \sqrt{\mu_s g r} = \sqrt{0.4 \times 9.8 \times 50}
v_{\max} = \sqrt{196} = 14 \text{ m/s}
\boxed{v_{\max} = 14 \text{ m/s} \approx 50 \text{ km/h}}

Why: the mass cancels — the maximum safe speed does not depend on the car's mass. A heavy truck and a light hatchback can both take this curve at the same maximum speed, because both the required centripetal force and the available friction scale linearly with mass.

Step 4. Interpret the result.

14 m/s is about 50 km/h. On a wet road where \mu_s drops to 0.2, the maximum speed falls to \sqrt{0.2 \times 9.8 \times 50} = \sqrt{98} \approx 9.9 m/s \approx 36 km/h. This is why speed limits on curves are lower when the road is wet — less friction means less centripetal force available.

Result: The maximum speed is 14 m/s (about 50 km/h). Beyond this, static friction cannot supply the required centripetal force, and the car skids outward along the tangent.

What this shows: On a flat road, friction is the only force that keeps a car turning. The maximum speed depends on \mu_s, g, and r — not on the car's mass. Banking the road (tilting the surface inward) provides additional centripetal force from the normal force's horizontal component, allowing higher speeds. That is the topic of banking of roads.

Common confusions

If you came here to understand what centripetal force is, identify it in different situations, and use F_c = mv^2/r, you have everything you need. What follows is for readers who want the vector derivation and the connection to non-uniform circular motion.

Vector derivation of centripetal acceleration

The derivation above used the magnitude of \Delta\vec{v}. Here is the full vector treatment.

Place the origin at the centre of the circle. At time t, the position vector of the object is:

\vec{r}(t) = r\cos(\omega t)\,\hat{i} + r\sin(\omega t)\,\hat{j}

Why: parametrise the circular path using the angular position \theta = \omega t, where \omega is the constant angular velocity.

Differentiate to get the velocity:

\vec{v}(t) = \frac{d\vec{r}}{dt} = -r\omega\sin(\omega t)\,\hat{i} + r\omega\cos(\omega t)\,\hat{j}

Why: apply the chain rule. The negative sine and positive cosine mean the velocity is perpendicular to the position vector (tangent to the circle), as expected.

Differentiate again to get the acceleration:

\vec{a}(t) = \frac{d\vec{v}}{dt} = -r\omega^2\cos(\omega t)\,\hat{i} - r\omega^2\sin(\omega t)\,\hat{j}
\vec{a}(t) = -\omega^2\left[r\cos(\omega t)\,\hat{i} + r\sin(\omega t)\,\hat{j}\right] = -\omega^2\vec{r}(t)

Why: the acceleration is -\omega^2 times the position vector. The minus sign means it points radially inward — toward the centre — opposite to \vec{r}. Its magnitude is \omega^2 r = v^2/r.

This is the cleanest statement of centripetal acceleration: \vec{a} = -\omega^2\vec{r}. It points toward the centre, has magnitude v^2/r, and requires a net force \vec{F} = m\vec{a} = -m\omega^2\vec{r} to sustain it.

When the speed changes — non-uniform circular motion

If the speed is not constant, the object has two components of acceleration:

  1. Centripetal (radial): a_c = v^2/r, directed inward, responsible for changing the direction.
  2. Tangential: a_t = dv/dt, directed along the tangent, responsible for changing the speed.

The total acceleration has magnitude:

a = \sqrt{a_c^2 + a_t^2} = \sqrt{\left(\frac{v^2}{r}\right)^2 + \left(\frac{dv}{dt}\right)^2}

The net force must now have both a radial component (mv^2/r inward) and a tangential component (m\,dv/dt along the tangent). A stone whirled in a vertical circle is a classic example: at every point, the centripetal direction changes, and the speed varies because gravity has a tangential component. The full treatment appears in Motion in a Vertical Circle.

Why mass cancels in the car-on-curve problem

The result v_{\max} = \sqrt{\mu_s g r} is mass-independent. This is not a coincidence — it is the same physics that makes all objects fall at the same rate in vacuum. Both the required centripetal force (mv^2/r) and the maximum available friction (\mu_s mg) are proportional to m. When you set them equal, the mass divides out. This means a motorcycle and a truck can take the same unbanked curve at the same maximum speed — provided their tyres have the same coefficient of friction with the road.

In practice, heavy vehicles have a higher centre of gravity and tend to topple before they skid, which is a different failure mode. The centripetal force analysis tells you when skidding starts; a separate torque analysis (beyond the scope of this article) tells you when rolling over starts. Trucks often roll before they skid — which is why truck speed limits on curves are lower than car speed limits.

Where this leads next