Centripetal Force

Centripetal force is the inward part of the net force that keeps an object moving in a circle. If that inward force disappears, the object stops curving and moves off along a tangent.

For motion on a circle of radius $r$ at speed $v$, the required inward force magnitude is

$$F_c = \frac{m v^2}{r}$$

This formula tells you how much inward force is needed. It does not tell you which physical force provides it.

You may also see the same relationship written as

$$F_c = m \omega^2 r$$

because $v = \omega r$ for circular motion.

What Centripetal Force Actually Means

Centripetal force is not an extra force you add to a free-body diagram. It is the name for the net force pointing toward the center. In one problem that force may be tension. In another, it may be gravity, friction, or the normal force.

If the speed is constant, the velocity still changes because its direction keeps turning. That directional change requires centripetal acceleration:

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

If the speed is changing too, then there is also a tangential component of acceleration. In that case, $m v^2 / r$ still gives the inward part of the net force, but not the entire net force.

Why The Force Points Inward

An object in motion tends to keep moving in its current direction. To keep bending that motion into a circle, the net force has to keep pointing inward.

A ball on a string is the standard example. The string tension pulls toward the center, so the ball's velocity keeps turning. If the string breaks, there is no new outward force. The ball simply continues in the direction it was already moving at that instant.

Worked Example: Car On A Flat Curve

Suppose a car of mass $m = 1200\ \mathrm{kg}$ moves around a flat circular curve of radius $r = 50\ \mathrm{m}$ at speed $v = 15\ \mathrm{m/s}$.

Start with the centripetal-force formula:

$$F_c = \frac{m v^2}{r}$$

Now substitute the values:

$$F_c = \frac{(1200)(15^2)}{50} = \frac{(1200)(225)}{50} = 5400\ \mathrm{N}$$

So the car needs $5400\ \mathrm{N}$ of net force toward the center of the curve. On a flat road, that inward force is usually provided by static friction between the tires and the road.

That last sentence is the real physics step. The formula gives the required inward force, but you still have to identify which real force supplies it.

Common Mistakes With Centripetal Force

Treating it like a separate force

Ask which real force points inward. Do not add a second force called "centripetal force" unless you mean the inward net result of the actual forces.

Calling circular motion balanced

If the path keeps curving, the net force is not zero. A nonzero inward force is required.

Ignoring the radius

For the same mass and speed, a smaller radius means a sharper turn, so it needs a larger centripetal force.

Inventing an outward force in an inertial frame

In an inertial frame, the required force points inward. The "outward" feeling in a turning car comes from your body's tendency to keep moving straight while the car changes direction.

Where You Use Centripetal Force

Centripetal force shows up whenever motion follows a circular path or a path that is well approximated as circular. Common examples include cars turning, objects on strings, roller-coaster loops, satellites, and planets in orbit.

In many problems, the main job is to connect circular motion to another force law. You may set tension equal to $m v^2 / r$, or gravity equal to $m v^2 / r$, depending on the situation.

Try A Similar Problem

Keep the same car and radius, but double the speed from $15\ \mathrm{m/s}$ to $30\ \mathrm{m/s}$. Because the force depends on $v^2$, the required centripetal force becomes four times larger, not two times larger.

If you want another case after that, try your own version with different numbers and check which real force would provide the inward pull.