Circular Motion: Formulas, Centripetal Force & Examples

Circular motion is motion along a circular path, and the key fact is that the velocity direction changes continuously even when the speed does not. Because velocity is changing, the object is always accelerating, and that acceleration points toward the center of the circle.

For an object moving at speed $v$ on a circle of radius $r$, the centripetal acceleration is

and the inward net force required to maintain the motion is $F_c = m v^2 / r$. Everything else in circular motion builds on these two relations.

The Core Quantities of Circular Motion

Circular motion problems use a small set of quantities that all connect to each other.

The conversion $v = \omega r$ is the bridge between the "angular" picture and the "linear" picture. If a problem gives you revolutions per minute, convert to $\omega$ in rad/s first, then move to linear quantities only if you need them.

Centripetal Acceleration and Centripetal Force

Even at constant speed, the velocity vector keeps turning. A turning velocity vector means acceleration, and for circular motion that acceleration points toward the center:

By Newton's second law, some real force must supply $F_c = m a_c$. Centripetal force is not a new kind of force — it is the inward role played by tension, gravity, friction, or a normal force, depending on the situation.

Uniform vs Non-Uniform Circular Motion

In uniform circular motion, the speed is constant. The acceleration is purely centripetal, and its magnitude $v^2/r$ never changes — only its direction does.

In non-uniform circular motion, the speed changes too. Then the acceleration has two perpendicular components: the centripetal part $v^2/r$ (turning the velocity) and a tangential part $a_t = \frac{dv}{dt}$ (changing the speed). The total acceleration magnitude is

A pendulum swinging through its arc and a car speeding up around a bend are both non-uniform cases.

Worked Example: Ball on a String

A $0.5\ \mathrm{kg}$ ball swings in a horizontal circle of radius $0.8\ \mathrm{m}$, completing one revolution every $0.4\ \mathrm{s}$.

Step 1 — find the angular velocity.

Step 2 — find the speed.

Step 3 — find the required inward force.

The string tension must supply roughly $99\ \mathrm{N}$ inward. Notice we never needed $v$ explicitly — $F_c = m\omega^2 r$ works directly when the period is given.

Worked Example: Satellite in Low Orbit

A satellite orbits Earth where gravity provides the centripetal force. Setting gravity equal to the required inward force:

For $r = 6.8 \times 10^6\ \mathrm{m}$ and $GM = 3.98 \times 10^{14}\ \mathrm{m^3/s^2}$:

The satellite's mass cancels — every object at that radius orbits at the same speed. This is the standard pattern: set the real inward force equal to $m v^2 / r$ and solve for the unknown.

Common Mistakes in Circular Motion

Adding a separate "centripetal force" to the diagram

Draw only the real forces (tension, gravity, friction, normal). Centripetal force is the inward component of their sum, not an extra arrow.

Drawing an outward force

In an inertial frame nothing pushes the object outward. The outward "feel" is your body trying to continue in a straight line while the path curves.

Mixing angular and linear units

Using rpm directly in $v^2/r$, or degrees instead of radians in $\omega$, silently breaks the formulas. Convert to rad/s before anything else.

Assuming constant speed means zero acceleration

If the direction of motion is changing, acceleration is nonzero. Constant speed only removes the tangential component.

Connecting Circular Motion to Other Topics

Circular motion is the entry point to rotation and orbital mechanics. The same $\omega$ appears in rotational kinematics and angular momentum; the satellite setup above is the first step toward Kepler's laws; and banked curves combine circular motion with free-body analysis. If you can confidently set "real inward force $= m v^2 / r$," most of those chapters open up.

As a quick self-check: take the ball-on-a-string example and halve the period. Since $F_c = m\omega^2 r$ and $\omega$ doubles, the required tension becomes four times larger.