Horizontal Circular Motion

Horizontal circular motion happens whenever an object travels at a steady speed around a circular path that lies entirely in a horizontal plane, like a ball swung around on a string above your head, a car turning on a flat road, or the bob of a conical pendulum tracing a circle beneath a fixed point. Even though the speed stays constant, the direction of the velocity is always changing, so the object is accelerating. That acceleration always points toward the center of the circle, which is why it is called centripetal acceleration.

This lesson focuses on the case where the circle itself is horizontal. If the circular path is oriented vertically instead, such as a ball on a string swinging in a vertical loop or a roller coaster going over a hill, the analysis changes because gravity acts along the direction of motion rather than only perpendicular to it — that situation is covered separately in vertical circular motion.

For any object moving at constant speed \(v\) around a circle of radius \(r\), the centripetal acceleration is:

\(a_c = \dfrac{v^2}{r}\)

Newton's second law then tells us that a net inward force must exist to produce that acceleration. This is the centripetal force:

\(F_c = m a_c = \dfrac{mv^2}{r}\)

It helps to remember that "centripetal force" is not a new, separate type of force. It is simply the name given to whatever combination of real forces (tension, friction, a component of the normal force, or a combination of these) happens to point toward the center and supply the required inward pull. In horizontal circular motion, gravity usually does not contribute directly to this inward force because gravity acts vertically while the circle lies flat.

If the object completes one full revolution in a period \(T\), the speed can also be written as \(v = \dfrac{2\pi r}{T}\), and using angular velocity \(\omega = \dfrac{2\pi}{T} = \dfrac{v}{r}\), the same formulas become \(a_c = \omega^2 r\) and \(F_c = m\omega^2 r\). These forms are useful when a problem gives you a rotation rate instead of a speed.

The figure below shows a top-down view of an object moving in a horizontal circle. The radius \(r\) points outward from the center \(O\) to the object of mass \(m\). The velocity \(v\) is always tangent to the circle, perpendicular to the radius, while the centripetal force \(F_c\) points inward along the radius, toward the center.

Because \(a_c = \dfrac{v^2}{r}\), doubling the speed at a fixed radius quadruples the required centripetal acceleration, and therefore quadruples the required centripetal force. The graph below shows centripetal acceleration as a function of speed for an object moving in a circle of radius 5 meters.

Object on a string over a frictionless horizontal table

A mass tied to a string and swung in a circle on a smooth horizontal surface has only tension acting horizontally, so tension alone supplies the centripetal force: \(T = \dfrac{mv^2}{r}\).

Conical pendulum

A conical pendulum is a mass on a string that swings in a horizontal circle while the string traces the surface of a cone, hanging below a fixed pivot at angle \(\theta\) from the vertical. Here tension has to do two jobs at once: its vertical component balances gravity, and its horizontal component supplies the centripetal force.

\(T\cos\theta = mg\) and \(T\sin\theta = \dfrac{mv^2}{r}\)

Dividing these two equations eliminates both \(T\) and \(m\), giving a compact relationship between the swing angle, speed, and radius:

\(\tan\theta = \dfrac{v^2}{rg}\)

A car turning on a flat road

When a car turns on a level (unbanked) road, friction between the tires and the road surface provides the centripetal force. The maximum speed before the car skids outward is found by setting the maximum available friction force equal to the required centripetal force:

\(\mu mg = \dfrac{mv_{max}^2}{r} \quad\Rightarrow\quad v_{max} = \sqrt{\mu g r}\)

Banked curve

On a banked curve, the road surface is tilted at an angle \(\theta\), so the normal force itself has a horizontal component that can help supply the centripetal force, reducing (or removing) the need for friction. For an ideally banked, frictionless curve, resolving the normal force into vertical and horizontal components gives the same relationship found for the conical pendulum:

\(\tan\theta = \dfrac{v^2}{rg}\)

This is why highway engineers bank curves more steeply on roads designed for higher speeds.

A 0.50 kg ball is swung in a horizontal circle of radius 0.80 m on a smooth table, completing one revolution every 1.2 seconds. Find the speed and the tension in the string.

First find the speed from the period: \(v = \dfrac{2\pi r}{T} = \dfrac{2\pi (0.80)}{1.2} \approx 4.19\ \)m/s\(\).

Then apply the centripetal force formula, remembering that tension is the only horizontal force here:

\(T = \dfrac{mv^2}{r} = \dfrac{(0.50)(4.19)^2}{0.80} \approx 11.0\ \)N\(\)

The string must supply about 11.0 N of tension to keep the ball moving in its circular path.

The core formulas for centripetal acceleration and force, \(a_c = v^2/r\) and \(F_c = mv^2/r\), apply to any circular path. What changes between the horizontal and vertical cases is how gravity interacts with the circle: in horizontal circular motion, gravity is usually balanced separately (as in the conical pendulum) rather than contributing to the centripetal force itself, whereas in vertical circular motion gravity adds to or subtracts from the required centripetal force depending on where the object is on the loop.

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