Horizontal Circular Motion

Circular motion refers to the movement of an object along a circular path at a constant radius from a fixed point. While the object's speed may remain constant, its velocity continuously changes due to the changing direction. As a result, the object experiences acceleration, requiring a force to sustain this motion.

Use this page to revise the following concepts within horizontal circular motion:

Centripetal acceleration and force

When you swing a ball on a rope in a circle above your head, what keeps it moving in that circular path?

A diagram showing a ball moving in a circular path (dashed line). The ball is shown at multiple points along the circle. A single point is shown where the ball is attached by a rope. It is labelled with two arrows: a velocity arrow pointing tangent to the circle, and a force arrow pointing toward the centre with a rope

The key is the inward pull toward the centre which is known as the centripetal force . Without this force – such as if the rope broke - the ball would fly off in a straight-line tangent to its circular path!

A diagram showing what happens when a ball's circular motion is interrupted. The ball is shown at multiple points in this motion. The ball's previous circular path is shown as a dashed curve, but the ball is now moving in a straight-line tangent to where it was in the circle when the rope broke. Velocity arrows point along the direction the ball is moving at each point.>

The velocity points in the direction the ball is trying to move at any instant. While the ball's speed remains constant throughout its circular motion, the velocity continuously changes as the ball's direction changes. This changing velocity means there must be an acceleration the centripetal acceleration . According to Newton's Second Law, since the centripetal force from the rope acts toward the centre of the circle, the resulting acceleration must also act toward the centre perpendicular to the velocity.

This applies to any object moving in a circle, not just a ball on a rope - the force and acceleration must point toward the centre of the circular path to maintain the motion. The magnitude of the centripetal acceleration can be determined from the object's speed and the radius of the circular path:

\[ a = \frac{v^2}{r} \]

Where:

  • \(a\) = magnitude of acceleration \((\mathrm{ms}^{-2})\)
  • \(v\) = speed or magnitude of velocity \((\mathrm{ms}^{-1})\)
  • \(r\) = radius of circular path \((\mathrm{m})\)

This equation shows that objects moving faster or in tighter circular paths require greater acceleration to maintain their motion. From Newton's Second Law, we can then find the centripetal force needed to produce this acceleration:

\[ \begin{aligned} F &= ma \\ &= \frac{mv^2}{r} \end{aligned} \]

Where:

  • \(F\) = magnitude of force \((\mathrm{N})\)

This force must be provided by some physical mechanism – tension in a rope, gravity, friction, or any other force that points to the centre of a circular path.

Horizontal motion scenarios

There are some key scenarios that demonstrate the use of different mechanisms to provide the force required to maintain motion on a circular path:

  • Vehicle moving around a circular road (friction)
  • Vehicle moving around a banked track (normal force )
  • Object at the end of a string (tension )

Vehicle moving around a circular road

When a vehicle moves around a flat circular road, the friction between the tyres and the road provide the centripetal force necessary to maintain circular motion. Without sufficient friction the car will skid.

Worked example

A car of mass \(1200\mathrm{~kg}\) travels around a circular track with a radius of \(50\text{ m}\) at \(40\text{ kmh}^{-1}\). Assuming no air resistance, calculate the friction force required to keep the car in motion around the track.

Identify the known variables:

\(m = 1200\mathrm{~kg}\)

\(r = 50\mathrm{~m}\)

\(v = 40\mathrm{~kmh}^{-1} = \frac{40}{3.6}\mathrm{~ms}^{-1} \approx\ 11.11\mathrm{~ms}^{-1}\)

The friction force required to keep the car in motion is equal to the centripetal force which can be calculated using:

\[F_f = \frac{mv^{2}}{r}\]

Substitute values:

\[\begin{align}F_f &= \frac{\left(1200\right)\left(11.11\right)^{2}}{50} \\ &\approx 2962.96 \mathrm{~N}\end{align}\]


Vehicle moving around a banked track

When a track is banked (tilted inward at an angle), part of the normal force from the road provides the centripetal force needed for circular motion. The banking creates a situation where even without friction, a vehicle can maintain its circular path at a speed that is called the design speed.

Worked example

A car drives around a circular banked track with a radius of \(50 \mathrm{~m}\). The track is banked at an angle of \(15^\circ\). Assuming no air resistance and \(g = 9.8 \mathrm{~ms}^{-2}\), what speed must the car maintain to prevent sliding down or up the bank?

Side section view of a car moving out of the page on a banked track that is \mathbf{15}° to the horizontal. Track radius is \mathbf{50}\ \mathbit{m}, labelled from the centre to the car.

Identify the known variables:
\(r = 50 \mathrm{~m}\)
\(g = 9.8 \mathrm{~ms}^{-2}\)
\(\theta = 15^\circ\)

Draw a force diagram to help with resolving components if required:

Phys-1-force-diagram_updated

Find the horizontal and vertical components of the normal force:

\[ \begin{aligned} F_{N,v} &= F_N\cos(\theta) \\ F_{N,h} &= F_N\sin(\theta) \end{aligned} \]

Note

The forces on a banked curve are different to that of an inclined plane. We do not resolve the horizontal and vertical components of the gravitational force. The normal force is greater than the gravitational force as it must also balance the centripetal force.

Resolve the vertical forces. These must be balanced because the vehicle has no vertical acceleration:

\[ \begin{aligned} F_{\text{net},v} &= F_{N,v} - mg \\ 0 &= F_{N,v} - mg \\ F_{N,v} &= mg \\ F_N\cos(\theta) &= mg \end{aligned} \]

Resolve the horizontal forces. The horizontal component of the normal force provides the centripetal force:

\[ \begin{aligned} F_{\text{net},h} &= F_{N,h} \\ F_{N,h} &= F_c \\ F_N\sin(\theta) &= \frac{mv^2}{r} \end{aligned} \]

Rearrange the vertical forces equation:

\[ F_N = \frac{mg}{\cos(\theta)} \]

Substitute into the horizontal forces equation:

\[ \frac{mg}{\cos(\theta)}\sin(\theta) = \frac{mv^2}{r} \]

Note

Notice that \(m\) is on both sides of the equation. The mass of the object does not matter in a banked track problem. As long as correct speed is maintained, the object can remain in the circular motion.

Simplify and substitute values:

\[ \begin{aligned} g\tan(\theta) &= \frac{v^2}{r} \\ v &= \sqrt{gr\tan(\theta)} \\ v &= \sqrt{(9.8)(50)\tan(15^\circ)} \\ &\approx 11.46 \mathrm{~ms}^{-1} \end{aligned} \]

A speed of \(11.46 \mathrm{~ms}^{-1}\) must be maintained to sustain the circular motion without sliding up or down the bank.


Object at the end of a string

An object at the end of a string experiences tension in the string which provides the centripetal force necessary to maintain circular motion.

Worked example

A \(2 \mathrm{~kg}\) mass is suspended by a string and swings in a horizontal circle with a radius of \(0.8 \mathrm{~m}\). The string makes a \(30^\circ\) angle with the vertical.

Side view of a conical pendulum setup. A \mathbf{2}\ \mathbit{kg} mass swings in a horizontal circle, suspended by a string. String makes a \mathbf{30}° angle with the vertical axis. Radius of circular path is marked as \mathbf{0}.\mathbf{8}\ \mathbit{m}. Tension force in string and weight force are shown as vectors.

Assuming no air resistance and \(g = 9.8 \mathrm{~ms}^{-2}\), calculate:

  1. The tension in the string
  2. The speed of the mass.

Identify known variables:

\(m = 2 \mathrm{~kg}\)

\(r = 0.8 \mathrm{~m}\)

\(\theta = 30^\circ\)

\(g = 9.8 \mathrm{~ms}^{-2}\)

To find the tension in the string, resolve the vertical forces. The vertical component of the tension balances the gravitational force:

\[ \begin{aligned} F_g &= T\cos(\theta) \\ mg &= T\cos(\theta) \\ T &= \frac{mg}{\cos(\theta)} \end{aligned} \]

Substitute values:

\[ \begin{aligned} T &= \frac{(2)(9.8)}{\cos(30^\circ)} \\ &\approx 22.63 \mathrm{~N} \end{aligned} \]

The tension in the string is \(22.63 \mathrm{~N}\).

To determine the speed, resolve the horizontal forces:

\[ \begin{aligned} F_c &= T\sin(\theta) \\ \frac{mv^2}{r} &= T\sin(\theta) \\ v &= \sqrt{\frac{Tr\sin(\theta)}{m}} \end{aligned} \]

Substitute values:

\[ \begin{aligned} v &= \sqrt{\frac{(22.63)(0.8)\sin(30^\circ)}{2}} \\ &\approx 2.13 \mathrm{~ms}^{-1} \end{aligned} \]

The speed of the mass is \(2.13 \mathrm{~ms}^{-1}\).