Module 5: Newtonian world and astrophysicsCentripetal force (5.2.2)

Centripetal force (5.2.2)

Centripetal acceleration, centripetal force, a = v²/r, F = mv²/r, and circular motion in horizontal and vertical circles in A-level Physics.
6 min

Newton’s first law states that an object will continue travelling at a constant speed in a straight line unless acted upon by a net force.

An object following a circular path changes direction, and therefore, a force must act on it. This force is directed towards the centre of the circle, perpendicular to the object’s velocity.

A centripetal (centre-seeking) force keeps an object moving at constant speed in a circle, but causes the direction of the object’s motion to change.

An illustration showing a car on a circular road with a roundabout featuring a tree and a pond. The image includes arrows labeled 'Linear velocity' pointing upwards and 'Centripetal force' pointing towards the center of the circle.

While the speed of an object undergoing uniform circular motion remains constant, the constantly changing direction means the object has a changing velocity. It is important to note that speed is a scalar quantity and velocity is a vector.

A change in velocity means an object is accelerating. The centripetal force provides this acceleration. Newton’s second law states that force and acceleration are proportional to each other; you cannot have one without the other.

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The centripetal force acts towards the centre of an object’s circular motion. Examples of centripetal forces are shown below:

Car going round a roundabout, centripetal force provided by friction. Moon orbiting Earth, centripetal force provided by gravity. Swing a ball on a string, centripetal force provided by tension.
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If an object is moving in a circular path and the centripetal force is removed, then the object will fly off at a tangent.

An example of this is cutting the string attached to a ball being swung in a circle parallel to the ground.

An illustration showing a person holding a rope attached to a green ball, with a dashed circular path around it. The text states: 'If the rope is cut, object will fly off at a tangent.' Below, it explains: 'Tension is providing the centripetal force. Without this, the ball will travel in a straight line.'

Once the centripetal force is removed, Newton’s first law applies again – the object will travel in a straight line at a constant speed unless another resultant force acts upon it.

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The centrifugal force is the name given to the fictitious outward pseudoforce you experience when you are turning.

For example, if you were sitting on the back seat of a car as it went round a corner you might feel like you are being pushed away from the centre of the turn and slide away from the centre of the turn.

However, this is just your inertia wanting you to continue on a straight path. When the car turns, the side of the bus pushes you towards the centre of the turn. This would be a real centripetal force caused by the reaction force between you and the side of the car.

An illustration showing two cars on a curved road. The car on the left is labeled with 'Fictitious centrifugal force' in red, and an arrow pointing to the left. The car on the right is labeled with 'Real centripetal force' in blue, and an arrow pointing downwards. The background includes a green area and a curved road.
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Be careful not to confuse centripetal force with centrifugal force in questions. Exam questions will only require you to perform calculations involving centripetal force.

Car going round a roundabout centipetal force provided by friction
Do

Make sure to use the force acting towards the centre of rotation. This is the centripetal force.

An illustration showing two cars navigating a curved road. The red arrow labeled 'Fictous centrifugal force' points outward from the curve, while the blue arrow labeled 'Real centripetal force' points inward towards the center of the curve.
Don't

Use the force which appears to be acting away from the centre of rotation. This is the centrifugal force, which is a fictitious force due to an object’s inertia.

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The linear speed of an object can be found by dividing the distance travelled by the time taken.

In the case of an object moving with circular motion, the speed can be found by dividing the circumference of the circle representing the object’s trajectory by the time taken to complete one full rotation (the period):

Where:

  • is linear speed, measured in metres per second (),
  • is the radius of the circle, measured in metres (),
  • is the time period, measured in seconds ().

The angular speed of an object moving with circular motion is given by:

Where is the angular speed, measured in radians per second ().

Combining these two equations using substitution leads to the relationship between linear speed and angular speed :

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The equation implies that the linear speed of an object moving with circular motion is proportional to the radius, if the angular speed is constant.

An example of this is two points at different distances from the hub on a bike wheel. They will both have the same angular speed because they will both take the same amount of time to complete one full revolution (the same time period).

However, their linear speeds will be different, as they have to travel different distances to complete one rotation.

A diagram showing two points, A and B, on a circular path. Point A is represented by a blue dot with an arrow indicating movement to the left, while point B is a red dot at the center of a smaller red circle with an arrow indicating movement to the left. The background includes a black circle and a grid.

In the diagram above, point A is further from the centre than point B and will need a greater linear speed in order to complete one full rotation in the same amount of time as point B.

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Question walkthrough

Linear Speed of Objects on a Rotating Disc

Use the shared angular speed of two children on a merry-go-round to find one's linear speed from the other's.

As an object travels in a circular path, its direction is constantly changing. Therefore, its velocity must be constantly changing as velocity is a vector. If the velocity of an object is changing, the object is accelerating.

This acceleration is known as a centripetal acceleration and is directed towards the centre of the circle and perpendicular to the velocity.

Car going round a roundabout, centripetal acceleration acts towards the center, in the same direction as the centripetal force provided by friction.
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The centripetal acceleration of an object moving with circular motion is given by:

Where:

  • is the centripetal acceleration, measured in metres per second squared (),
  • is the linear speed, measured in metres per second (),
  • is the radius, measured in metres ().

Combining the equation above with the equation that links linear speed to angular speed produces an alternative way to find the centripetal acceleration :

combined with returns

Where is the angular speed, measured in radians per second ().

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Question walkthrough

Comparing Centripetal Acceleration on a Turntable

Use the shared angular speed of two objects on a turntable to find one's centripetal acceleration from the other's.

Centripetal force it is not a force itself. It is a measure of the pull of another force towards the centre of a circle.

In circular motion, the centripetal force is provided by a force acting towards the centre, such as friction, gravity, tension, or a combination of forces.

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An example of a combination of forces constituting centripetal force is a ball spun in a vertical circle on a string. The centripetal force () is a combination of the object’s weight () and the tension in the string ().

At the top of the circle CF = T + W. At the bottom of the circle CF = T - W.

The weight of the object acts downwards and is constant (the mass does not change).

The centripetal force is also constant if the radius and speed stay constant (true for both linear and angular speed).

The tension in the string is the force that can change in size and direction:

  • When the ball is at the bottom, the tension supports the weight and provides the centripetal force.
  • When the ball is at the top, its weight points to the centre, providing some centripetal force; the tension provides the rest.

When calculating tension or weight, remember that the resultant centripetal force is always directed towards the centre.

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Circular motion can be investigated using a whirling bung experiment:

  1. A rubber bung is attached to a thin piece of string, which is passed through a hollow glass tube.
  2. A weight is suspended from the other end of the string.
  3. The student holds the glass tube and whirls the rubber bung above their head horizontally with circular motion.
A person holding a glass tube connected to a bung, with a weight hanging from the tube.

The suspended weight creates tension in the string as the bung is whirled, providing the centripetal force to keep it moving in a circular path.

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In the whirling bung experiment, the suspended weight causes a tension in the string as the bung is whirled around. This provides the centripetal force to keep it moving in a circular path.

The relationship between the force of the suspended weight and the centripetal force determines how the weight will move.

A table displaying conditions and outcomes related to the equation Mg = mv²/r. The first row states that when Mg = mv²/r, the weight remains stationary. The second row indicates that when Mg < mv²/r, the weight will ascend. The third row shows that when Mg > mv²/r, the weight will descend.

Where:

  • is the mass of the bung and is the mass of the weight, measured in kilograms (),
  • is the linear speed, measured in metres per second (),
  • is the radius, measured in metres (), and
  • = 9.81 is the gravitational acceleration.

You can also determine the linear speed of the bung by measuring the time taken for one complete rotation:

Or find the angular velocity of the bung:

Then use the equation that links linear speed to angular speed:

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Question walkthrough

Frictional Force for Cars on Different Radii

Calculate the frictional force needed for two cars at the same speed to take corners of different radii.

Question walkthrough

String Tension in Vertical Circular Motion

Calculate the tension in a string at the top and bottom of a vertical circle for a ball undergoing circular motion.