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Centripetal Force (SL IB Physics)

Revision Note

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Centripetal Force

  • Velocity and acceleration  are both vector  quantities
  • An object in uniform circular motion is continuously changing direction, and therefore is constantly changing velocity
    • The object must therefore be accelerating

  • This is called the centripetal acceleration and is perpendicular to the direction of the linear speed
    • Centripetal means it acts towards the centre of the circular path

  • From Newton's second law, this must mean there is a resultant force acting upon it
    • This is known as the centripetal force and is what keeps the object moving in a circle
    • This means the object changes direction even if its magnitude of velocity remains constant

  • The centripetal force (F) is defined as:

The resultant force perpendicular to the velocity required to keep a body in a uniform circular motion which acts towards the centre of the circle

 

  •  The magnitude of the centripetal force F can be calculated using:

 

F space equals fraction numerator space m v squared over denominator r end fraction space equals space m r omega squared

  • Where:
    • F = centripetal force (N)
    • v = linear speed (m s1)
    • ⍵ = angular speed (rad s−1)
    • r = radius of the orbit (m)

Force and acceleration direction, downloadable AS & A Level Physics revision notes

Centripetal force is always perpendicular to the linear velocity (i.e, the direction of travel)

  • The direction of the centripetal force is:
    • In the same direction as the centripetal acceleration (towards the centre of the circle) 
      • This is due to Newton's Second Law
    • Perpendicular to the linear velocity

  • The centripetal force is not a separate force of its own
  • It can be any type of force, depending on the situation, which keeps an object moving in a circular path
    • For example, tension, friction, gravitational, electrostatic or magnetic

Examples of centripetal forceTable showing examples of centripetal force, downloadable AS & A Level Physics revision notes

  • When solving circular motion problems involving one of these forces, the equation for centripetal force can be equated to the relevant force equation
  • For example, for a mass orbiting a planet in a circular path, the centripetal force is provided by the gravitational force

  • When an object travels in circular motion, there is no work done
    • This is because there is no change in kinetic energy

Horizontal Circular Motion

  • An example of horizontal circular motion is a vehicle driving on a curved road
  • The forces acting on the vehicle are:
    • The friction between the tyres and the road
    • The weight of the vehicle downwards

  • In this case, the centripetal force required to make this turn is provided by the frictional force
    • This is because the force of friction acts towards the centre of the circular path

  • Since the centripetal force is provided by the force of friction, the following equation can be written:

fraction numerator m v squared over denominator r end fraction space equals space mu m g

  • Where:
    • m = mass of the vehicle (kg)
    • v = speed of the vehicle (m s–1)
    • r = radius of the circular path (m)
    • μ = static coefficient of friction
    • g = acceleration due to gravity (m s–2)

  • Rearranging this equation for v gives:

 

v squared space equals space mu g r

v subscript m a x end subscript space equals space square root of mu g r end root

  •  This expression gives the maximum speed at which the vehicle can travel around the curved road without skidding
    • If the speed exceeds this, then the vehicle is likely to skid
    • This is because the centripetal force required to keep the car in a circular path could not be provided by friction, as it would be too large

6-1-4-horizontal-circular-motion_sl-physics-rn

The frictional force provides the centripetal force

  • Therefore, in order for a vehicle to avoid skidding on a curved road of radius r, its speed must satisfy the equation

v space less than space square root of mu g r end root

  • A mass attached to a string rotating around is another example of horizontal circular motion
  • In this case, the tension is the centripetal force as it acts towards the centre of the circle
  • This time, the weight of the mass will be acting as well as the tension of the string

1-2-19-horizontal-circular-motion-

A mass attached to a string will have its weight acting meaning the string is at an angle

  • The weight mg of the mass needs to be balanced by the vertical component of the tension

F subscript t cos theta space equals space m g

 

  • This means the string will always be at an angle and never perfectly horizontal
  • The ball's linear velocity, v is still perpendicular to the tension and its weight, mg points downward
  • All three forces are perpendicular to each other, so no other component contributes to the centripetal force, just the tension
  • The centripetal force is still towards the centre of the circle, but now is just the horizontal component of the tension

F subscript t sin theta space equals fraction numerator space m v squared over denominator r end fraction

  • This is an important example of resolving vectors properly. The vertical component does not always have 'sinθ', it depends on what θ is defined as

Banking

  • A banked road, or track, is a curved surface where the outer edge is raised higher than the inner edge
    • The purpose of this is to make it safer for vehicles to travel on the curved road, or track, at a reasonable speed without skidding

  • When a road is banked, the centripetal force no longer depends on the friction between the tyres and the road
  • Instead, the centripetal force depends solely on the horizontal component of the normal force

6-1-4-banking_sl-physics-rn

During banking, the horizontal component of the normal reaction force provides the centripetal force

Worked example

A 300 g ball is made to travel in a circle of radius 0.8 m on the end of a string. If the maximum force the ball can withstand before breaking is 60 N, what is the maximum speed of the ball?

Answer:

Step 1: List the known quantities

  • Mass, m = 300 g = 300 × 10-3 kg
  • Radius, r = 0.8 m
  • Resultant force, F = 60 N

Step 2: Rearrange the centripetal force equation for v

F subscript m a x end subscript space equals fraction numerator space m v squared subscript m a x end subscript over denominator r end fraction

v subscript m a x end subscript space equals square root of fraction numerator r F subscript m a x end subscript over denominator m end fraction end root

Step 3: Substitute in the values

v subscript m a x end subscript equals space square root of fraction numerator 0.8 space cross times space 60 over denominator 300 space cross times space 10 to the power of negative 3 end exponent end fraction end root space equals space12.6 m s–1

Examiner Tip

The linear speed, v is sometimes referred to as the ‘tangential’ speed.

The centripetal force equation is not given in your data book, but you are given in the equations for centripetal acceleration. You just need to multiply them by mass m since the centripetal force Fma.

It is important you understand the foundations of circular motion, especially how to use the equations. This will heavily link with kepler's laws and magnetic fields.

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Ashika

Author: Ashika

Expertise: Physics Project Lead

Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.