Edexcel A Level (IAL) Physics-4.10 - 4.11 Maintaining Circular Motion & Centripetal Force- Study Notes- New Syllabus
Edexcel A Level (IAL) Physics -4.10 – 4.11 Maintaining Circular Motion & Centripetal Force- Study Notes- New syllabus
Edexcel A Level (IAL) Physics -4.10 – 4.11 Maintaining Circular Motion & Centripetal Force- Study Notes -Edexcel A level Physics – per latest Syllabus.
Key Concepts:
- understand that a resultant force (centripetal force) is required to produce and maintain circular motion
- be able to use the equations for centripetal force
Centripetal Force: Resultant Force in Circular Motion
For an object to move in a circular path at constant speed, it must experience a resultant force directed towards the centre of the circle. This force is called the centripetal force.
Why a Resultant Force Is Needed
- In circular motion, the speed may be constant but the direction of velocity continuously changes.
- A change in velocity means the object is accelerating.
- According to Newton’s laws, acceleration requires a resultant force.
- Therefore, a force must act towards the centre of the circular path.
This inward force keeps the object moving in a circle.
Centripetal Force
The centripetal force is the resultant force directed towards the centre of the circle.
Key points:
- It is not a new type of force.
- It is provided by real forces such as tension, friction, gravity, or a normal reaction.
- Its direction is always radially inward.
Relationship Between Centripetal Force and Acceleration
Centripetal acceleration is given by:
\( a = \dfrac{v^2}{r} = r\omega^2 \)
Using Newton’s second law:
\( F = ma \)
The centripetal force is therefore:
\( F = \dfrac{mv^2}{r} = mr\omega^2 \)
Nature of the Centripetal Force
- If the centripetal force is removed, the object moves off in a straight line tangential to the circle.
- The force does not act in the direction of motion.
- It changes the direction of velocity, not the speed.
Examples of Forces Acting as Centripetal Force
| Situation | Centripetal Force Provided By |
|---|---|
| Stone on a string | Tension in the string |
| Car turning a corner | Friction between tyres and road |
| Satellite orbiting Earth | Gravitational force |
| Object on a rotating turntable | Friction |
Consequences of Insufficient Centripetal Force
- If the centripetal force is too small, circular motion cannot be maintained.
- The object will move in a tangential straight line.
- This explains skidding of cars on bends at high speed.
Example (Easy)
Why does a stone attached to a string move in a circle when the string is pulled?
▶️ Answer / Explanation
- The string provides an inward tension force.
- This force acts towards the centre of the circle.
- It supplies the centripetal force needed for circular motion.
Example (Medium)
A car of mass \( 1000\ \mathrm{kg} \) travels around a circular bend of radius \( 50\ \mathrm{m} \) at \( 15\ \mathrm{m\,s^{-1}} \). Calculate the centripetal force.
▶️ Answer / Explanation
\( F = \dfrac{mv^2}{r} = \dfrac{1000 \times 15^2}{50} = 4500\ \mathrm{N} \)
Example (Hard)
Explain why increasing the speed of an object moving in a circle makes it harder to maintain circular motion.
▶️ Answer / Explanation
- Centripetal force required is proportional to \( v^2 \).
- Increasing speed greatly increases the required inward force.
- If available force is insufficient, circular motion cannot be maintained.
Equations for Centripetal Force
For an object moving in a circular path, a resultant force directed towards the centre is required to produce and maintain circular motion. This inward force is known as the centripetal force.
Centripetal Force from Newton’s Second Law
Newton’s second law states:
\( F = ma \)
For circular motion, the acceleration is the centripetal acceleration:
\( a = \dfrac{v^2}{r} = r\omega^2 \)
Substituting into \( F = ma \):
\( F = m\dfrac{v^2}{r} = mr\omega^2 \)
Forms of the Centripetal Force Equation
The centripetal force can be written in several equivalent forms:
\( F = ma \)
\( F = \dfrac{mv^2}{r} \)
\( F = mr\omega^2 \)
- \( F \) = centripetal force (N)
- \( m \) = mass (kg)
- \( v \) = linear speed (m s⁻¹)
- \( r \) = radius of circular path (m)
- \( \omega \) = angular velocity (rad s⁻¹)
Choosing the Correct Equation
- Use \( F = \dfrac{mv^2}{r} \) when linear speed is known.
- Use \( F = mr\omega^2 \) when angular velocity is known.
- Use \( F = ma \) when centripetal acceleration is already calculated.
Important Physical Meaning
- The centripetal force always acts towards the centre.
- It changes the direction of velocity, not the speed.
- It is provided by real forces (tension, friction, gravity, normal reaction).
- If the force is removed, the object moves off tangentially.
Typical Situations
| Situation | Centripetal Force Provided By |
|---|---|
| Stone on a string | Tension |
| Car turning a bend | Friction |
| Satellite in orbit | Gravitational force |
| Object on a turntable | Friction |
Example (Easy)
A \( 2.0\ \mathrm{kg} \) mass moves in a circle of radius \( 1.5\ \mathrm{m} \) with speed \( 3.0\ \mathrm{m\,s^{-1}} \). Calculate the centripetal force.
▶️ Answer / Explanation
\( F = \dfrac{mv^2}{r} = \dfrac{2.0 \times 3.0^2}{1.5} = 12\ \mathrm{N} \)
Example (Medium)
A particle of mass \( 0.50\ \mathrm{kg} \) moves in a circle of radius \( 0.80\ \mathrm{m} \) with angular velocity \( 4.0\ \mathrm{rad\,s^{-1}} \). Find the centripetal force.
▶️ Answer / Explanation
\( F = mr\omega^2 = 0.50 \times 0.80 \times 4.0^2 = 6.4\ \mathrm{N} \)
Example (Hard)
A car of mass \( 1200\ \mathrm{kg} \) travels around a circular track of radius \( 60\ \mathrm{m} \). If the maximum available centripetal force is \( 4800\ \mathrm{N} \), calculate the maximum safe speed.
▶️ Answer / Explanation
Use \( F = \dfrac{mv^2}{r} \):
\( v^2 = \dfrac{Fr}{m} = \dfrac{4800 \times 60}{1200} = 240 \)
\( v = \sqrt{240} \approx 15.5\ \mathrm{m\,s^{-1}} \)