Edexcel IAL - Mechanics 3- 4.2 Radial Acceleration- Study notes - New syllabus
Edexcel IAL – Mechanics 3- 4.2 Radial Acceleration -Study notes- New syllabus
Edexcel IAL – Mechanics 3- 4.2 Radial Acceleration -Study notes -Edexcel A level Maths- per latest Syllabus.
Key Concepts:
- 4.2 Radial Acceleration
Radial Acceleration in Circular Motion
When a particle moves in a circular path, its velocity is constantly changing in direction, even if its speed is constant. This change in direction gives rise to an acceleration called radial acceleration.
Radial acceleration is always directed towards the centre of the circle and is sometimes called centripetal acceleration.
Direction of Radial Acceleration
For circular motion:
- Velocity is tangential to the circle
- Radial acceleration acts towards the centre of the circle
This inward acceleration keeps the particle moving along a circular path.
Formula for Radial Acceleration
If a particle moves in a circle of radius \( r \) with angular speed \( \omega \), the radial acceleration is
\( a = r\omega^2 \)
Since the linear speed \( v \) is related to angular speed by \( v = r\omega \), the radial acceleration can also be written as
\( a = \dfrac{v^2}{r} \)
Both forms are required and may be used interchangeably, depending on the information given.
Radial Force
By Newton’s second law, the resultant force required to produce radial acceleration is
\( F = ma = mr\omega^2 = \dfrac{mv^2}{r} \)
This force must act towards the centre of the circular path.
Example :
A particle moves in a circle of radius \( 2 \) m with angular speed \( 5 \,\text{rad s}^{-1} \). Find its radial acceleration.
▶️ Answer/Explanation
Using the formula
\( a = r\omega^2 \)
\( a = 2 \times 5^2 = 50 \)
Conclusion: The radial acceleration is \( 50 \,\text{m s}^{-2} \).
Example :
A particle moves with constant speed \( 10 \,\text{m s}^{-1} \) in a circle of radius \( 4 \) m. Find the radial acceleration.
▶️ Answer/Explanation
Using the formula
\( a = \dfrac{v^2}{r} \)
\( a = \dfrac{10^2}{4} = 25 \)
Conclusion: The radial acceleration is \( 25 \,\text{m s}^{-2} \).
Example :
A particle of mass \( 0.5 \) kg moves in a horizontal circle of radius \( 0.8 \) m with angular speed \( 6 \,\text{rad s}^{-1} \). Find the magnitude of the force required to maintain the circular motion.
▶️ Answer/Explanation
First find the radial acceleration:
\( a = r\omega^2 = 0.8 \times 6^2 = 28.8 \)
Using Newton’s second law:
\( F = ma = 0.5 \times 28.8 = 14.4 \)
Conclusion: The force required is \( 14.4 \) N.