recall and use F = mrω2 and F = mv2 / r
Centripal Motion – Cambridge IGCSE/A‑Level (9702) – Topic 12
Learning Outcomes
- Define the radian and angular speed \( \omega \).
- Relate angular speed \( \omega \) to period \(T\), frequency \(f\) and linear (tangential) speed \(v\).
- Derive and use the expressions for centripetal (radial) acceleration
\[
a_c = r\omega^{2}= \frac{v^{2}}{r}
\]
and for the required centripetal force
\[
Fc = m ac = m r\omega^{2}= \frac{m v^{2}}{r}.
\]
- Identify the physical force that supplies \(F_c\) (tension, friction, normal reaction, weight component, gravity, etc.) and choose the appropriate form of the equation for any given problem.
Key Definitions
| Radian | The angle subtended at the centre of a circle by an arc whose length equals the radius. |
| Angular speed \( \omega \) | Rate of change of angular displacement. \[ \omega = \frac{2\pi}{T}=2\pi f\quad\text{(rad s}^{-1}\text{)} \] where \(T\) is the period (s) and \(f\) the frequency (Hz). |
| Linear (tangential) speed \(v\) | Magnitude of the velocity tangent to the circular path. \[ v = r\omega \qquad\text{or}\qquad \omega = \frac{v}{r}. \] |
| Centripetal acceleration \(a_c\) | Radial acceleration required to keep an object moving in a circle: \[ a_c = r\omega^{2}= \frac{v^{2}}{r}. \] |
| Centripetal force \(F_c\) | Net radial force that provides the required centripetal acceleration. \[ Fc = m ac = m r\omega^{2}= \frac{m v^{2}}{r}. \] |
Quick‑Conversion Table
| Quantity | Symbol | Conversion |
|---|---|---|
| Revolutions per minute | \(n\) (rpm) | \(\displaystyle \omega = 2\pi\frac{n}{60}\;\text{rad s}^{-1}\) |
| Period | \(T\) (s) | \(\displaystyle \omega = \frac{2\pi}{T}\) |
| Frequency | \(f\) (Hz) | \(\displaystyle \omega = 2\pi f\) |
Key Concepts
- Uniform circular motion – motion at constant speed along a circular path; the direction of the velocity changes continuously.
- Direction of the force – the centripetal force always points towards the centre of the circle (radially inwards). “Centrifugal” is a pseudo‑force that appears only in a rotating reference frame.
- Sources of centripetal force – see the table below for the most common physical origins.
Typical Sources of the Required Centripetal Force
| Situation | Physical source of \(F_c\) | One‑sentence justification |
|---|---|---|
| Mass on a light string (horizontal circle) | Tension in the string | The string pulls the mass towards the centre, providing the required inward force. |
| Car turning on a flat road | Static friction between tyres and road | Friction opposes the tendency of the tyres to slide outward, acting radially inward. |
| Car on a banked curve (no friction) | Component of the normal reaction | The inclined road surface directs part of the normal force towards the centre of curvature. |
| Pendulum at the lowest point of a vertical circle | Weight component + tension | Both gravity (downwards) and the string tension combine to give the net inward force. |
| Satellite in circular orbit | Gravitational attraction of the Earth | Earth’s gravity provides exactly the inward force needed for the orbital motion. |
Derivation of the Formulas
1. Vector change in velocity
Consider an object of mass \(m\) moving with constant speed \(v\) around a circle of radius \(r\). In a short time \(\Delta t\) it sweeps an angle \(\Delta\theta\) (radians). The velocity vectors \(\mathbf v1\) and \(\mathbf v2\) have the same magnitude but differ in direction by \(\Delta\theta\).
Tip‑to‑tail construction of \(\Delta\mathbf v\); the resultant points towards the centre.
For small \(\Delta\theta\) the magnitude of the change in velocity is
\[
\Delta v = v\,\Delta\theta .
\]
Since \(\omega = \dfrac{\Delta\theta}{\Delta t}\),
\[
\frac{\Delta v}{\Delta t}=v\frac{\Delta\theta}{\Delta t}=v\omega .
\]
Taking the limit \(\Delta t\to0\) gives the radial (centripetal) acceleration
\[
ac = \lim{\Delta t\to0}\frac{\Delta v}{\Delta t}=v\omega = r\omega^{2}= \frac{v^{2}}{r}.
\]
2. Newton’s II law
The net inward force required is
\[
Fc = m ac = m r\omega^{2}= \frac{m v^{2}}{r}.
\]
3. Inter‑conversion
Whenever one of the quantities \(v\), \(\omega\) or \(r\) is known, the others follow from \(v = r\omega\). This makes the two force forms interchangeable.
When to Use Which Form
- \(F_c = m r\omega^{2}\) – most convenient when \(\omega\) (or period \(T\) or frequency \(f\)) is given.
- \(F_c = \dfrac{m v^{2}}{r}\) – use when the linear speed \(v\) is supplied directly.
- Both give identical results; always ensure that the radius used is the distance from the axis of rotation to the line of action of the force.
Variables, Units & Typical A‑Level Ranges
| Symbol | Quantity | SI Unit | Typical A‑Level range |
|---|---|---|---|
| \(F_c\) | Centripetal force | newton (N) | 0.1 – 10⁴ |
| \(m\) | Mass | kilogram (kg) | 0.01 – 10 |
| \(r\) | Radius of path | metre (m) | 0.1 – 5 |
| \(\omega\) | Angular speed | rad s⁻¹ | 1 – 100 |
| \(v\) | Linear speed | m s⁻¹ | 0.5 – 200 |
| \(a_c\) | Centripetal acceleration | m s⁻² | 0.1 – 10⁴ |
Worked Example 1 – Tension in a Horizontal Circle
Problem: A 0.50 kg mass is attached to a light string and whirled in a horizontal circle of radius 0.75 m at a constant speed of 4.0 m s⁻¹. Find the tension in the string.
- Linear speed is given → use \(F_c = \dfrac{m v^{2}}{r}\).
- Substitute: \[
F_c = \frac{(0.50\;\text{kg})(4.0\;\text{m s}^{-1})^{2}}{0.75\;\text{m}}.
\]
- Calculate: \[
F_c = \frac{0.50 \times 16.0}{0.75}= \frac{8.0}{0.75}\approx 10.7\;\text{N}.
\]
- Interpretation – the string must supply a centripetal (tension) force of ≈ 11 N directed towards the centre.
Worked Example 2 – Satellite in Circular Orbit
Problem: A satellite of mass 500 kg orbits the Earth at an orbital radius \(r = 7.0\times10^{6}\) m. The gravitational acceleration at that radius is \(g' = 8.7\) m s⁻². Find the orbital speed and the required centripetal force.
- For a circular orbit, gravity provides the centripetal force: \(\dfrac{m v^{2}}{r}= m g'\).
- Solve for speed: \[
v = \sqrt{g' r}= \sqrt{(8.7)(7.0\times10^{6})}\approx 7.8\times10^{3}\;\text{m s}^{-1}.
\]
- Centrepital force: \[
F_c = m g' = (500)(8.7)=4.35\times10^{3}\;\text{N}.
\]
- Interpretation – Earth’s gravity supplies a centripetal force of ≈ 4.4 kN, keeping the satellite in its circular orbit.
Worked Example 3 – Car on a Banked Curve (no friction)
Problem: A car of mass 1200 kg travels round a banked curve of radius 50 m that is inclined at \(\theta = 30^{\circ}\). Find the speed at which the car can negotiate the curve without relying on friction.
- Resolve the normal reaction \(N\) into components: vertical \(N\cos\theta = mg\) and horizontal \(N\sin\theta = F_c\).
- From the vertical equilibrium: \(N = \dfrac{mg}{\cos\theta}\).
- Insert into the horizontal component: \[
F_c = N\sin\theta = \frac{mg}{\cos\theta}\sin\theta = mg\tan\theta.
\]
- Set \(F_c = \dfrac{m v^{2}}{r}\) and solve for \(v\): \[
\frac{m v^{2}}{r}= mg\tan\theta \;\Longrightarrow\; v = \sqrt{r g \tan\theta}.
\]
- Numerical result: \[
v = \sqrt{(50)(9.81)\tan30^{\circ}} \approx 12.0\;\text{m s}^{-1}.
\]
- Interpretation – at about 12 m s⁻¹ the component of the normal reaction alone supplies the required centripetal force.
Common Mistakes & How to Avoid Them
- Confusing centripetal and centrifugal forces – remember the real force is always inward; centrifugal is only an apparent force in a rotating frame.
- Using the wrong speed–angle relation – when \(\omega\) is given, always use \(v = r\omega\). Do not replace \(\omega\) by \(2\pi\) unless the frequency is exactly 1 Hz.
- Incorrect conversion of rpm to rad s⁻¹ – apply \(\omega = 2\pi n/60\). Forgetting the factor \(2\pi\) leads to errors of more than an order of magnitude.
- Omitting the radius in the denominator – the formula is \(\dfrac{m v^{2}}{r}\); a missing “\(r\)” gives a result that is too large.
- Not identifying the source of \(F_c\) – always state which physical interaction (tension, friction, normal, weight component, gravity) is providing the inward force.
Links to Other Syllabus Topics
- Dynamics (Topic 3) – Newton’s II law underpins the derivation of \(F_c\).
- Energy (Topic 5) – In uniform circular motion the kinetic energy is \(\tfrac12 m v^{2}\); the centripetal force does no work because it is perpendicular to the displacement.
- Gravitational fields (Topic 12 – vertical circular motion) – For satellites and planets, gravity is the centripetal force, leading to \(v = \sqrt{GM/r}\).
- Friction (Topic 3) – On flat or banked curves the required centripetal force is supplied by static friction, \(Ff = \mus N\).
Suggested Revision Diagram