recall and use F = mrω2 and F = mv2 / r

Centripetal Acceleration

Learning Objective

Recall and use the equations

$$F = m r \omega^{2}$$

and

$$F = \frac{m v^{2}}{r}$$

to solve problems involving objects moving in a circular path.

Key Concepts

• Uniform circular motion: motion at constant speed along a circular path.
• Centripetal force: the net force directed towards the centre of the circle that keeps the object in circular motion.
• Centripetal acceleration: the acceleration associated with the change in direction of the velocity vector, given by $a_c = r\omega^{2} = \dfrac{v^{2}}{r}$.
• Angular velocity ($\omega$): the rate of change of angular displacement, measured in rad s\(^{-1}\).
• Linear speed ($v$): the magnitude of the tangential velocity, related to $\omega$ by $v = r\omega$.

Derivation of the Formulas

Consider an object of mass $m$ moving in a circle of radius $r$ with constant speed $v$.

The change in velocity over a small time $\Delta t$ points towards the centre and has magnitude $\Delta v = v\Delta\theta$, where $\Delta\theta = \omega\Delta t$.

Thus the average acceleration is

$$a_c = \frac{\Delta v}{\Delta t} = \frac{v\omega\Delta t}{\Delta t}=v\omega = r\omega^{2}.$$

Multiplying by the mass gives the required centripetal force:

$$F_c = m a_c = m r \omega^{2}.$$

Using $v = r\omega$ we can also write

$$F_c = m\frac{v^{2}}{r}.$$

When to Use Which Form

• Use $F = m r \omega^{2}$ when the angular speed $\omega$ is known or more convenient.
• Use $F = \dfrac{m v^{2}}{r}$ when the linear speed $v$ is given.
• Both forms are interchangeable via $v = r\omega$.

Variables and Units

Symbol	Quantity	SI Unit	Typical \cdot alues (A‑Level)
$F$	Centripetal force	newton (N)	0.1 – 10⁴
$m$	Mass of the object	kilogram (kg)	0.01 – 10
$r$	Radius of the circular path	metre (m)	0.1 – 5
$\omega$	Angular velocity	radian per second (rad s⁻¹)	1 – 100
$v$	Linear speed	metre per second (m s⁻¹)	0.5 – 200
$a_c$	Centripetal acceleration	metre per second squared (m s⁻²)	0.1 – 10⁴

Worked Example

A 0.50 kg mass is attached to a 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.

1. Identify the required formula: $F = \dfrac{m v^{2}}{r}$ because $v$ is given.
2. Substitute the known values: $$F = \frac{(0.50\ \text{kg})(4.0\ \text{m s}^{-1})^{2}}{0.75\ \text{m}}$$
3. Calculate: $$F = \frac{0.50 \times 16.0}{0.75} = \frac{8.0}{0.75} \approx 10.7\ \text{N}$$
4. Interpretation: The tension in the string must provide a centripetal force of about 11 N (to 2 sf).

Common Mistakes

• Confusing centripetal (towards centre) with centrifugal (apparent outward) force.
• Using $v = 2\pi r$ instead of $v = r\omega$ when angular speed is given.
• Forgetting to convert revolutions per minute (rpm) to rad s⁻¹: $\omega = 2\pi \times \text{rpm}/60$.
• Omitting the radius in the denominator of $F = \dfrac{m v^{2}}{r}$.