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:

\$Fc = m ac = 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

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}\$.