3.6.1.1
Centripetal acceleration has three equivalent forms
Circular Motion — AQA A-Level Physics
Key Definition
Centripetal acceleration — The acceleration of an object towards the centre of a circle when the object is in uniform circular motion. Always directed towards the centre and perpendicular to the velocity.
$$\begin{aligned}
a &= \frac{\(v^{2}\)}{r} \\
&= \omega^2 r
\end{aligned}$$
$$\begin{aligned}
a &= \frac{\(v^{2}\)}{r} \\
&= r\omega^2 \\
&= v\omega
\end{aligned}$$
- $a$: centripetal accelerationThe rate of change of velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹.. A vector quantity. Measured in m s⁻².The accelerationThe rate of change of velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹.. A vector quantity. Measured in m s⁻². directed towards the centre of the circular path, required to maintain circular motion. (m \(s^{-2}\))
- $v$: linear speed (m \(s^{-1}\))
- $r$: radius of circular orbit (m)
- $\omega$: angular speed (rad \(s^{-1}\))
- The first form $a = v^2/r comes directly$ from the definition.
- Substituting $v = r \omega gives a = r \omega^2$.
- Substituting $r = v/\omega gives a = v \omega$.
- Choose whichever form matches the quantities given in the question.
Worked Example
A ball on a string has radius 1.5 m and angular speed 3.5 rad \(s^{-1}\). Calculate the centripetal accelerationThe rate of change of velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹.. A vector quantity. Measured in m s⁻².The acceleration directed towards the centre of the circular path, required to maintain circular motion. if both the radius and angular speed are doubled.
Show Solution
1
Write the modified equation
$$a = (2r) \times (2\omega)^2 = 2r \times 4\omega^2 = 8r\omega^2$$
2
Substitute values
$$a = 8 \times 1.5 \times 3.5^2 = 147 \text{ m s}^{-2}$$
Answer
$a = 147$ m s$^{-2}$ (8 times the original)