If a mass \begin{align*} m \end{align*}

\begin{align*}r = \frac{m v^2}{|\vec{F}|}\text{ [1]}\end{align*}Alternatively, to keep this mass moving at this velocity in a circle of this radius, one needs to apply a centripetal force of

\begin{align*}\vec{F_c} = \frac{mv^2}{r} \text{ [2]}\end{align*}By Newton's Second Law, this is equivalent to a centripetal acceleration of

\begin{align*}\vec{F_c} =\cancel{m}\vec{a_c} = \cancel{m}\frac{v^2}{r} \text{ [3]}\end{align*}

Centripetal Force\begin{align*} F_C = \frac{mv^2}{r} \begin{cases} m & \text{mass (in kilograms, kg)}\\ v & \text{speed (in meters per second, m/s}\text{)}\\ r & \text{radius of circle} \end{cases}\end{align*}

Centripetal Acceleration

\begin{align*} a_C = \frac{v^2}{r} \begin{cases} v & \text{speed (in meters per second, m/s}\text{)}\\ r & \text{radius of circle} \end{cases}\end{align*}

#### Example

If you are 4m from the center of a Merry-Go-Round that is rotating at 1 revolution every 2 seconds, what is your centripetal acceleration?

First, we need to find your tangential velocity. We can do this using the given angular velocity.

\begin{align*} \omega&=\frac{2\pi\text{ rad}}{2\text{ s}}\\ \omega&=\pi\text{ rad/s}\\ \omega&=\frac{v}{r}\\ v&=\omega r\\ v&=\pi\;\text{rad/s}*4\;\text{m}\\ v&=4\pi\;\text{m/s} \end{align*}

Now we can find your centripetal acceleration using the radius of your rotation.

\begin{align*} a_c&=\frac{v^2}{r}\\ a_c&=\frac{(4\pi\;\text{m/s})^2}{4\;\text{m}}\\ a_c&=4\pi^2\;\text{m/s}^2\\ \end{align*}

### Interactive Simulations

### Review

- A 6000 kg roller coaster goes around a loop of radius 30m at 6 m/s. What is the centripetal acceleration?

- For the Gravitron ride above, assume it has a radius of 18 m and a centripetal acceleration of 32 m/s
^{2}. Assume a person is in the graviton with 180 cm height and 80 kg of mass. What is the speed it is spinning at? Note you may not need all the information here to solve the problem.

**Review (Answers)**

1. 1.2 m/s^{2}

2. 24 m/s