# At how many revolutions per minute the ride should spin in order for the rider to feel a centripetal acceleration of about 1.5 times Earth’s gravitational acceleration?

## A flying-saucer shaped fairground ride is rotating in a horizontal plane. If the rider’s circular path has a radius of 8 m R = 8m

Feb 21, 2017

$\approx 13$ per minute.

#### Explanation:

For circular motion the magnitude of the centripetal force on an object of mass $m$ moving along a path with radius $r$ with tangential velocity $v$ is

$F = m {a}_{c} = \setminus \frac{m {v}^{2}}{r}$ ......(1)
where ${a}_{c}$ is the centripetal acceleration.

The angular velocity $\omega$ of the object about the center of the circle is related to the tangential velocity by the expression

$v = r \omega$ and
$\omega = 2 \pi f$
where $f$ is frequency in number of cycles per second.

So that
${a}_{c} = {v}^{2} / r = r {\omega}^{2}$ ......(2)
$\implies {a}_{c} = r {\left(2 \pi f\right)}^{2}$
$\implies 4 {\pi}^{2} {f}^{2} r = {a}_{c}$.....(3)

We are required to find $f$ per minute for a given centripetal acceleration which is $1.5 \times g$.
Solving (3) for $f$ per minute and inserting given values we get
$f = \sqrt{{a}_{c} / \left(4 {\pi}^{2} r\right)} \times 60$
$f = \sqrt{\frac{1.5 \times 9.81}{4 {\pi}^{2} \times 8}} \times 60 \approx 13$