# If the horizontal circular path the riders follow has a 7.00 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration whose magnitude is 2.25 times that due to gravity?

## A fairgrounds ride spins its occupants inside a flying saucer-shaped container. rev/min

Apr 9, 2017

centripetal acceleratrion can be expressed as:
$a = r {\omega}^{2}$

Where $\omega$ is angular velocity . Equating this to the question:
$r {\omega}^{2} = 2.25 \times g$

Substituting in values:
$7.00 \times {\omega}^{2} = 2.25 \times 9.81$

solving for $\omega$ we get:
$\omega = 1.78$

Angular velocity is the rate of change of angle
$\omega = \frac{2 \cdot \pi}{T}$

Where T = time period (time for one revolution)

Hence we can solve for T

$T = \frac{2 \cdot \pi}{\omega} = 3.5 s$

So it completes one revolution every 3.5s. In one minute it will complete:

$\frac{60}{3.5} = 17.1$ revolutions