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?

Question

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

19211 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-09T21:00:18

centripetal acceleratrion can be expressed as:
$a=romega^2$

Where $omega$ is angular velocity . Equating this to the question:
$romega^2=2.25timesg$

Substituting in values:
$7.00timesomega^2=2.25times9.81$

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

Angular velocity is the rate of change of angle
$omega=(2*pi)/T$

Where T = time period (time for one revolution)

Hence we can solve for T

$T=(2*pi)/omega=3.5s$

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

$60/3.5=17.1$ revolutions