Question

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

Answer 1

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