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

1 Answer
Apr 9, 2017

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