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

1 Answer
A08
Feb 21, 2017

#~~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=ma_{c}=\frac {mv^{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 = romega# and
#omega=2pif#
where #f# is frequency in number of cycles per second.

So that
#a_c=v^2/r=romega^2# ......(2)
#=>a_c=r(2pif)^2#
#=>4pi^2f^2r=a_c#.....(3)

We are required to find #f# per minute for a given centripetal acceleration which is #1.5xxg#.
Solving (3) for #f# per minute and inserting given values we get
#f=sqrt(a_c/(4pi^2r))xx60#
#f=sqrt((1.5xx9.81)/(4pi^2xx8))xx60~~13#

Related questions
See all questions in Circular Motion
Impact of this question
2279 views around the world
You can reuse this answer
Creative Commons License