Question

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

Answer 1

`~~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`