Question

How many revolutions per minute would a 20 m -diameter ferris wheel need to make for the passengers to feel "weightless" at the topmost point?

Answer 1

The general equation for the forces acting on the passengers at the topmost point of the ferris wheel is

`mg - R = m \omega^2 r`
where
mg is the weight of the passengers
R is the normal reaction of the cabin

`m \omega^2 r` is the centripetal force

In order to feel weightless, the normal reaction felt by the passengers should be zero. Therefore, the equation becomes:

`mg=m \omega^2 r`
or

`g=\omega^2 r`
where
`\omega` is the angular frequency of the wheel and r is its radius. Since we know its radius,

`r= (20 m)/(2)=10 m`
we can calculate the angular frequency:

`\omega= \sqrt{ (g)/(r) } = \sqrt{ (9.81 m/s^2)/(10 m) } =0.99 rad/s`
From which we find the frequency at which the ferris wheel should rotate:

`f= (\omega)/(2 \pi)= (0.99 rad/s)/(2 \pi)=0.158 s^(-1)`
This is the number of revolutions per second, so the number of revolutions per minute will be

`f=0.158 s^(-1) \cdot 60 = 9.48 min^(-1)`