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, 
$r= \frac{20 m}{2}=10 m$
we can calculate the angular frequency:
$\omega= \sqrt{ \frac{g}{r} } = \sqrt{ \frac{9.81 m/s^2}{10 m} } =0.99 rad/s$
From which we find the frequency at which the ferris wheel should rotate:
$f= \frac{\omega}{2 \pi}= \frac{0.99 rad/s}{2 \pi}=0.158 s^{-1}$
This is the number of revolutions per second, so the number of revolutions per minute will be
$f=0.158 s^{-1} \cdot 60 = 9.48 min^{-1}$