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?

The general equation for the forces acting on the passengers at the topmost point of the ferris wheel is[tex]mg - R = m \omega^2 r[/tex]wheremg is the weight of the passengersR is the normal reaction of the cabin[tex]m \omega^2 r[/tex] is the centripetal forceIn order to feel weightless, the normal reaction felt by the passengers should be zero. Therefore, the equation becomes:[tex]mg=m \omega^2 r[/tex]or[tex]g=\omega^2 r[/tex]where [tex]\omega[/tex] is the angular frequency of the wheel and r is its radius. Since we know its radius, [tex]r= \frac{20 m}{2}=10 m [/tex]we can calculate the angular frequency:[tex]\omega= \sqrt{ \frac{g}{r} } = \sqrt{ \frac{9.81 m/s^2}{10 m} } =0.99 rad/s[/tex]From which we find the frequency at which the ferris wheel should rotate:[tex]f= \frac{\omega}{2 \pi}= \frac{0.99 rad/s}{2 \pi}=0.158 s^{-1} [/tex]This is the number of revolutions per second, so the number of revolutions per minute will be[tex]f=0.158 s^{-1} \cdot 60 = 9.48 min^{-1}[/tex]