Question

How many revolutions per minute would a 25 m -diameter ferris wheel need to make for the passengers to feel "weightless" at the topmost point?

Answer 1

Refer to the diagram shown below.

Let ω =  the angular velocity (rad/s) of the wheel.

At the topmost point, the passenger, with mass m,  will feel weightless if the passenger's weight matches the centripetal force.
The passenger's weight is mg.
The centripetal force is mω²r.

Therefore
mω²r = mg
ω = √(g/r)

Because g = 9.8 m/s² and r = 25/2 = 12.5 m, therefore
ω = √(9.8/12.5) = 0.8854 rad/s

Because 1 revolution per second is 2π rad/s, therefore
$\omega = (0.8854 \, \frac{rad}{s} )*( \frac{1}{2 \pi} \, \frac{rev}{rad} )*(60 \, \frac{s}{min} ) = 8.455 \, rev/min$

Answer: 8.455 rev/min