Question

How many revolutions per minute would a 23 m -diameter Ferris wheel need to make for the passengers to feel "weightless" at the topmost point?

Answer 1

Approximately
`6.2\; {\rm rpm}`, assuming that the gravitational field strength is
`g = 9.81\; {\rm m\cdot s^(-2)}`.

Step-by-step explanation:

Let
`\omega` denote the required angular velocity of this Ferris wheel. Let
`m` denote the mass of a particular passenger on this Ferris wheel.

At the topmost point of the Ferris wheel, there would be at most two forces acting on this passenger:

• Weight of the passenger (downwards),
  `m\, g`, and possibly
• Normal force
  `F_\text{normal}` that the Ferris wheel exerts on this passenger (upwards.)

This passenger would feel "weightless" if the normal force on them is
`0`- that is,
`F_\text{normal} = 0`.

The net force on this passenger is
`(m\, g - F_\text{normal})`. Hence, when
`F_\text{normal} = 0`, the net force on this passenger would be equal to
`m\, g`.

Passengers on this Ferris wheel are in a centripetal motion of angular velocity
`\omega` around a circle of radius
`r`. Thus, the centripetal acceleration of these passengers would be
`a = \omega^(2)\, r`. The net force on a passenger of mass
`m` would be
`m\, a = m\, \omega^(2)\, r`.

Notice that
`m\, \omega^(2) \, r = (\text{Net Force}) = m\, g`. Solve this equation for
`\omega`, the angular speed of this Ferris wheel. Since
`g = 9.81\; {\rm m\cdot s^(-2)}` and
`r = 23\; {\rm m}`:

`\begin{aligned} \omega^(2) = (g)/(r)\end{aligned}`.

`\begin{aligned} \omega &= \sqrt{(g)/(r)} \\ &= \sqrt{\frac{9.81\; {\rm m \cdot s^(-2)}}{23\; {\rm m}}} \\ &\approx 0.653\; {\rm rad \cdot s^(-1)} \end{aligned}`.

The question is asking for the angular velocity of this Ferris wheel in the unit
`{\rm rpm}`, where
`1\; {\rm rpm} = (2\, \pi\; {\rm rad}) / (60\; {\rm s})`. Apply unit conversion:

`\begin{aligned} \omega &\approx 0.653\; {\rm rad \cdot s^(-1)} \\ &= 0.653\; {\rm rad \cdot s^(-1)} * \frac{1\; {\rm rpm}}{(2\, \pi\; {\rm rad}) / (60\; {\rm s})} \\ &= 0.653\; {\rm rad \cdot s^(-1) * \frac{60\; {\rm s}}{2\, \pi\; {\rm rad}} * 1\; {\rm rpm} \\ &\approx 6.2\; {\rm rpm} \end{aligned}`.