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

Answer:

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

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 $r = 23\; {\rm m}$:

$\begin{aligned} \omega^{2} = \frac{g}{r}\end{aligned}$.

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

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:

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