Respuesta :
Answer:
Approximately [tex]6.2\; {\rm rpm}[/tex], assuming that the gravitational field strength is [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].
Explanation:
Let [tex]\omega[/tex] denote the required angular velocity of this Ferris wheel. Let [tex]m[/tex] 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), [tex]m\, g[/tex], and possibly
- Normal force [tex]F_\text{normal}[/tex] that the Ferris wheel exerts on this passenger (upwards.)
This passenger would feel "weightless" if the normal force on them is [tex]0[/tex]- that is, [tex]F_\text{normal} = 0[/tex].
The net force on this passenger is [tex](m\, g - F_\text{normal})[/tex]. Hence, when [tex]F_\text{normal} = 0[/tex], the net force on this passenger would be equal to [tex]m\, g[/tex].
Passengers on this Ferris wheel are in a centripetal motion of angular velocity [tex]\omega[/tex] around a circle of radius [tex]r[/tex]. Thus, the centripetal acceleration of these passengers would be [tex]a = \omega^{2}\, r[/tex]. The net force on a passenger of mass [tex]m[/tex] would be [tex]m\, a = m\, \omega^{2}\, r[/tex].
Notice that [tex]m\, \omega^{2} \, r = (\text{Net Force}) = m\, g[/tex]. Solve this equation for [tex]\omega[/tex], the angular speed of this Ferris wheel. Since [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex] and [tex]r = 23\; {\rm m}[/tex]:
[tex]\begin{aligned} \omega^{2} = \frac{g}{r}\end{aligned}[/tex].
[tex]\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}[/tex].
The question is asking for the angular velocity of this Ferris wheel in the unit [tex]{\rm rpm}[/tex], where [tex]1\; {\rm rpm} = (2\, \pi\; {\rm rad}) / (60\; {\rm s})[/tex]. Apply unit conversion:
[tex]\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}[/tex].
