Question

How many revolutions per minute would a 26m -diameter Ferris wheel need to make for the passengers to feel "weightless" at the topmost point?

Answer 1

How many revolutions per minute would a 26m -diameter Ferris wheel need to make for the passengers to feel "weightless" at the topmost point?

Solution:

$D = 26m -$ diameter of the wheel;

$F_{c} -$ centrifugal force;

$V -$ velocity at the topmost point;

Newton's second law for passengers at the topmost point ("weightless" means that the resultant of all forces is zero)

$$\overrightarrow{\mathrm{mg}} + \overrightarrow{\mathrm{F_{c}}} = \overrightarrow{0}$$ $$y: -mg + F_{c} = 0$$ $$F_{c} = mg$$

Formula of the centrifugal force $(D = 2R)$:

$$F_{c} = m \frac{V^{2}}{R} = \frac{2mV^{2}}{D}$$

(2)in(1):

$$\frac{2mV^{2}}{D} = mg$$ $$2mV^{2} = Dmg$$ $$V = \sqrt{\frac{Dg}{2}}$$

Formula for the frequency:

$$\nu = \frac{1}{T} = \frac{1}{\frac{2\pi R}{V}} = \frac{V}{2\pi R} = \frac{V}{\pi D}$$

(2)in(3):

$$\nu = \sqrt{\frac{Dg}{2}} \cdot \frac{1}{\pi D} = \sqrt{\frac{g}{2\pi D}} = \sqrt{\frac{9.8 \frac{m}{s^{2}}}{2\pi \cdot 26m}} = 0.245 \frac{\text{revolutions}}{\text{sec}} = 14.7 \frac{\text{revolutions}}{\text{min}}$$

Answer: Ferris wheel need to make $14.7\frac{\text{revolutions}}{\text{min}}$ for the passengers to feel "weightless" at the topmost point.