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?

From Newton’s second law, the net force acting in the vertical direction of the Ferris wheel can be equated to the centripetal force.

At the topmost point on the Ferris wheel, the passengers will experience weightlessness as there is no contact between them and the seat. So, the normal reaction on the Ferris wheel will be zero.

The given data can be listed below as,

• The diameter of the Ferris wheel is, $d=25\mathrm{m}$.
• The acceleration due to gravity is, $g=9.81\mathrm{m}/{\mathrm{s}}^2$.

The forces on the Ferris wheel can be shown as,

Here, mgis the total weight of the Ferris wheel and passengers, $F_{\mathrm{N}}$is the normal reaction on the Ferris wheel.

From the Newton’s second law, the equation can be expressed as,

$\begin{array}{c}\sum F_y=F_{\mathrm{c}}\\ mg-F_{\mathrm{N}}=m{a}_{\mathrm{c}}\\ \frac{m{v}^2}{r}=mg-F_{\mathrm{N}}\\ v=\sqrt{\frac{r\left(mg-F_{\mathrm{N}}\right)}{m}}\end{array}$

Here, $F_{\mathrm{c}}$is the centripetal force, $a_{\mathrm{c}}$is the centripetal acceleration of the Ferris wheel.

At the topmost point on theFerris wheel, the passengers feel weightlessness.So, the normal reaction will become zero.

Now, the above equation can be written as,

$\begin{array}{c}v=\sqrt{\frac{r\left(mg-0\right)}{m}}\\ =\sqrt{\frac{rmg}{m}}\\ =\sqrt{rg}\\ =\sqrt{\frac{d}{2}\times g}\end{array}$

Substitute the values in the above equation.

localid="1654878994240" $\begin{array}{c}v=\sqrt{\frac{25 \mathrm{m}}{2}\times \left(9.81 \mathrm{m}/{\mathrm{s}}^2\right)}\\ v=\sqrt{122.625} \mathrm{m}/\mathrm{s}\\ v=11.074 \mathrm{m}/\mathrm{s}\end{array}$

The speed of the Ferris wheel can be expressed as,

$\begin{array}{c}v=\frac{2\pi r}{T}\\ v=\left(\frac{2\pi r}{\frac{1}{N}}\right)\\ \frac{1}{N}=\left(\frac{2\pi \left(\frac{d}{2}\right)}{v}\right)\\ N=\left(\frac{v}{\pi d}\right)\end{array}$

Here, N is the number of rotations of the Ferris wheel per unit time, T is the time taken by the Ferris wheel to do one rotation.

Substitute the values in the above equation.

$\begin{array}{c}N=\frac{11.074\mathrm{m}/\mathrm{s}\left(\frac{60\mathrm{s}}{1\min }\right)}{\pi \times 25\mathrm{m}}\\ =8.459\mathrm{rpm}\end{array}$

Thus, the rotation of a Ferris wheel is $8.459\mathrm{rpm}$.