In an amusement park ride called The Roundup, passengers stand inside a 16-m-diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane, as shown in FIGURE P8.51.
a. Suppose the ring rotates once every 4.5 s. If a rider’s mass is 55 kg, with how much force does the ring push on her at the top of the ride? At the bottom?
b. What is the longest rotation period of the wheel that will prevent the riders from falling off at the top?

Given that ring rotates once every 4.5 s.

And mass of rider is 55 kg

Diameter of ring = 16 m

First draw free body diagram and the equate forces

From the diagram we can get

$$\begin{gathered} T_{Top}=\frac{m{v}^2}{r}-mg........................\left(1\right) \\ {T}_{Bottom}=\frac{m{v}^2}{r}+mg........................\left(2\right) \end{gathered}$$

We can find v by using v=ω r

We have period as 4.5 s So

$\omega =\frac{2\pi }{T}$

And $v=\frac{2\pi r}{T}=\frac{(2\times 3.14)\left(8m\right)}{4.5s}=11.17m/s$

Now substitute the given values in equation (1) and (2)

role="math" localid="1649077252870" $T_{Top}=\frac{\left(55kg\right){(11.17m/s)}^2}{8m}-\left(55kg\right)(9.8m/s^2)=318.24N=0.32N$

Similarly

$T_{Top}=\frac{\left(55kg\right){(11.17m/s)}^2}{8m}+\left(55kg\right)(9.8m/s^2)=1397.34N=1.4KN$

Given that ring rotates once every 4.5 s.

Mass of rider = 55 kg

Diameter of ring = 16 m

We have calculated velocity = 11.17 m/s

For longest rotation period prevent riders to fall means resultant force is zero

$$\begin{gathered} \frac{m{v}^2}{r}-mg=0 \\ v=\sqrt{rg} \end{gathered}$$

Substitute the values given we get

$v=\sqrt{\left(8m\right)(9.8m/s^2)}=8.86m/s$

We know

$T=\frac{2\pi r}{v}$

Substitute values we get

$T=\frac{(2\times \mathrm{\pi }\times 8)}{8.86}=5.67s$