Problem 6-10: The angular velocity of a person in a circular fairground ride

A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow has an 8.00 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration whose magnitude is 1.50 times that due to gravity?

Solution:

Centripetal acceleration $a_{c}$ is the acceleration experienced while in uniform circular motion. It always points toward the center of rotation. The relationship between the centripetal acceleration $a_{c}$ and the angular velocity $\omega$ is given by the formula

a_{c}=r\omega^{2}

Now, taking the formula and solving for the angular velocity:

\omega = \sqrt{\frac{a_{c}}{r}}

From the given problem, we are given the following values: $r=8.00\ \text{m}$ and $a_{c}=1.50\times 9.81 \ \text{m/s}^2=14.715\ \text{m/s}^2$ . If we substitute these values in the formula, we can solve for the angular velocity.

\begin{align*} \omega & = \sqrt{\frac{a_{c}}{r}} \\ \\ \omega & = \sqrt{\frac{14.715\ \text{m/s}^2}{8.00\ \text{m}}} \\ \\ \omega & = 1.3561\ \text{rad/sec} \\ \\ \end{align*}

Then, we can convert this value into its corresponding value at the unit of revolutions per minute.

\begin{align*} \omega & = 1.3561\ \frac{\text{rad}}{\text{sec}} \times \frac{60\ \text{sec}}{1\ \text{min}}\times \frac{1\ \text{rev}}{2\pi \ \text{rad}} \\ \\ \omega & = 12.9498\ \text{rev/min} \\ \\ \omega & = 13.0 \ \text{rev/min} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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