Question

A fairgrounds ride spins its occupants inside a flying-saucer-shaped container. If the horizontal circular path the riders follow has a 7.00 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration 1.85 times that of gravity? rev/min

Answer 1

We know the centripetal acceleration is given by:

$a_c=\omega^2r$

In this case we know that the radius is 7 m and the we want the centripetal acceleration to be 1.85 times that of gravity, then we have:

$\begin{gathered} 1.85g=7\omega^2 \\ \omega^2=\frac{1.85g}{7} \\ \omega=\sqrt{\frac{1.85g}{7}} \\ \omega=\sqrt{\frac{1.85(9.8)}{7}} \\ \omega=1.61 \end{gathered}$

Hence, the angular speed needed is 1.61 rad/s. To determine the revolutions per minute we just need to convert the angular speed we found:

$1.61\frac{rad}{s}\cdot\frac{1\text{ rev}}{2\pi\text{ rad}}\cdot\frac{60\text{ s}}{1\text{ min}}=15.37\frac{rev}{min}$

Therefore, at 15.37 rpm the riders will be subjected to 1.85 times the acceleration of gravity.