Worked example 7.3: Amusement park ride

Question: An amusement park ride consists of a vertical cylinder that spins about a vertical axis. When the cylinder spins sufficiently fast, any person inside it is held up against the wall. Suppose that the coefficient of static friction between a typical person and the wall is $\mu=0.25$. Let the mass of an typical person be $m=60 {\rm kg}$, and let $r=7 {\rm m}$ be the radius of the cylinder. Find the critical angular velocity of the cylinder above which a typical person will not slide down the wall. How many revolutions per second is the cylinder executing at this critical velocity?

Answer: In the vertical direction, the person is subject to a downward force $m g$ due to gravity, and a maximum upward force $f=\mu R$ due to friction with the wall. Here, $R$ is the normal reaction between the person and the wall. In order for the person not to slide down the wall, we require $f>m g$. Hence, the critical case corresponds to

$$f=\mu R=m g.$$

In the radial direction, the person is subject to a single force: namely, the reaction $R$ due to the wall, which acts radially inwards. If the cylinder (and, hence, the person) rotates with angular velocity $\omega$, then this force must provided the acceleration $r \omega^2$ towards the axis of rotation. Hence,

$$R = m r \omega^2.$$

It follows that, in the critical case,

$$\omega = \sqrt{\frac{g }{\mu r}} = \sqrt{\frac{9.81}{0.25\times 7}} = 2.37 {\rm rad/s}.$$

The corresponding number of revolutions per second is

$$f = \frac{\omega}{2 \pi} = \frac{2.37}{2\times 3.1415} = 0.38 {\rm Hz}.$$