Question

In an old-fashioned amusement park ride, passengers stand inside a$5.0m$-diameter hollow steel cylinder with their backs against the wall. The cylinder begins to rotate about a vertical axis. Then the floor on which the passengers are standing suddenly

drops away! If all goes well, the passengers will “stick” to the wall and not slide. Clothing has a static coefficient of friction against steel in the range $0.60$to $1.0$and a kinetic coefficient in the range$0.40$to $0.70.$A sign next to the entrance says “No children under $30kg$allowed.” What is the minimum angular speed, in rpm, for which the ride is safe?

Short answer

The minimum angular speed is$24rpm$

Step-by-step solution

Given information

diameter of cylinder $d=5m$

mass $m=30kg$

static coefficient of friction${\mu }_{s,min}=0.60;{\mu }_{s,max}=1.0$

kinetic coefficient of friction${\mu }_{k,min}=0.40;{\mu }_{k,max}=0.70$

calculation of minimum angular speed

The force of static friction must be no lower than the weight of the person.

The normal force that cylinder wall applies to person is equal to the centripetal force and it is given by, $N=\frac{m{V}^2}{r}$

The friction force that keeps the person from sliding down is given by,$f_f={\mu }_{s}N$

The weight of the person is given by $F_G=mg$

When the net force is zero, we got,

role="math" localid="1649596502522" $$\begin{gathered} f_f=F_G \\ \Rightarrow {\mu }_{s}N=mg \\ \Rightarrow {\mu }_s\frac{m{V}^2}{r}=mg \\ \Rightarrow V=\sqrt{\frac{rg}{{\mu }_s}}\left(I\right) \end{gathered}$$

Use the lowest coefficient of static friction in the above equation $\left(I\right)$, to get the minimum speed.

$V_{min}=\sqrt{\frac{rg}{{\mu }_{s,min}}}=\sqrt{\frac{2.5\times 9.81}{0.60}}=6.39m/s$

The minimum angular speed is given by,${\omega }_{min}=\frac{V_{min}}{r}=\frac{6.39}{2.5}=2.6rad/s=2.6\times 9.5493\left(rpm\right)=24rpm$