Question

In an old-fashioned amusement park ride, passengers stand inside a 5.0-m-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 30 kg allowed.” What is the minimum angular speed, in rpm, for which the ride is safe?

Short answer

The minimum angular speed for safe ride in rpm is 24.446 rpm

Step-by-step solution

Given Information

µ =0.60

r =2.5 m

Explanation

The frictional force is given by

$f=\mu Fn$

Now equate force components

$\frac{\mu m{v}^2}{r}=mg..........................\left(1\right)$

So

$v=\sqrt{\frac{rg}{\mu }}\dots \dots \dots \dots \dots .\left(2\right)$

Substitute v=ωr , and solve for ω, we get

$\omega =\sqrt{\frac{g}{\mu r}}...............\left(3\right)$

Substitute the given values in equation(3) we get

$\omega =\sqrt{\frac{9.8m/s^2}{(0.6)(2.5m)}}=2.556rad/sec$

Convert in rpm

ω=24.446 rpm