Question

Consider an amusement park ride in which participants are rotated about a vertical axis in a cylinder with vertical walls. Once the angular velocity reaches its full value, the floor drops away and friction between the walls and the riders prevents them from sliding down. How many revolutions does the cylinder make each second, given that the minimum coefficient of static friction that is needed to keep the riders from sliding down is 0.695 and the diameter of the cylinder is 13.5 m?

Answer 1

Answer:

0.231 rev/s

Explanation:

$m$ = Mass of cylinder

$d$ = Diameter of the cylinder = $13.5\ \text{m}$

$r$ = Radius = $\dfrac{13.5}{2}=6.75\ \text{m}$

$\mu$ = Coefficient of static friction = $0.695$

$N$ = Normal force = $mr\omega^2$

$g$ = Acceleration due to gravity = $9.81\ \text{m/s}^2$

The force equation of the system will be

$\mu N=mg\\\Rightarrow \mu\times mr\omega^2=mg\\\Rightarrow \omega=\sqrt{\dfrac{g}{\mu r}}\\\Rightarrow \omega=\sqrt{\dfrac{9.81}{0.695\times 6.75}}\\\Rightarrow \omega=1.45\ \text{rad/s}$

Converting to rev/s

$\dfrac{1.45}{2\pi}=0.231\ \text{rev/s}$

The number of revolutions of the cylinder is 0.231 rev/s.