Answer to Question #140681 in Mechanics | Relativity for Zach William Ament

Question

The rotor is an amusement park ride where people stand against the inside of a cylinder. Once the cylinder is spinning fast enough, the floor drops out.

If the coefficient of static friction between a person and the wall of the cylinder is 0.600 and the cylinder has a radius of 3.80 m, what is the minimum angular speed of the cylinder so that the people don’t fall out?

Answer 1

Given:-

Coefficient of static friction $\mu=0.600$

Radius of cylinder r=3.80m

Let $\omega$ be the angular speed of rotor

To avoid slipping the frictional force must be balanced by the weight of the person

$\therefore mg=\mu\dfrac{mv^2}{r}\\\to g=\mu\dfrac{r^2\omega^2}{r}$ ( as v=r$\omega)$

$\to \omega^2=\dfrac{g}{\mu r}$

$\to w=\sqrt{\dfrac{9.8}{0.600\times3.80}}$

$\to \omega=\sqrt{\dfrac{9.8}{2.28}}$

$\to \omega=\sqrt{4.71}=2.071rad s^{-1}$