Question

A 5 kg mass is riding on a horizontal platform that is rotating with a period of 8 seconds. The mass is held in place by the friction at a distance of 5.20 meters from the center of rotation. The coefficient of friction between the mass and the platform must at least be?

Answer 1

For the equilibrium:

$$\mu mg=m\omega^2r$$

$$\mu =\frac{4\pi^2r}{gT^2}$$

$$\mu =\frac{4\pi^2(5.2)}{(9.8)(8)^2}=0.327$$