Question

A cat dozes on a stationary merry-go-round, at a radius of 4.6 m from the center of the ride. then the operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 5.5 s. what is the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding?

Answer 1

Radius = 4.6 m
 Time for one complete rotation t = 5.5 s.
 Distance = 2 x 3.14 x R = 2 x 3.14 x 4.6 m = 28.888.
 Velocity V = distance / time = 28.888 / 5.5 s = 5.25 m/s
 Force exerted by cat Fc = mV^2 / R = (mx 5.25^2) / 4.6 m
  Force of the cat Fc = 6m, m being the mass.
 Normal force = Us x m x g = Us x m x 9.81 = Us9.81m
 equating the both forces => Us9.81m = 6m => Us = 6 / 9.81 => Us = 0.6116 So coefficient of static friction = 0.6116