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Vika [28.1K]
3 years ago
7

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 rid

e 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?
Physics
1 answer:
ioda3 years ago
4 0
<span>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</span>
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