Question

A cat dozes on a stationary merry-go-round, at a radius of 5.7 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

Answer:

0.76

Explanation:

we are given:

radius (r) =5.7 m

speed (s) = 1 revolution in 5.5 seconds

acceleration due to gravity (g) = 9.8 m/s^{2}

coefficient of friction (Uk) = ?

 we can get the minimum coefficient of friction from the equation below

centrifugal force = frictional force

m x r x ω^{2} = Uk x m x g

r x ω^{2} = Uk x g

Uk = $\frac{ r x ω^{2} }{g}$

where ω (angular velocity) = $\frac{2π}{time}$

= $\frac{2π }{5.5}$ = 1.14

Uk = $\frac{ 5.7 x 1.14^{2} }{9.8}$ = 0.76