Answer:
The coefficient of static friction is 0.29
Explanation:
Given that,
Radius of the merry-go-round, r = 4.4 m
The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.
We need to find 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. For this the centripetal force is balanced by the frictional force.
[tex]\mu mg=\dfrac{mv^2}{r}[/tex]
v is the speed of cat, [tex]v=\dfrac{2\pi r}{t}[/tex]
[tex]\mu=\dfrac{4\pi^2r}{gt^2}\\\\\mu=\dfrac{4\pi^2\times 4.4}{9.8\times (7.7)^2}\\\\\mu=0.29[/tex]
So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.
