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3 years ago

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A cat dozes on a stationary merry-go-round, at a radius of 5.5 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 6.3 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

1 answer:

bixtya [17]3 years ago

6 0

Answer:

$u_{s}=0.56$

Explanation:

For the cat to stay in place on the merry go round without sliding the magnitude of maximum static friction must be equal to magnitude of centripetal force

$F_{s.max}=\frac{mv^2}{r} \\$

Where the r is the radius of merry-go-round and v is the tangential speed

but

$F_{s.max}=u_{s}F_{N}=u_{s}mg$

So we have

$u_{s}mg=\frac{mv^2}{r}\\ u_{s}=\frac{v^2}{gr}\\ Where\\v=\frac{2\pi R}{T} \\So\\u_{s}=\frac{(\frac{2\pi R}{T} )^2}{gr} \\u_{s}=\frac{4\pi^2 r}{gT^2}$

Substitute the given values

So

$u_{s}=\frac{4\pi^2 5.5m}{(9.8m/s^2)(6.3s)^2} \\u_{s}=0.56$

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