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vova2212 [387]
3 years ago
8

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
Physics
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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                                7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂²  = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂²  = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂²  + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

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x₂ ₁  =  451.27*10∧5/1160

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