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A cat dozes on a stationary merry-go-round in an amusement park, at a radius of 5.4 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 6.0 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 (or the cat clinging with its claws)?

Respuesta :

Answer:

μ = 0.604

Explanation:

For the cat to stay in place on the merry go round, the maximum static frictional force must be equal in magnitude to that of the centripetal force.

Now, Centripetal force is given as;

Fc = mv²/r

Where r is radius and v is tangential speed and m is mass.

We also know that maximum static frictional force is given by;

F_static = μmg

Where μ is coefficient of friction

Now, equating both forces, we have;

mv²/r = μmg

Divide through by m;

v²/r = μg

Now, tangential speed can be expressed as;

v = circumference/period

Thus, v = 2πr/T

Where T is period of rotation and

2πr is the circumference of the merry go round.

Thus,

v²/r = μg is now;

(2πr/T)²/r = μg

Making μ the subject, we have;

(2πr/T)²/rg = μ

μ = [(2π x 5.4)/6]²/(5.4 x 9.8)

μ = 0.604

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