Question

A cat dozes on a stationary merry-go-round, at a radius of 4.3 m from the center of the ride. The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 6.6 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

The coefficient of static friction is 0.39

Step-by-step explanation:

It is given that,

Radius of merry go round, r = 4.3 m

Time for one complete rotation, t = 6.6 s

The tangential speed of the operator is calculated as :

`v=(2\pi r)/(t)`

`v=(2\pi * 4.3)/(6.6)`

v = 4.09 m/s

When the cat is moving in a circular path it will posses centripetal force which is balanced by the frictional force as :

`\mu mg=(mv^2)/(r)`

`\mu=(v^2)/(rg)`

`\mu=((4.09)^2)/(4.3* 9.8)`

`\mu = 0.39`

So, the coefficient of static friction between the cat and the merry-go-round is 0.39. Hence, this is the required solution.