Answer:
The coefficient of static friction is 0.39
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 :
[tex]v=\dfrac{2\pi r}{t}[/tex]
[tex]v=\dfrac{2\pi \times 4.3}{6.6}[/tex]
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 :
[tex]\mu mg=\dfrac{mv^2}{r}[/tex]
[tex]\mu=\dfrac{v^2}{rg}[/tex]
[tex]\mu=\dfrac{(4.09)^2}{4.3\times 9.8}[/tex]
[tex]\mu = 0.39[/tex]
So, the coefficient of static friction between the cat and the merry-go-round is 0.39. Hence, this is the required solution.
