Answer :
Answer:
[tex]u_{s}=0.56[/tex]
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
[tex]F_{s.max}=\frac{mv^2}{r} \\[/tex]
Where the r is the radius of merry-go-round and v is the tangential speed
but
[tex]F_{s.max}=u_{s}F_{N}=u_{s}mg[/tex]
So we have
[tex]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}[/tex]
Substitute the given values
So
[tex]u_{s}=\frac{4\pi^2 5.5m}{(9.8m/s^2)(6.3s)^2} \\u_{s}=0.56[/tex]