Answer:0.53
Explanation:
Given
Radius of merry-go-round [tex]r=4.6 m[/tex]
Time Period [tex]T=5.9 s[/tex]
Velocity of merry-go-round[tex]=\frac{distance\ moved}{time}[/tex]
[tex]v=\frac{2\pi \cdot r}{T}[/tex]
[tex]v=\frac{2\pi \cdot 4.6}{5.9}[/tex]
[tex]v=4.89 m/s[/tex]
Now Friction Force will nullify the centripetal Force
i.e. [tex]\frac{mv^2}{r}=f_r[/tex]
[tex]\frac{mv^2}{r}=\mu mg[/tex] , where [tex]\mu [/tex]coefficient of static friction
[tex]\mu g=\frac{v^2}{r}[/tex]
[tex]\mu =\frac{v^2}{rg}[/tex]
[tex]\mu =\frac{4.89^2}{4.6\times 9.8}[/tex]
[tex]\mu =0.53[/tex]
