Physics question?

A 50.0 kg child stands at the rim of a merry-go-round of radius 1.65 m, rotating with an angular speed of 2.75 rad/s





(c) What minimum coefficient of static friction is required?





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Physics question?
force of friction=mass*acceleration


(u)n=m(V^2/R)


n=mg


so umg=m(V^2/r)


u=V^2/gr


v^2/r=rw^2





u=rw^2/g=(1.65m)(2.75rad/s)^2/9.8m/s^2


where u is the coefficient of static friction


u=1.273