A() 61 kg child stands at the rim of a merrygo-round of radius 1.05 m, rotating with an angular speed of 1.87 rad/s The acceleration of gravity is 9.8 m/s^2 What is the child’s centripetal acceleration? Answer in units of m/s What minimum force between her feet and the ﬂoor of the carousel is required to keep her in the circular path? Answer in units of N What minimum coeﬃcient of static friction is required? 3. The attempt at a solution 1) I found the centripetal acceleration since a = v^2/r and v = 1.9635 and r = 1.05. The acceleration is 3.671745 2) I found the force which is F=ma F = 61(3.671745) = 223.976445 N 3) I'm confused as to how I should solve for the minimum of coefficient of friction. Help is appreciated!!