A 50.0 kg child stands at the rim of a merry-go-round of radius2.05 m, rotating

Question

A 50.0 kg child stands at the rim of a merry-go-round of radius2.05 m, rotating with an angular speedof 3.25 rad/s. (a) What is the child's centripetalacceleration?
1 m/s2

(b) What is the minimum force between her feet and the floor of thecarousel that is required to keep her in the circular path?
2 N

(c) What minimum coefficient of static friction is required?
3
(a) What is the child's centripetalacceleration?
1 m/s2

(b) What is the minimum force between her feet and the floor of thecarousel that is required to keep her in the circular path?
2 N

(c) What minimum coefficient of static friction is required?
3

Explanation / Answer

   a.   Centripetalacceleration      a   =   r* 2          given   radius   r   =   2.05   m                angularspeed      =   3.25   rad/s          a   =   2.05* 3.252                =   21.65   m/s2                =   21.65   m/s2    b.   Force between floor andchild'sfeet   F   =   centripetalforce                                                                               =   m* a                                                                               =   50.0* 21.65                                                                               =   1082.5   N    c.   Force offriction   Ff   =   F           *m * g   =   1082.5           coefficient of staticfriction      =   1082.5/ 50.0 * 9.8                                                             =   2.209

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