A child of mass 50 kg is stand
A child of mass 50 kg is standing at the edge of a playgroundmerry-go-round of diameter 3 m. Other children push themerry-go-round faster and faster until the first child’s feet beginto slip. (a) If the coefficient of friction between the shoes andthe rotating surface is 0.2, how long does it take themerry-go-round to make one revolution? (b) How fast is the firstchild traveling when he begins to slip? (c) What is the anglebetween the first child’s body and a vertical line before he beginsto slip?
Answer:
Solution:
Given The mass of first child on merry go round 
Radius of merry go round 
Coefficient of friction 
When the merry go round starts rotating a centripetal forceequal to 
acts radially outward on the child, and to balance this forcefrictional force acts radially inward. But this frictional forcehas a limiting value, and as the rotating speed 
increases,the centripetal force exceeds the frictional force and slippingstarts.
Frictional force 
whereN in normal reaction on the child
As there in no acceleration in upward direction,
So normal Reaction 
So, 
At the time of slipping 
So, 
So, time required for One revolution 
ANS
(b) So tangential velocity of child when hebegins to slip
ANS
(c) Child’s body makes an angle of around 90 deg with verticaltowards mery go round. This is to conterbalance the torque offrictional force.
Please comment if all the answers are right.
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