Question

A child of mass 50 kg is standing at the edge of a playground merry-go-round of diameter 3 m. Other children push the merry-go-round faster and faster until the first child's feet begin to slip. (a) If the coefficient of friction between the shoes and the rotating surface is 0.2, how long does it take the merry-go-round to make one revolution? (b) How fast is the first child traveling when he begins to slip? (c) What is the angle between the first child's body and a vertical line before he begins to slip?

Answer 1

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 force equal to acts radially outward on the child, and to balance this force frictional force acts radially inward. But this frictional force has a limiting value, and as the rotating speed increases, the centripetal force exceeds the frictional force and slipping starts.

Frictional force     where N 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 he begins to slip

  ANS

(c) Child's body makes an angle of around 90 deg with vertical towards mery go round. This is to conterbalance the torque of frictional force.

Please comment if all the answers are right.