A child puts a penny of mass 2.5g is on a merry-go-round with a steel flour. The

Question

A child puts a penny of mass 2.5g is on a merry-go-round with a steel flour. The coefficients of kinetic and static friction between copper and steel are uk= 0.41 and us= 0.53. At the moment, the merry-go-round is spinning at 0.5 radians per second and the penny is 1.5 m from the center. A. Draw a free body diagram of the top view of the penny. Draw in the radial and tangential axis. Draw in the forces on the penny that are visible in this picture. B. Draw a free body diagram of the side view of the penny at the same moment in time as part a. Draw in the radial and z axis. Draw in the forces of the penny that are visible in this picture. C. What is the current force of static friction on the penny? D. What is the maximum tangential speed of the merry-go-round so that the penny does not slide? Would changing either of the following change your answer to part d: changing the mass of the coin; changing the material of which the coin is made. Explain.

Explanation / Answer

frictional force acts radially weight acts downwards normal force acts upwards frictional force provides the centripetal action maximum static friction=mus*mg=12.99 N centripetal force=m*r*w^2=0.9375 N so current force of static friction on the penny=0.9375 N F(max)=12.99=m*r*w^2 12.99=2.5*1.5*w^2 w=1.86 rad/s v(vax)=w*r=1.86*1.5=2.79 m/s changing the mass and material both affect the answer because w is inversely proportional to mass and F(max) depends upon coefficient of friction

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