While merry-go-rounds are seen as dangerous playground equipment by some, by oth

Question

While merry-go-rounds are seen as dangerous playground equipment by some, by others they are a source of a great deal of physics. At a nearby playground, you come upon an empty merry-go-ronud that is rotating about its central (assumed to be frictionless) axis at 3 rad/s. Despite the bars sticking out of the merry-go-round (drawn as white rectangles), we will treat it as a uniform cylinder with a diameter of 1.8m. The initial rotational inertia of the merry-go-round is 200 kgm2.

a. While you watch, three kids of equal mass (60kg) who were standing by the side of the merry-go-round jump in the radial directiononto the merry-go-round. The children can be treated as point masses at a distance of 0.9m from the center ofthe merry-go-round. What is the new rotational velocity of the merry-go-round?

Explanation / Answer

moment of inertia of a point mass at a distance of d

from the axis of rotation=mass*distance^2

hence moment of inertia of each kid=60*0.9^2=48.6 kg.m^2

initial moment of inertia=200 kg.m^2

final moment of inertia (after the kids have jumped onto it)

=200+3*48.6=345.8 kg.m^2

using conservation of angular momentum principle:

initial moment of inertia*initial angular speed=final moment of inertia*final angular speed

==>200*3=345.8*final angular veloicty

==>final angular velocity=200*3/345.8=1.7351 rad/sec

part b:

friction force should balance the centripetal force

hence friction force=m*angular speed^2*radius

=60*9^2*0.9=4374 N

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