Question

A merry-go-round rotates at the rate of 0.38 rev/s with an 98 kg man standing at a point 1.8 m from the axis of rotation. What is the new angular speed when the man walks to a point 0 m from the center? Consider the merry-go-round is a solid 93 kg cylinder of radius of 1.8 m. Answer in units of rad/s. 024 (part 2 of 2) 10.0 points What is the change in kinetic energy due to this movement? Answer in units of J.

Answer 1

In this problem the angular momentum of the system is conserved. So the angular momentum before and after is equal. The initial angular momentum is (1, +12) = The moment of inertia of merry go round is 1, = MR = 5x93x1.8? = 151 kgm? The moment of inertia of man is 1, = mR2 = 98x1.82 = 317.5 kgm? As the person moves to the centre, the moment of inertia of man becomes zero. So only the moment of inertia of merry go round remains. Now using conservation of angular momentum we get ( Mr?+ mx") a = 5 MR?o=0=(1+2) a ==(1+299 +0.38 m=1.18 rad /s 98 The change in kinetic energy is sx= ( me ]os -}{} MR? +mk?) ax=[x93x1.8).18 -4 (193x1.5 +98 21.5* )0.38° AK =71.1 These are the answers./