A merry-go-round of radius 2 m is rotating about a frictionless pivot. It makes one revolution every 5 s. The moment of inertia of the merry-go-round (about an axis through its center) is 500 kg·m². A child of mass 25 kg, originally standing at the exact center, walks out to the rim. The child can be considered as a point mass.

a) What is the new angular velocity of the merry-go-round?

b) Does the total kinetic energy of the system increase or decrease? By how much?

c) Where does this change in kinetic energy go to or come from?

We have no net external torques on the system, so we can apply conservation of angular momentum. Thus we have

L = I = I_newnew

or

_new = I/I_new

The child when at the center of the merry-go-round contributes nothing to the moment of inertia, but when she is at the rim she contributes

I_child = m_childR²

Thus we have

I = 500 kg·m²

I_new = 500 kg·m² + (25 kg)(2 m)²

         = 600 kg·m²

Converting the units of :

= [1 rev/(5 s)](2 rad/rev) = (2/5) rad/s

Therefore

_new = [(500 kg·m²)/(600 kg·m²)]( (2/5) rad/s)

      = /3 rad/s = 1.047 rad/s

b) Rotational kinetic energy:

K = (1/2)I² = (1/2)(500 kg·m²)((2/5) rad/s)² = 395 J

K_new = (1/2)I_newnew² = (1/2)(600 kg·m²)((1/3) rad/s)² = 329 J

Thus the energy decreases by 395 J = 329 J = 66 J

.

c) The loss is energy is due to work done by friction between the child's feet and the surface of the platform. If there was no friction, the child would slip as the merry-go-round rotates under her feet. The child's muscle's end up dissipating most of the energy as heat.

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