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 rim, walks in to the exact center. 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 energy go to or come from?

a) This is just like problem 10-49, except this time the child starts at the outside and walks inward instead of vice-versa. Apply conservation of angular momentum (there are no net external torques on the system of merry-go-round and child). Thus we have

L = constant = I_{i i} = I_{f f}

or

_f = I_{i i}/ I_f

so all that is left to do is to find the initial angular velocity and the initial and final moments of inertia. We are told the period (T = 5 seconds) so the initial angular velocity is

_i = 2/T = 1.257 rad/s

The initial moment-of-inertia is that of the merry-go-round plus that of the child located at the rim

I_i = 500 Kg·m² + mR²

= 500 Kg·m² + (25 kg)(2 m)²

= 600 kg·m²

Since the child ends up at the center (r=0), she/he contributes no rotational inertia in the final situation, so the I_f is just that of the merry-go-round, i.e.

I_f = 500 Kg·m²

Plugging these in gives

_f = (600 kg·m²)(1.257 rad/s)/(500 Kg·m²)

= 1.51 rad/sec

b) The initial rotational kinetic energy is

K_i = ½I_ii²

= ½(600 kg·m²)(1.257 rad/s)²

= 474 J

the final rotational kinetic energy is

K_f = ½I_ff²

= ½(500 kg·m²)(1.51 rad/s)²

= 568 J

Thus the energy has increased by 94 J.

c) The extra energy had to have come from somewhere, in this case from stored (chemical) energy in the child's muscles - the child did work on the merry-go-round while walking in to the center.