Question

A 39.0kg child runs with a speed of 2.70m/s tangential to the rim of a stationary merry-go-round . The merry-go-round has a moment of inertia of 506 kg m^2 and a radius of 2.51m . When the child jumps onto the merry-go-round, the entire system begins to rotate.
Calculate the final kinetic energy of the system.

Answer 1

Angular momentum is conserved. The boy has an initial angular momentum about the merry-go-round's axis of
m V R. V is the boy's initial velocity

Let the final angular velocity of boy and merry-go-round be w.

m V R = (I + m R^2) w
w = mVR/(I + m R^2)

Final KE = (1/2)(I + mR^2)w^2
= (1/2) m^2 V^2 R^2/(I + mR^2)
= (1/2) m V^2 /{[I/(mR^2)] + 1]}

Note that KE must be lost since the denominator [I/(mR^2) + 1]
is greater than 1