A disk-shaped merry-go-round of radius 2.63 m and mass 155 kg rotates freely with an angular speed of 0.615 rev/s. A 59.4 kg person running tangential to the rim of the merry-go-round at 3.99 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round's rim. (b) Calculate the initial and final kinetic energies for this system. Ki = Kf = So I am stuck on Kf, but for Ki i thought it should just be the sum of the kinetic energy of the man plus the kinetic energy of the disk I set up my equation like this: 0.5mv^2(KE of person) + 0.5Iw^2(KE of disk) = 0.5(59.4)(3.99^2) + (0.5)(0.5(155)(2.63^2))(0.615^2) I solved for this and did not get the correct answer which means I have probably set up the equation wrong...can anyone help me out with this?