A disk-shaped merry-go-round of radius 2.63 m and mass 155 kg rotates freely wit

Question

A disk-shaped merry-go-round of radius 2.63 m and mass 155 kg rotates freely with an angular speed of 0.603 rev/s. A 59.4-kg person running tangential to the rim of the merry-go-round at 3.61 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. A)Does the kinetic energy of the system increase, decrease, or stay the same when the person jumps on the merry-go-round? B)Calculate the initial and final kinetic energies for this system.

Explanation / Answer

For the device: I = .5mr² = .5*155*2.63² = 536.06 kg.m² ? = .564 rev/sec * 2p rad/rev = 3.5437 rad/sec....so, Ld = I? = 536.06*3.5437 = 1899.64 kg.m²/sec For the man: I = mr² = 59.4*2.63² = 410.86 kg.m² ? = v/r = 2.73/2.63 = 1.038 rad/sec so, Lm = I? = 410.86*1.038 = 426.47 kg.m²/sec The initial momentum is therefore 1899.64 + 426.47 = 1946.11 The momentum is conserved, so: 1946.11 = (Id +Im)*?, from which we can solve for the final ?: = 2.0552 rad/sec

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