Question

IP A disk-shaped merry-go-round of radius 2.84 m and mass 155 kg rotates freely with an angular speed of 0.62 rev/s . A 61.1 kg person running tangential to the rim of the merry-go-round at 3.17 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.

Part B Calculate the initial kinetic energy for this system.

Part C Calculate the final kinetic energy for this system.

Answer #1

For the device: I = .5mr² = .5*155*2.84² = 625.08 kg.m²

ω = .62 rev/sec * 2π rad/rev = 3.8936 rad/sec....so,

Ld = Iω = 625.08*3.8936 = 2415.05 kg.m²/sec

For the man: I = mr² = 61.1*2.84² = 492.80kg.m²

ω = v/r = 3.17/2.84 = 1.116 rad/sec so,

Lm = Iω = 492.80*1.116 = 549.96 kg.m²/sec

The initial momentum is therefore 2415.05 +549.96 = 2965.01

The momentum is conserved, so:

2965.01 = (Id +Im)*ω, from which we can solve for the final ω in
rad/sec.

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