A disk-shaped merry-go-round of radius 2.63 m and mass 155 kg rotates freely wit
ID: 1367691 • Letter: A
Question
A disk-shaped merry-go-round of radius 2.63 m and mass 155 kg rotates freely with an angular speed of 0.570 rev/s. A 59.4-kg person running tangential to the rim of the merry-go-round at 3.82 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
Moment of inertia of the merry-go-round = mr^2/2
Moment of inertia of the merry-go-round = 155 * 2.63^2 / 2
Now,
mr^2 = 155 * 2.63^2 / 2
m = 155 / 2 = 77.5 kg
Tangential velocity of this single point mass
= 0.570 * 2 * pi * 2.63
= 9.42 m/s
Momentum of this single point mass
= 9.42 * 77.5
= 730.1 Kg.m/s
Momentum of the person while running
= 59.4 * 3.82
= 227.0 Kg.m/s
Total momentum before the person jumps on the merry-go-round
= 730.1 + 227.0
= 957.1 Kg.m/s
Momentum is conserved, so
(77.5 + 59.4) * v = 957.1
v = 957.1 / 136.9
v = 7.0 m/s when the person has jumped on the merry-go-round
Tangential KE of this point mass before the person jumps
= mv^2/2
= (77.5 * 9.42^2)/2 J
= 3438.5 J
KE of person before jumping
= (59.4 * 3.82^2) / 2
= 433.4 J
Total linear KE before the person jumps on the merry-go-round
Ki = 3438.5 + 433.4
Ki = 3871.9 J
Total Mass of point mass and the person
= 77.5 + 59.4
= 136.9 kg
Total KE of point mass and the person when together
Kf = 136.9 * 7.0^2 / 2
Kf = 3354.05 J
The KE decreases when the person jumps on the merry-go-round.
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