IP A disk-shaped merry-go-round of radius 2.57 m and mass 170 kg rotates freely
ID: 1609183 • Letter: I
Question
IP A disk-shaped merry-go-round of radius 2.57 m and mass 170 kg rotates freely with an angular speed of 0.684 rev/s . A 61.6 kg person running tangential to the rim of the merry-go-round at 3.5 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.
Does the kinetic energy of the system increase, decrease, or stay the same when the person jumps on the merry-go-round?
increase?decrease?same?
Calculate the initial kinetic energy for this system.
Ki =
Calculate the final kinetic energy for this system.
Kf =Explanation / Answer
Angular momentum = I *
I = ½ * m * r^2 = ½ * 170 * 2.57^2 = 561.4165
As the merry-go-round rotates one time, it rotates an angle of 2 radians. To convert rev/s to rad/s, multiply by 2
= 0.684 * 2 = 4.3 rad/s.
Initial angular momentum = 561.4165 * 4.3 = 2412.8
For the person, I = m * r^2 = 61.6 * 2.57^2 = 406.86
Total I = 561.4165 + 406.86 = 968.27
Person’s initial momentum = 61.6 * 3.5 = 215.6
To convert to angular momentum, divide the radius of the merry-go-round.
Person’s angular momentum = 215.6÷ 2.57 = 83.9
Total initial momentum = 2412.8 + 215.6
Final angular momentum = 968.27 * f
968.27 * f = 2412.8 + 215.6
f = 2.714 rad/s.
KE = ½ * I * ^2
Initial KE = ½ * 561.4165 * (4.3 )^2
= 5190.3 J
For the person, initial KE = ½ * 61.6 * 3.5^2 = 377.3 J
Decrease
Total initial KE = 5190.3 + 377.3 = 5567.6 J
Final KE = ½ * 968.27 *[(2412.8 + 215.6) /968.27]^2
This is approximately 3567.4 J.
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