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 0.526 rev/s. A 59.4-kg person running tangential to the rim of the merry-go-round at 4.09 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?

increase decrease     stay the same

(b) Calculate the initial and final kinetic energies for this system.

Explanation / Answer

angular momentum is conserved. So,

Li = Lf

I(initial) = Mr^2/2,

I(final) = 0.5*Mr^2 + mr^2

0.5*Mr^2*wi + m*Vp*r = ( 0.5*Mr^2 + mr^2)*wf

wf = (M*r*wi + 2*m*Vp)/(M*r + 2*m*r)

wi = 2*3.14*0.526 = 3.303 rad/sec

wf = (155*2.63*3.303 + 2*59.4*4.09)/(155*2.63 + 2*59.4*2.63) = 2.544 rad/sec

As this is inelastic collision

dKE < 0

KEi = 0.5*Ii*wi^2 + 0.5*m*v^2 = 0.5*(0.5*155*2.63^2)*3.303^2 + 59.4*0.5*4.09^2 = 3420.97 J = 3.42 kJ

KEf = 0.5*If*wf^2 = 0.5*(0.5*155*2.63^2 + 59.4*2.63^2)*2.544^2 = 3064.21 = 3.064 kJ

dKE = 3.064 - 3.42 = -0.356 J which is negative.

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