Find the KINETIC ENERGY in KJ please A disk-shaped merry-go-round of radius 2.65

Question

Find the KINETIC ENERGY in KJ please A disk-shaped merry-go-round of radius 2.65m and mass 165kg rotates freely with an angular speed of 0.666rev/s . A 64.7kg person running tangential to the rim of the merry-go-round at 3.2m/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.) Calculate the initial kinetic energy for this system. (in KJ) b.) Calculate the final kinetic energy for this system. (in KJ)

Explanation / Answer

A) K.Ei = (1/2)*I*w^2....

w =0.666*2*3.142 rad/sec =4.185 rad/sec

K.Ei =0.5*(1/2)*m*R^2*w^2 = 0.5*0.5*165*2.65^2*4.185^2 = 5.073 kJ....
K.Ei2 = (1/2)*m*v^2 = 0.5*64.7*3.2^2 = 331.264 = 0.331 kJ..
total initial K.E =5.404 kJ
B) I1*w1 = (I1+I2)*w2...
w2 = I1*w1/(I1+I2)

I1 = (1/2)*m*R^2 = 579.35 kg m^2....
I2 = m*r^2 = 64.7*2.65^2 = 454.35 kg m^2.....

w2 = 579.35*4.185/(579.35+454.35) = 2.345 rad/sec = 0.373 rev/sec

KEf = (1/2)*(I1+I2)*w2^2 = 0.5*(579.35+454.35)*2.345^2 = 2.842 kJ

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