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 0.679 rev/s. A 59.4 kg person running tangential to the rim of the merry-go-round at 3.10 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

Here,

radius ,r = 2.63 m

mass ,m1 = 155 Kg

angular speed , wi = 0.679 rev/s

wi = 4.26 rad/s

v1 = 3.1 m/s

m2 = 59.4 Kg

a) as the collision is not elastic

kinetic energy after the person jump on the Merry go round is less than the initial kinetic energy of system .

b)

let the final angular speed is wf

using conservation of angular momentum

0.5 * 155 * 2.63^2 * 4.26 + 59.4 * 3.1 * 2.63 = (0.5 * 155 * 2.63^2 + 59.4 * 2.63^2) * wf

wf = 2.92 rad/s

Now ,

Initial kinetic energy = kinetic energy of disk + kinetic energy of person

Initial kinetic energy = 0.5 * 155 * 2.63^2 * 4.26^2 + 0.5 * 59.4 * 3.1^2

Initial kinetic energy = 10013.6 J

the Initial kinetic energy is 10013.6 J

Final kinetic energy = 0.5 * I * wf^2

Final kinetic energy = 0.5 * (0.5 * 155 * 2.63^2 + 59.4 * 2.63^2) * 2.92^2

Final kinetic energy = 4037 J

the Final kinetic energy is 4037 J

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