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 0.532 rev/s. A 59.4-kg person running tangential to the rim of the merry-go-round at 3.55 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-o-round? O decrease increase stay the same (b) Calculate the initial and final kinetic energies for this system. K,=3.37 Kf- Your response differs from the correct answer by more than 10%, Double check your calculations.

Explanation / Answer

The first problem you have already solved. Therefore, I am solving the second part of the problem.

Here for the merry-go-round, I = (1/2)*m*r² = 0.5*155*2.63² = 536.06 kg.m²

= 0.532 rev/sec * 2 rad/rev = 3.342 rad/sec

So, angular momentum Ld = I = 536.06*3.342 = 1791.5 kg.m²/sec

Now, for the man: I = mr² = 59.4*2.63² = 410.86 kg.m²
= v/r = 3.55/2.63 = 1.35 rad/sec

so,
Lm = I = 410.86*1.35 = 554.66 kg.m²/sec

The initial momentum is therefore 1791.5 + 554.66 = 2346.16 kg.m²/sec

Now according to conservation of momentum -  

2346.16 = (Id +Im)*,

=> = 2346.16 / (536.06 + 410.86) = 2.48 rad/s

therefore, the final kinetic energy of the system = (1/2)*(Id +Im)*^2

= 0.5*(536.06+410.86)*2.48^2 = 2912 J = 2.912 kJ

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