Adisk-shaped merry-go-round of radius 2.63 m and mass 155 kg rotates freely with an angular speed of 0.718 rev/s. a 59.4 kg person running tangential to the rim of the merry-go-round at 3.34 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? stay the same increase decrease (b) calculate the initial and final kinetic energies for this system. ki = kjkf = kj

The kinetic energy of the system decrease

Ki = 5.78 KJ

Kf = 4.55 KJ

Explanation:

For answer this question we will use the law of the conservation of the angular momentum so,

Li = Lf

Where Li is the inicial momentum of all the system, and Lf is the final momentum of the system.

also, the angular momentum L can be calculated in two ways

L = IW

where I is the momentum of inertia and the W is the Angular velocity.

or,

L = MVD

where M is the mass, V is the lineal velocity and the D is the lever arm.

Therefore,

Li = Ld ( merry-go-round) + Lp ( person )

Lf = Ls

Where Ld is the angular momentum of the merry go round, Lp is the angular momentum of the person and Ls is the angular momentum of the sistem (merry-go-round +  person)

so,

Ld =  

Ld =

Ld = 2418.43

and,

Lp = (59.4)(3.34)(2.63)

Lp = 521.78

then,

Lf = Ls

L_d=I_sW_s

Lf =

so, solving for Ws

Lf = Li

Ws =  3.1 rad/s

Finally, the inicial and the final Kinetic energy

Ki =

Ki = 5786.284 J = 5.78 KJ

Kf =

Kf =  4549.97 J = 4.55 KJ

Then, The kinetic energy of the system decrease because Kf < Ki

e) none of the above answers are correct

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y