Question

A disk-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

Answer 1

Answer:

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,

$L_d=I_dW_d$

Ld =  $\frac{1}{2}M_dR^{2}W_d$

Ld = $\frac{1}{2}(155) (2.63)^{2}(0.718*2\pi)$

Ld = 2418.43

and,

$L_p=M_pV_pD$

Lp = (59.4)(3.34)(2.63)

Lp = 521.78

then,

Lf = Ls

L_d=I_sW_s

Lf = $(\frac{1}{2}(155)(2.63)^{2}+(59.4)(2.63^2))(W_s)$

$Lf = 946.92W_s$

so, solving for Ws

Lf = Li

$946.92W_s = 521,78 + 2418.43$

Ws =  3.1 rad/s

Finally, the inicial and the final Kinetic energy

Ki = $\frac{1}{2}I_d(W_d)^2 + \frac{1}{2}M_p(V_p)^2$

Ki = 5786.284 J = 5.78 KJ

Kf = $\frac{1}{2}I_s(W_s)^2$

Kf =  4549.97 J = 4.55 KJ

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