I need this question please A merry go round of diameter 5.0 m and moment of ine

Question

I need this question please

A merry go round of diameter 5.0 m and moment of inertia I=310 kg m^2 is initially spinning at 20 rpm counterclockwise. A 38 kg girl is running with a velocity of 3.0 m/s on a pat tanget to the merry go round . Once at the merry go around she jumps on hold on to the rim of the merry go round . Assume that the merry go around is a uniform solid cylinder and that the girls velocity is opposite the velocity of the outer rim of the merry go round at the point of contact .

1- what is the initial angluar momentum of the merry go round ?
2- What is the magnitude of the initial angular momentum of the girl just before she jumps on the merry go round ?
3- what is the angular speed of the system ( in rpm ) after the girls jumps on ?
4- calculate the force the girls needs to hold on ?

Explanation / Answer

1)

initial angluar momentum of the merry go round = L1 = I*w1

w1 = 20 rpm = 20*2*pi/60 = 2.1 rad/s

L1 = 310*2.1 = 651 kg m^2/s

+++++++++++++

magnitude of the initial angular momentum of the girl

just before she jumps on the merry go round L2 = -m*v2*R = -38*3*2.5

L2 = 285 kg m^2/s

++++++++++++++++

initial angular momentum Li = L1 + L2

after the child jumps the total moment of inertia = I2 = I + m*R^2

final angular momentum Lf = I2*w2

from conservationof angular moemntum

Lf =Li

(310 + (38*2.5^2))*wf = 651-285

wf = 0.66 rad/s

+++++++

F = m*R*wf^2

F = 38*2.5*0.66^2 = 41.4 N

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