A person with mass m p = 78 kg stands on a spinning platform disk with a radius

Question

A person with mass mp = 78 kg stands on a spinning platform disk with a radius of R = 1.62 m and mass md = 180 kg. The disk is initially spinning at ? = 1.6 rad/s. The person then walks 2/3 of the way toward the center of the disk (ending 0.54 m from the center).

1)What is the total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk? kg-m2

2)What is the total moment of inertia of the system about the center of the disk when the person stands at the final location 2/3 of the way toward the center of the disk? kg-m2

3)What is the final angular velocity of the disk? rad/s

4)What is the change in the total kinetic energy of the person and disk? (A positive value means the energy increased.) J

5)What is the centripetal acceleration of the person when she is at R/3? m/s2

6)If the person now walks back to the rim of the disk, what is the final angular speed of the disk? rad/s

Explanation / Answer

1)moment of inertia=0.5MR^2+mR^2

=0.5*180*1.62^2 + 78*1.62^2

=440.9 kg m^2

2)moment of inertia=0.5MR^2+ md^2

=0.5*180*1.62^2 + 78*0.54^2

=258.941 kg m^2

3)conserving momentum

I1w1=I2w2

or 440.9*1.6=258.941*w

or w=2.72 rad/s

4)change=0.5*258.941*2.72^2-0.5*440.9*1.6^2

=393.523 J

5)CENTRIPETAL ACCELERATION=w^2/R

=2.72^2*0.54

=4 m/s^2

6)again conserving momentum,

440.9*w=258.941*2.72

or w=1.6 rad/s

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