A person with mass mp = 74 kg stands on a spinning platform disk with a radius of R = 2.31 m and mass md = 183 kg. The disk is initially spinning at ω = 1.8 rad/s. The person then walks 2/3 of the way toward the center of the disk (ending 0.77 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

1) 883 kgm2

2) 532 kgm2

3) 2.99 rad/s

4) 944 J

5) 6.87 m/s2

6) 1.8 rad/s

Explanation:

1)Suppose the spinning platform disk is solid with a uniform distributed mass. Then its moments of inertia is:

If we treat the person as a point mass, then the total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk:

2) Similarly, he 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 (1/3 of the radius from the center):

3) Since there's no external force, we can apply the law of momentum conservation to calculate the angular velocity at R/3 from the center:

4)Kinetic energy before:

Kinetic energy after:

So the change in kinetic energy is: 2374 - 1430 = 944 J

5)

6) If the person now walks back to the rim of the disk, then his final angular speed would be back to the original, which is 1.8 rad/s due to conservation of angular momentum.