A person with mass m p = 77 kg stands on a spinning platform disk with a radius

Question

A person with mass mp = 77 kg stands on a spinning platform disk with a radius of R = 2.25 m and mass md = 196 kg. The disk is initially spinning at = 2 rad/s. The person then walks 2/3 of the way toward the center of the disk (ending 0.75 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?

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?

3) What is the final angular velocity of the disk?

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

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

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

PLEASE SHOW STEPS

Explanation / Answer

1) I1 = Id + Ip = mdR2/2 + mpR2 = (md/2 + mp)R2 = (196/2 + 77) * 2.252 = 885.94 kg-m2

2) I2 = Id + Ip = mdR2/2 + mp(R/3)2 = (md/2 + mp/9)R2 = (196/2 + 77/9) * 2.252 = 539.44 kg-m2

3) Angular momentum of the (disk + person) system is conserved about the centre of the disk.

So, I11 = I22

=> 885.94 * 2 = 539.44 * 2

=> 2 = 885.94 * 2 / 539.44 = 3.285 rad/s

4) Change in KE = I222/2 - I112/2 = (539.44 * 3.2852 / 2) - (885.94 * 22 / 2) = 1138.73 J

Hence, KE of system increased by 1138.73 J.

5) Centripetal acceleration (ap) of the person is given by,

ap = 22(R/3) = 3.2852 * (2.25/3) = 8.1 m/s2

6) Since angular momentum of the disk is conserved, final angular speed of the person when walking back to the rim of the disk will be back to the same initial value, i.e. 2 rad/s

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