A person with mass m p = 72 kg stands on a spinning platform disk with a radius

Question

A person with mass mp = 72 kg stands on a spinning platform disk with a radius of R = 1.8 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 0.6 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

PLEASE, CLEARLY WORK IT OUT CORRECTLY..THNKS

Explanation / Answer

Here ,

mp = 72 Kg

R = 1.8 ,

md = 196 Kg

w = 2 rad/s

1)

TOtal moment of inertia of system initially = 0.5 *md * R^2 + mp * R^2

TOtal moment of inertia of system initially = 0.5* 196 * 1.8^2 + 72 * 1.8^2

TOtal moment of inertia of system initially = 550.8 Kg.m^2

2)

after the person stands at 2/3 distance

TOtal moment of inertia of system = 0.5 *md * R^2 + mp * R^2

TOtal moment of inertia of system = 0.5* 196 * 1.8^2 + 72 * 0.6^2

TOtal moment of inertia of system = 343.44 Kg.m^2

3)

let the final angular speed is wf

Using conservation of angular momentum

550.8 * 2 = wf * 343.44

wf = 3.21 rad/s

the final angular velocity of disk is 3.21 rad/s

4)

the change in kinetic energy of the system = final kinetic energy - initial kinetic energy

change in kinetic energy of the system = 0.5 * 343.44 * 3.21^2 - 0.5 * 550.8 * 2^2

change in kinetic energy of the system = 667.8 J

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