Person On Disk A person with mass mp-76 kg stands on a spinning platform disk wi

Question

Person On Disk A person with mass mp-76 kg stands on a spinning platform disk with a radius of R-1.92 m and mass md-191 kg. The disk is initially spinning at w-1.6 rad/s. The person then walks 2/3 of the way toward the center of the disk (ending 0.64 m from the center). 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? 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? What is the final angular velocity of the disk? rad/s Semt "What is the change in the total kinetic energy of the person and disk? (A positive value means the energy increased.) OWhat is the centripetal acceleration of the person when she is at R/3? m/s 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 of the system I1 = (1/2)*md*R^2 + mp*R^2

moment of inertia of the system I1 = (1/2)*191*1.92^2 + 76*1.92^2 = 632.22 kg m^2

(2)

moment of inertia of the system I2 = (1/2)*md*R^2 + mp*r^2

moment of inertia of the system I2 = (1/2)*191*1.92^2 + 76*0.64^2 = 383.2 kg m^2

(c)

from comservation of angular momentum

initial angular momentum = final angular momentum

Li = Lf

I1*w1 = I2*w2

632.2*1.6 = 383.2*w2

final angular velocity w2 = 2.64 rad/s

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4)

change in KE = (1/2)*I2*w2^ - (1/2)*I1*w1^2

change in KE = (1/2)383.2*2.64^2 - (1/2)*632.2*1.6^2 = 526.2 J

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5)

centripetal acceleration a2 = r*w2^2 = 0.64*2.64^2 = 4.46 m/s^2

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6)

final angular speed wf = w1 = 1.92 rad/s

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