A person with mass m p = 79 kg stands on a spinning platform disk with a radius

Question

A person with mass mp = 79 kg stands on a spinning platform disk with a radius of R = 1.77 m and mass md = 182 kg. The disk is initially spinning at = 1.7 rad/s. The person then walks 2/3 of the way toward the center of the disk (ending 0.59 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

(Survey Question)

Explanation / Answer

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?

The moment of inertia of the uniform disk is
I = ½mr²

The moment of inertia of the person acts like a point mass so is

I = mr²

The total moment of inertia

I = ½mdr² + mpr²
I = r²(½md + mp)
I = 1.77²(182/2 + 79)
I = 532.598 kg•m²


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?

I = ½mdr² + mpr²
I = ½mdr² + mp(r/3)²
I = r² (md/2 + mp/9)

I = 1.77 ²(182/2 + 79/9)
I = 312.5938 kg•m²


What is the final angular velocity of the disk?

angular momentum will be conserved

I = I
= I / I

= 532.598(1.7) / 312.5938  
= 3.015..
= 3.015 rad/s

What is the change in the total kinetic energy of the person and disk?

KE = ½I²

KE = ½I²
KE = ½(532.598)1.77²
KE =834.288 J

KE = ½(312.5938)3²
KE = 1406 J

KE = KE - KE
KE = 1406 - 834
KE = 572 J

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

ac = ²r
ac = 3.0²(1.77/3)
ac = 5.31 m/s²


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

conservation of momentum will still occur so original velocity will be restored.

= 1.7 rad/s

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