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Law Incorporation [45]

1 year ago

7

A disk of mass m and radius r is initially held at rest just above a larger disk of mass M and radius R that is rotating at angu

lar speed ωi. What is the final angular speed of the disks after the top one is dropped onto the bottom one and they stop slipping on each other? Note: the moment of inertia of a disk of mass M and radius R about an axis through its center and perpendicular to the disk is I = (1/2)MR .

1 answer:

Leto [7]1 year ago

6 0

Answer:

Explanation:

Moment of inertia of larger disk I₁ = 1/2 MR²

Moment of inertia of smaller disk I₂ = 1/2 m r ²

Initial angular velocity

We shall apply law of conservation of angular momentum .

initial total momentum = final angular momentum

I₁ X ωi = ( I₁ + I₂ )ωf

1/2 MR² x ωi = 1/2 ( m r² + MR² ) ωf

ωf = ωi / ( 1 + m r²/MR² )

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Given:

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