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A cylinder with moment of inertia 12.6 kg m2 rotates with angular velocity 9.28 rad/s on a frictionless vertical axle. A second cylinder, with moment of inertia 41.8 kg m2 , initially not rotating, drops onto the first cylinder and remains in contact. Since the surfaces are rough, the two eventually reach the same angular velocity.Calculate the final angular velocity.

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Answer:

Final angular velocity = 2.15rad/s

Explanation:

The final angular velocity can be determined using the equation:

L=Ii × wi

L= (Ii + II) × wf

Where I is the moment of inertia of the 2nd cylinder and wf is the final angular velocity.

Given:

Ii=12.6kgm^2

I=41.8kgm^2

Wi=9.28rad/s

Ii ×wi= (Ii + I)×wf

12.6×9.28 = (12.6+41.8)×wf

116.93= 54.4wf

Wf=116.93/54.4

Wf= 2.15rad/s

Explanation:

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