A cylinder with moment of inertia I 1 rotates with angular velocity 0 about a fr

Question

A cylinder with moment of inertia I1 rotates with angular velocity 0 about a frictionless vertical axle. A second cylinder, with moment of inertia I2, initially not rotating, drops onto the first cylinder. Because the surfaces are rough, the two cylinders eventually reach the same angular speed .

(a) Calculate . (Use the following as necessary: I1, I2, and 0. )

Calculate the ratio of the final to the initial kinetic energy. (Use the following as necessary: I1, I2, and 0.)

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Explanation / Answer

From conservation of angular momentum

Li = Lf

Iiwi =Ifwf

I1wo = (I1+I2)w

Final angular velocity w =( I1wo)/(I1+I2)

(b) Ki = (1/2)Iiwi^2 = (1/2)I1wo^2

Kf = (1/2)Ifwf = (1/2) (I1+I2)w^2

Substitute w value from (a)

= (1/2) (I1+I2)[( I1wo)/(I1+I2)]2

Kf= (1/2) [I12wo2/ (I1+I2)]

Kf/Ki = {(1/2) [I12wo2/(I1+I2)]} /[(1/2)I1wo2]

Kf/Ki = I1/ (I1+I2)

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