A cylinder with moment of inertia I 1 rotates about a vertical, frictionless axl

Question

A cylinder with moment of inertiaI1rotates about a vertical, frictionless axle with angular velocity?i. A second cylinder; this one having a moment of inertia ofI2and initially not rotating, drops onto the first cylinder. Because of friction between the surfaces, the two eventually reach the same angular speed?f.

KEi = KEf = KEf/ KEi =

Explanation / Answer

from conservation of angular momentum I1*wi =(I1+I2)*wf =>wf = (I1/(I1+I2))*wi initial kinetic energy KEi = 0.5*I1*wi^2 final kinetic energy KEf = 0.5*(I1+I2)*wf^2 = 0.5*(I1^2/(I1+I2))*wi^2 dividing (2) by (1) we get KEf/KEi = (I1/(I1+I2))

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