Problem on Rotational mechanics / Conservation of angular momentum.

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SUMMARY

The discussion centers on the problem of determining the angular speeds of two cylinders after slipping ceases due to friction. The equations governing the system include the conservation of angular momentum and the relationship between angular speeds and radii, expressed as w1' * r1 = w2' * r2. The user successfully derived an expression for the new angular speeds using angular impulse but seeks clarification on achieving the correct final expressions for w1' and w2'. The correct formula for w1' is given as ω1' = (I1ω1r2 + I2ω2r1) / (I1r2² + I2r1²) r2.

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Homework Statement


The figure shows two cylinders rotating about their axes, all the variables involved are shown.
The two cylinders are moved closer to touch each other keeping the axes parallel. The cylinders first slip over each other at the point of contact but slipping finally ceases due to friction between them.

Find the angular speeds of the cylinders after the slipping ceases.
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Homework Equations



  1. After slipping ceases, if w1' and w2' are new angular speeds then...
    w1' * r1 = w2' * r2
  2. Angular momentum = Iw.

The Attempt at a Solution


I have succeeding in solving the problem by equating the angular impulse due to friction to the change in angular momentums of both cylinders and eliminating friction force from the two equations.

But I'd like to solve this using conservation of angular momentum as well since it should be pretty straightforward and easy.

Since no external torque acts on the system, conserving angular momentum, I get:

I_{1}\omega_{1}+I_{2}\omega_{2}=I_{1}\omega_{1}' + I_{2}\omega_{2}'

Using this with w1' * r1 = w2' * r2, I get

\omega_{1}' = \frac{I_{1}\omega_{1}+I_{2}\omega_{2}}{I_{1}r_{2}+I_{2}r_{1}} r_{2}
and a similar expression for w2'.

The answer, however, is \omega_{1}' = \frac{I_{1}\omega_{1}r_{2}+I_{2}\omega_{2}r_{1}}{I_{1}r_{2}^{2}+I_{2}r_{1}^{2}} r_{2}
 
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