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## Homework Statement

The figure shows two cylinders of radius r

_{1}and r

_{2}rotating about their axes (along the length) with speeds [tex]\omega_1[/tex] and [tex]\omega_2[/tex]

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.

## Homework Equations

- After slipping ceases,
*if [tex]\omega_{1}'[/tex] and [tex]\omega_{2}'[/tex] are new angular speeds*then...

w1' * r1 = w2' * r2 .................(1) - 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:

[tex]I_{1}\omega_{1}+I_{2}\omega_{2}=I_{1}\omega_{1}' + I_{2}\omega_{2}'[/tex]

Using this with w1' * r1 = w2' * r2 (the equation 1 above), I get

[tex]\omega_{1}' = \frac{I_{1}\omega_{1}+I_{2}\omega_{2}}{I_{1}r_{2}+I_{2}r_{1}} r_{2}[/tex]

and a similar expression for w2'.

The answer, however, is [tex]\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}[/tex]