Figure shows two cylinders of radii `r_1 and r_2` having moments of inertia `I_1 and I_2` about their respective axes. Initially the cylinders rotate about their axes with angular speed `omega_1 and omega_2` as shown in the figure. The cylinders are moved closer to touch each other keeping the axes parallel. The cylinders first slip over each other at the contact but the slipping finally ceases due to the friction between them. Find the angular speeds of the cylinders after the slipping ceases.

Explanation

When slipping ceases, the linear speeds of the points of contact of the two cylinders will be equal. If `omega_(1)' and omega_(2) ` ' be the respective angular speeds, we have
`omega_(1)'r_(1) = omega_(2)'r_(2) " ". . . (i) `
The change in the angular speed is brought about by the frictional force which acts as long as the slipping exists. If this force f acts for a time t, the torque on the first cylinder is `fr_(1)` and that on the second is `f r_(2)` . Assuming`omega_(1) r_(1) gt omega_(2) r_(2)` , the corresponding angular impluses are –` f r_(1) t and f r_(2)` t. We, therefore, have
` - f r_(1) t = I_(1) (omega_(1)'= omega_(1))`
and `f r_(2) t = I_(2) (omega_(2)' = omega_(2))`
or, `- (l_(1))/(r_(1)) (omega_(1)' - omega_(1)) = (l_(2))/(r_(2))(omega_(2)'-omega_(2))" ". . . (ii)`
Solving (i) and (ii)
`omega_(1)' = (l_(1) omega_(1)r_(2) + l_(2)omega_(2)r_(1))/(l_(2)r_(1)^(2)+l_(1)r_(2)^(2))r_(2) "and" omega_(2)' = (l_(1) omega_(1) r_(2) + l_(2) omega_(2) r_(1))/(l_(2) r_(1)^(2) + l_(1)r_(2)^(2))r_(1)`