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Question

There are two cylinders of radii R1andR2 having moments of inertia I1andI2 about their respective axes as shown in the figure. Initially, the cylinders rotate about their axes with angular speed α1 and α2 as shown in the figure. The cylinders are moved closed 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. Calculate the angular speeds of the cylinders after the slipping ceases.
301134_ef103fb57d8c4757901fa56e9baa2cd9.png

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Solution

When slipping ceases, the linear speeds of the points of contact of the two cylinders will be equal. If α1andα2 be the respective angular speeds, we have
α1R1andα2R2 ....................(1)
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 fR1 and that on the second is fR2. Assuming α1>α2, the corresponding angular impulses are - fR1t and fR2t, We, therefore, have

fR1t=I1(α1α1)andfR2t=I2(α2α2)
or, I1R1(α1α1)=I2R2(α2α2) ........................(2)
Solving (1)(2) α1=I1ω1R2+I2ω2R1I2R21+I1R22R2 and α2=I1ω1R2+I2ω2R1I2R21+I1R22R1


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