Two uniform cylinders are spinning independently about their axes, which are parallel. One has radius R 1 mass M 1 , the other R 2 and M 2 . Initially they rotate in the same sense with angular speeds Ω 1 and Ω 2 respectively as shown in fig . They are then displaced until they touch along a common tangent. After a steady state is reached, what is the final angular velocity of each cylinder ?
Text Solution
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Sol. Let ω 1 , ω 2 be the final angular velocities of the two cylinders respectively after steady state is reached. Then
ω 1 R 1 = ω 2 R 2
Let J 1 and J 2 be the time-integrated torque 2 exerts 1 & 2, then
,
J 1 = I 1 ( ω 1 – Ω 1 ), J 2 = I 2 ( ω 2 – Ω 2 ),
or 
As I 
MR 2 , the last equation becomes.
M 1 R 1 ( ω 1 – Ω 1 ) = M 2 R 2 ( ω 2 – Ω 2 )
i.e. M 1 R 1 ω 1 – M 2 R 2 ω 2 = M 1 R 1 Ω 1 – M 2 R 2 Ω 2
Hence
ω 1 = 
ω 2 = 
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