Three identical cylinders rotate with the same angular velocity Ω about parallel central axes. They are brought together until they touch, keeping the axes parallel. A new steady state is achieved when, at each contact line, a cylinder does not slip with respect to its neighbor as shown in fig. How much of the original spin kinetic energy is now left ?
(The precise order in which the first and second touch, and the second and third touch, is irrelevant.)
Text Solution
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Sol. As there is no slipping, if Ω ' is the final angular velocity of cylinder 1, then cylinders 2 and 3 have final angular velocities – Ω ' and Ω ' respectively. Let I be the moment of inertia of each cylinder about its axis of rotation, M ij be the angular impulse imparted to i th cylinder with respect to its axis of rotation by the j th cylinder. Newton's third law requires that, as the cylinders have the same radius ,
M ij = M ji (i, j = 1, 2, 3, i ≠ j)
Dynamical considered give
I ( Ω ' – Ω ) = M 12 (1)
I (– Ω ' – Ω ) = M 21 + M 23 (2)
I ( Ω ' – Ω ) = M 32 (3)
(1) + (2) + (3) gives
I (3 Ω ' – Ω ) = M 32
or 
The ratio of the spin kinetic energies after and before touching is
Ans.
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