The two uniform discs rotate separately on parallel axles. The upper disc (radius `a` and momentum of inertia `I_1`) is given an angular velocity `omega_0` and the lower disc of (radius `b` and momentum of inertia `I_2`) is at rest. Now the two discs are moved together so that their rims touch. Final angular velocity of the upper disc is.
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Options

A. `((I_1 omega_0))/([I_1 + (a^2 I_2//b^2)])`
B. `((I_2 omega_0))/([I_2 + (a^2 I_1//b^2)])`
C. `((I_1 omega_0))/([I_1 + (b^2 I_2//a^2)])`
D. `((I_2 omega_0))/([I_2 + (b^2 I_2//a^2)])`

Answer

A

Explanation

(a) The two discs exert equal and opposite forces on each other when in contact. The torque due to these forces changes the angular momentum of each disc. From angular impulse-angular momentum theorem, we have
`f a Delta t = I_1 (omega_0 - omega_1)` ...(i)
and `f b Delta t = I_2 omega_2` ...(ii)
From eqns. (1) and (2), we get `(a)/(b) = (I_1 (omega_0 - omega_1))/(I_2 omega_2)` ...(iii)
When slipping ceases between the discs, the contact points of the two discs have the same linear velocity, i.e.,
`a omega_1 = b omega_2` ...(iv)
On substituting `omega_2` is eqn. (iii), we get
`omega_1 = ((I_1 omega_0))/([I_1 + (a^2 I_2//b^2)]]`.
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