Two discs of moments of inertia I-1 and I_2 about their respective axes , rotaiting with angular frequencies omega_1 and omega_2 respectively, are brought into contact face to face with their axes of rotation coincident. The angular frequency of the composite disc will be

Two discs of moments of inertia I-1 and I_2 about their respective axe

1.	Two discs of moments of inertia I-1 and I_2 about their respective axes , rotaiting with angular frequencies omega_1 and omega_2 respectively, are brought into contact face to face with their axes of rotation coincident. The angular frequency of the composite disc will be
Answer» `I_(1)omega_1 + (I_(1)omega_2)/I_1 + I_2` `I_(2)omega_1 + (I_(1)omega_2)/I_1 + I_2` `I_(1)omega_1 - (I_(2)omega_2)/I_1 - I_2` `I_(2)omega_1 - (I_(1)omega_2)/I_1 - I_2` Solution :Total initial angular momentum of the TWO discs is: `L_(i) = I_(1)omega_(1)+I_(2)omega_(2)` When two discs are brought into contact face to face (one on top of the other) and their axes of rotation coincide, the moment of inertia, i.e., `I=I_(1)+I_(2)` Let `OMEGA` be the final angular speed of the system. The final angualr momentum of the system is `L_(f) = Iomega = (I_(1)+I_(2))omega` As no EXTERNAL torque acts on the system, therefore according to the law of conservation of angular momentum, we get `L_(i)= L_(f)` `I_(1)omega_(1)+ I_(2)omega_(2) = (I_(1)+I_(2))omega` `therefore omega = (I_(1)omega_(1) + I_(2)omega_(2))/(I_(1)+I_(2))`