question_answer 55)

                  Two discs of moments of inertia I1 and I2 about their respective axes (normal to the disc and passing through the centre), and rotating with angular speed \[{{\omega }_{1}}\] and \[{{\omega }_{2}}\] are brought into contact face to face with their axes of rotation coincident.                 (a) Does the law of conservation of angular momentum apply to the situation? Why?                 (b) Find the angular speed of the two-disc system.                 (c) Calculate the loss in kinetic energy of the system in the process.                 (d) Account for this loss.                

Answer:

                  (a) Law of conservation of angular momentum applies in the situation, when no net external torque acts on the system. In this case, as no net external torque acts, so the law of conservation of momentum will apply to this situation.                 (b) Let \[{{I}_{1}}\] and \[{{I}_{2}}\] be the M.I. of two discs respectively about the given axis of rotation \[{{\omega }_{1}}\]and \[{{\omega }_{2}}\] be the respective angular velocities of the discs. Let I be the moment of inertia of two disc system and \[\omega \] is its angular velocity.                 According to the law of conservation of angular momentum,                 \[{{I}_{1}}\,{{\omega }_{1}}+{{I}_{2}}{{\omega }_{2}}=I\omega \]                 \[\therefore \] \[\omega =\,\frac{{{I}_{1}}{{\omega }_{1}}+{{I}_{2}}{{\omega }_{2}}}{I}\]               ?. (i)                 (c) Total initial rotational K.E. of two discs                 \[{{K}_{1}}=\,\frac{1}{2}\,{{I}_{1}}\omega _{1}^{2}\,+\frac{1}{2}\,{{I}_{2}}\omega _{2}^{2}\]                 Total rotational K.E. of the two discs system,                 \[{{K}_{2}}=\frac{1}{2}\,I\,{{\omega }^{2}}\]                 \[=\,\frac{1}{2}\,({{I}_{1}}+\,{{I}_{2}})\,\frac{{{({{I}_{1}}{{\omega }_{1}}+\,{{I}_{2}}{{\omega }_{2}})}^{2}}\,\,}{{{({{I}_{ & 1}}+{{I}_{2}})}^{2}}}\]                                                                 \[(\because \,\,I=\,{{I}_{1}}+\,{{I}_{2}})\]                 \[=\,\frac{I_{1}^{2}\,\omega _{1}^{2}+I_{2}^{2}\,\omega _{2}^{2}\,+2{{I}_{1}}{{I}_{2}}\,{{\omega }_{1}}{{\omega }_{2}}}{2({{I}_{1}}+\,{{I}_{2}})}\]                 \[\therefore \] Loss in K.E. \[=\,{{K}_{2}}-\,{{K}_{1}}\]                 \[=\,\frac{I_{1}^{2}\,\omega _{1}^{2}\,+\,I_{2}^{2}\,\omega _{2}^{2}+\,2{{I}_{1}}{{I}_{2}}\,\,{{\omega }_{1}}{{\omega }_{2}}}{2({{I}_{1}}+\,{{I}_{2}})}\]                 \[\left( \frac{1}{2}\,{{I}_{1}}\omega _{2}^{2}+\frac{1}{2}\,{{I}_{2}}\omega _{2}^{2} \right)\]                 \[=\,\frac{-{{I}_{1}}{{I}_{2}}\,(\omega _{1}^{2}+\omega _{2}^{2}-2{{\omega }_{1}}{{\omega }_{2}})}{2({{I}_{1}}+{{I}_{2}})}\]                 \[=\,\frac{-{{I}_{1}}{{I}_{2}}{{({{\omega }_{1}}-{{\omega }_{2}})}^{2}}}{2({{I}_{1}}+{{I}_{ & 2}})}\]                 (d) The loss in kinetic energy of the system is due to the work done against the force of friction between two discs when they move one over the other.