• # question_answer Two discs of moment of inertia ${{I}_{1}}$ and ${{I}_{2}}$ are rotated about their geometrical axis with angular velocity ${{\omega }_{1}}$and ${{\omega }_{2}}$ respectively. If the two discs are joint face to face coinciding their axes, then the kinetic energy of system: A)  $\frac{1}{2}\,({{I}_{1}}+{{I}_{2}})\,{{({{\omega }_{1}}+{{\omega }_{2}})}^{2}}$B)  $\frac{1}{2}\,({{I}_{1}}+{{I}_{2}})\,({{\omega }_{1}}+{{\omega }_{2}})$C)  $\frac{1}{2}\frac{{{({{I}_{1}}{{\omega }_{1}}+{{I}_{2}}{{\omega }_{2}})}^{2}}}{({{I}_{1}}+{{I}_{2}})}$D)  $\frac{1}{16}\,({{I}_{1}}+{{I}_{2}})\,{{({{\omega }_{1}}+{{\omega }_{2}})}^{2}}$

From the law of conservation of Angular momentum $(L)={{I}_{1}}{{\omega }_{1}}+{{I}_{2}}{{\omega }_{2}}=({{I}_{1}}+{{I}_{2}})\omega$ Rotational kinetic energy $=\frac{1}{2}\frac{{{L}^{2}}}{({{I}_{1}}+{{I}_{2}})}$ $=\frac{1}{2}\frac{{{({{I}_{1}}{{\omega }_{1}}+{{I}_{2}}{{\omega }_{2}})}^{2}}}{({{I}_{1}}+{{I}_{2}})}$