• # question_answer A square of mass M and sides of length L has a moment of inertia Io when rotated about an axis perpendicular to its surface and passing through its center, as shown. Now a lump of clay, also of mass M is attached to one corner of the square as shown. What is the new moment of inertia of the masses about the same axis of rotation?   A) ${{I}_{0}}+\frac{M{{L}^{2}}}{4}$ B) ${{I}_{0}}+\frac{M{{L}^{2}}}{2}$ C) ${{I}_{0}}+\frac{\sqrt{2}M{{L}^{2}}}{2}$    D) ${{I}_{0}}+2M{{L}^{2}}$

 The moment of inertia for the system can be calculated by adding the two individual moments of inertia as following ${{I}_{total}}={{I}_{0}}+{{I}_{clay}}$ ${{I}_{clay}}=M{{R}^{2}}$ $R=\sqrt{2}\frac{L}{2}$ ${{I}_{clay}}=M{{\left( \sqrt{2}\frac{L}{2} \right)}^{2}}=\frac{M{{L}^{2}}}{2}$ ${{I}_{total}}={{I}_{0}}+\frac{M{{L}^{2}}}{2}$