For a uniform rectangular sheet shown in the figure, the ratio of moments of inertia about the axes perpendicular to the sheet and passing through O (the centre of mass) and O' (corner point) is :

A uniform rod of length ‘$$l$$’ is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed $$\omega $$ the rod makes an angle $$\theta $$ with it (see figure). To find $$\theta $$ equate the rate of change of angular momentum (direction going into the paper) $${{m{l^2}} \over {12}}{\omega ^2}\sin \theta \cos \theta $$ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces F_H and F_V about the CM. The value of $$\theta $$ is then such that :

Moment of inertia of a cylinder of mass M, length L and radius R about an axis passing through its centre and perpendicular to the axis of the cylinder is
I = $$M\left( {{{{R^2}} \over 4} + {{{L^2}} \over {12}}} \right)$$. If such a cylinder is to be made for a given mass of a material, the ratio $${L \over R}$$ for it to have minimum possible I is

Two uniform circular discs are rotating independently in the same direction around their common axis passing through their centres. The moment of inertia and angular velocity of the first disc are 0.1 kg-m² and 10 rad s^–1 respectively while those for the second one are 0.2 kg-m² and 5 rad s^–1 respectively. At some instant they get stuck together and start rotating as a single system about their common axis with some angular speed. The kinetic energy of the combined system is :