JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Rotational Mechanics

For a rolling object, $a=\frac{g\sin \theta }{1+I/(M{R}^2)}$. Cylinder $P$ has larger moment of inertia (mass near surface), so smaller acceleration. Cylinder $Q$ has smaller moment of inertia (mass near axis), so larger acceleration and reaches the ground first. Using $v=\sqrt{2a\ell }$, cylinder $Q$ has larger linear velocity. Since $\omega =v/R$, cylinder $Q$ also has larger angular speed. Both have the same total mechanical energy (same initial height), but different distributions between translational and rotational kinetic energy.