Rotational Motion Part 3

Concept:Conservation of mechanical energy is used as the system is isolated and the string does not slip.Explanation:When the $2m$ mass descends by height $h = 3.6$ m, its potential energy decreases by $2mgh$.The $m$ mass rises by the same height $h$, so its potential energy increases by $mgh$.Net change in potential energy: $\Delta U = mgh - 2mgh = -mgh$.The system loses $mgh$ of potential energy, which converts into kinetic energy.Initial kinetic energy is zero because the system starts from rest.Final kinetic energy has three parts:• Translational KE of mass $m$: $\frac{1}{2} m v^2$• Translational KE of mass $2m$: $\frac{1}{2} (2m) v^2 = m v^2$• Rotational KE of pulley (solid disc, mass $30m$, radius $r$): $I = \frac{1}{2} (30m) r^2 = 15 m r^2$Angular speed $\omega = v / r$ (no slip), so rotational KE = $\frac{1}{2} I \omega^2 = \frac{1}{2} (15 m r^2) \left( \frac{v}{r} \right)^2 = 7.5 m v^2$.Total final KE: $K_f = 0.5 m v^2 + 1 m v^2 + 7.5 m v^2 = 9 m v^2$.Energy conservation: $|\Delta U| = K_f$ $\Rightarrow$ $mgh = 9 m v^2$ $\Rightarrow$ $v^2 = \frac{gh}{9}$.Substitute $g=10$, $h=3.6$: $v^2 = \frac{10 \times 3.6}{9} = 4$ $\Rightarrow$ $v = 2$ m/s.Answer:$2$ m/s (Option A).