Rotational Mechanics: Moment of Inertia and Angular Momentum
Rotational Mechanics is one of the highest-weight topics in JEE Advanced Physics Paper 1 and Paper 2. Questions require fluency with moment of inertia (I), torque (τ), angular momentum (L), and the parallel and perpendicular axis theorems.
Moment of Inertia for standard objects: solid sphere = 2MR²/5; hollow sphere = 2MR²/3; solid cylinder/disc about central axis = MR²/2; hollow cylinder/ring about central axis = MR²; thin rod about centre = ML²/12; thin rod about end = ML²/3. The Parallel Axis Theorem: I_new = I_cm + Md², where d is the distance from the centre of mass to the new axis. The Perpendicular Axis Theorem (planar bodies only): I_z = I_x + I_y, where z is perpendicular to the plane of the body.
Angular momentum conservation: L = Iω is conserved when net external torque = 0. JEE Advanced problems test: (1) A person on a rotating platform extends/retracts arms — I changes, ω changes such that Iω = constant. (2) A bullet hits a rod pivoted at one end and embeds in it — use conservation of angular momentum about the pivot (no external torque during the brief impact), then energy methods for subsequent motion. (3) A disc rolling on a rough surface — distinguish between torque of friction and angular momentum about the contact point.
Rolling without slipping: the condition a_cm = αR connects translational and rotational dynamics. For rolling on an inclined plane: a = g sinθ / (1 + I_cm/MR²). For a solid sphere: a = 5g sinθ/7. For a ring: a = g sinθ/2. The solid sphere rolls fastest down an incline because it has the smallest moment of inertia ratio I_cm/MR² = 2/5.