# Rotational Motion > Rotational Motion for NEET-UG Physics — moment of inertia of standard bodies, torque, angular momentum and rolling from NCERT Class 11. - Canonical URL: https://prepelephant.com/topics/neet-ug/physics/rotational-motion-ncert - Exam / course: NEET-UG · Subject: Physics - Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team - First published: 2026-10-02 - Last updated: 2026-10-02 - How to cite: "Rotational Motion", PrepElephant, https://prepelephant.com/topics/neet-ug/physics/rotational-motion-ncert ## Direct answer Rotational mechanics rewrites translation symbol by symbol: torque replaces force, moment of inertia replaces mass, and τ = Iα replaces F = ma. Angular momentum L = Iω is conserved whenever external torque vanishes — the physics behind the spinning skater and behind half the NEET-UG questions this chapter generates. Rolling without slipping adds the rolling-energy split, which decides the famous race of the ring, disc and sphere down an incline. ## What you must remember - Moment of inertia I = Σmr^2 about a specified axis: ring MR^2 about its central axis, disc ½MR^2, solid sphere 2/5 MR^2 about a diameter, hollow sphere 2/3 MR^2, rod ML^2/12 about its centre and ML^2/3 about one end. - Parallel-axis theorem I = I_cm + Md^2 (d is the shift from the centre-of-mass axis); perpendicular-axis theorem for plane lamina: I_z = I_x + I_y. - Torque τ = r × F with magnitude rF sinθ; rotational second law τ = Iα; angular momentum L = Iω, conserved when τ_ext = 0. - Rolling without slipping: v_cm = ωR; total kinetic energy = ½Mv^2(1 + k^2/R^2) — for a ring Mv^2, for a disc ¾Mv^2, for a solid sphere 7/10 Mv^2. - Acceleration down an incline a = g sinθ/(1 + k^2/R^2): sphere 5/7 g sinθ beats disc 2/3 g sinθ beats ring ½ g sinθ — independent of both mass and radius. - Angular kinematics mirror the linear set: ω = ω0 + αt, θ = ω0t + ½αt^2, ω^2 = ω0^2 + 2αθ for constant α. - Angular momentum is also mvr for a particle (planetary orbits) and is conserved in its presence under a central force. ## Why the sphere wins the race Release a ring, a disc and a solid sphere together at the top of the same incline, all rolling without slipping. Energy conservation gives Mgh = ½Mv^2(1 + k^2/R^2), so v = sqrt(2gh/(1 + k^2/R^2)). The sphere carries the smallest rotational baggage (k^2/R^2 = 2/5), the disc the middle (1/2), the ring the heaviest (1) — a ring must deposit exactly half of its energy into rotation before it rolls. So the sphere arrives first, the disc next, the ring last, and no amount of mass or radius changes the order, because those factors cancel: the race is decided purely by the shape factor k^2/R^2. The acceleration hierarchy follows the same logic, sphere 5/7 g sinθ down to ring ½ g sinθ. The skater closes the argument from the other end. She pulls her arms in, cutting her moment of inertia; with no external torque, L = Iω stays fixed, so her angular velocity rises as I falls. Her rotational kinetic energy ½Iω^2 actually increases — paid for by the muscular work of pulling the arms in against the centrifugal tendency. Conservation of energy and conservation of angular momentum are doing different jobs in the same act, and distinguishing them is precisely what the depth of this chapter requires. ## Where students slip Moment of inertia is not a property of the body alone but of the body-axis pair — "the moment of inertia of a disc" is an incomplete sentence until the axis is named, and NCERT's own figures vary the axis deliberately. The parallel-axis theorem starts from the centre-of-mass axis, never from an arbitrary one; applying it from a random starting axis is the commonest computational error. In rolling problems, v = ωR holds only when there is no slipping, and the kinetic energy splits as ½Mv^2 + ½Iω^2 — forgetting the second term quietly turns every disc answer into a block answer. When a rotating disc is dropped onto another and they couple, angular momentum is conserved but kinetic energy is not (the loss appears at the rubbing faces); choosing the wrong conserved quantity flips the answer. Finally, L = Iω in that form belongs to rigid bodies about fixed axes; for a particle, use mvr. ## Frequently asked questions ### What is the moment of inertia of a uniform disc about its central axis? ½MR^2, half the ring's MR^2 — mass nearer the axis contributes less, since I depends on r^2 of every element. ### Why does an ice skater spin faster when she pulls her arms in? Her moment of inertia falls while angular momentum L = Iω stays conserved (no external torque), so ω must rise. The extra kinetic energy comes from the work her muscles do. ### Which body rolls down an incline fastest, and why? The solid sphere: its k^2/R^2 = 2/5 is the smallest of the standard bodies, leaving the largest share of energy in translation. Mass and radius are irrelevant. ### What is the total kinetic energy of a ring rolling without slipping? Mv^2 — exactly half translational (½Mv^2) and half rotational (½Iω^2 = ½Mv^2 with I = MR^2 and v = ωR). ### State the perpendicular-axis theorem. For a plane lamina in the xy plane, the moment of inertia about the z-axis equals the sum of the moments about the x and y axes: I_z = I_x + I_y.