# Moment of Inertia of Standard Bodies

> Moment of inertia for NEET Physics: values for ring, disc, rod, sphere, the two theorems, radius of gyration and rolling-body comparisons.

- Canonical URL: https://prepelephant.com/topics/neet-ug/physics/moment-of-inertia-standard-bodies
- 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: "Moment of Inertia of Standard Bodies", PrepElephant, https://prepelephant.com/topics/neet-ug/physics/moment-of-inertia-standard-bodies

## Direct answer

Moment of inertia measures how mass is spread about an axis: I = Σ m r^2, so for the same mass a ring (MR^2) resists rotation twice as much as a disc (MR^2/2) because all its mass sits at the full radius. It enters dynamics through torque τ = Iα and rotational kinetic energy (1/2) Iω^2. Two theorems relocate the axis cheaply: the parallel-axis theorem I = I_cm + Md^2 for any parallel axis, and the perpendicular-axis theorem I_z = I_x + I_y, valid only for planar bodies. The radius of gyration K, defined by I = MK^2, is where all the mass could sit for the same rotational inertia; the standard values are quoted directly in NEET questions.

## What you must remember

- **Standard values (uniform bodies, symmetry axes):** ring MR^2; disc or cylinder MR^2/2; rod about centre ML^2/12, about one end ML^2/3; hollow sphere or shell 2MR^2/3; solid sphere 2MR^2/5.
- **Parallel-axis theorem:** I = I_cm + Md^2; moving the axis away from the centre of mass always increases I — the rod about its end is ML^2/12 + M(L/2)^2 = ML^2/3, the theorem in action.
- **Perpendicular-axis theorem:** I_z = I_x + I_y for planar laminas only; using it for a sphere is a classic error.
- **Radius of gyration:** I = MK^2; K_ring = R, K_disc = R/√2 — a number the options like to quote directly.
- **Rolling kinetic energy:** KE = (1/2) Mv^2 [1 + I/MR^2]; for a ring the split is 50:50, for a disc 2/3 translational and 1/3 rotational, for a solid sphere 5/7 translational.
- **Rolling race:** acceleration down an incline a = g sin θ/(1 + I/MR^2), so sphere beats disc beats ring, independent of mass and radius.

## Two bodies, one hill, every fraction accounted

Release a ring and a solid disc of equal mass and radius from rest on a 30-degree incline of height h = 1.4 m (take g = 9.8 m s^-2). Energy conservation for a rolling body reads Mgh = (1/2)Mv^2(1 + I/MR^2). For the disc, I/MR^2 = 1/2, so v = √[2gh/1.5] = √(2 × 9.8 × 1.4/1.5) = √18.3 ≈ 4.3 m s^-1. For the ring, I/MR^2 = 1, giving v = √(gh) = √13.7 ≈ 3.7 m s^-1 — the ring is markedly slower at the bottom although both stored identical potential energy Mgh. Where did the ring's energy go? Nowhere: it is split. The ring carries half its energy as rotation, the disc only a third, the sphere (2/5) would carry barely 29 per cent — which is precisely why the sphere wins any fair rolling race, mass and radius cancelling out of the verdict entirely.

## What the question setter reaches for

The highest-frequency device is the parallel-axis shortcut applied where it does not belong: the theorem links an arbitrary axis only to the parallel axis through the centre of mass, so jumping from a rod's end value to a random perpendicular axis a distance d away requires going through I_cm — ML^2/3 minus M(L/2)^2 returns you to ML^2/12 before you can go anywhere else. The second device is mass-versus-distribution: doubling the mass of a disc doubles I, but doubling the radius at fixed mass quadruples it — r-squared leverage that surprises candidates expecting linearity. Third, the perpendicular-axis trap: from a disc's Iz = MR^2/2 and symmetry, Ix = Iy = MR^2/4 about a diameter — a result derivable only because the disc is flat, and options include it precisely because the sphere's diameter value 2MR^2/5 tempts those who skip the "planar only" condition. Finally, rolling questions hide a static-friction truth: the friction providing rotation does no net work on a perfectly rolling body, which is why energy conservation applies cleanly.

## Frequently asked questions

### Why does a ring have a larger moment of inertia than a disc of equal mass and radius?

Every particle of the ring lies at distance R from the axis, whereas the disc's mass is spread from the centre outward, lowering Σmr^2 to MR^2/2.

### When can the perpendicular-axis theorem be used?

Only for a flat, planar body, relating the moment about an axis perpendicular to the plane to the two in-plane axes: I_z = I_x + I_y.

### What is the radius of gyration?

The distance K = √(I/M) at which the entire mass could be concentrated without changing the moment of inertia.

### In what order do bodies finish a rolling race down an incline?

Solid sphere first, then disc or cylinder, then ring — the order of I/MR^2, which alone decides it; mass and radius cancel out.

### Does friction do work on a body rolling without slipping?

No — the contact point is instantaneously at rest relative to the surface, so static friction acts with no displacement there and no work is done, leaving mechanical energy conserved.