
The moment of inertia (MI) measures an object’s resistance to rotational motion about a specific axis. It determines resistance to angular acceleration and depends not only on the mass of the object but also on the distribution of that mass relative to the axis of rotation. A ring (or circular hoop) is a thin circular object with all its mass concentrated at a fixed radius from the centre.
Consider a thin ring of mass M, radius R, and negligible thickness.
Considering the ring rotates about an axis perpendicular to its plane and passing through the centre (central axis), the mass element of the ring is at the same distance R from the axis.
By the definition of moment of inertia:
I = ∫ r² dm
Since r = R for all mass elements dm, the integral simplifies to:
Icentral = ∫ R² dm
= R² ∫ dm
= R² M
Icentral = MR²
This shows that the moment of inertia of a thin ring is directly proportional to its mass and the square of its radius.

For rotation about an axis lying in the plane of the ring and passing through the centre, the perpendicular axis theorem is applied.

If the axis of rotation is parallel to the central axis but does not pass through the centre of mass, the parallel axis theorem is used:
I = ICM + Md²
Where:
ICM – moment of inertia about the central axis through the centre of mass
d – distance between the axes
| Shape | Axis | Formula |
|---|---|---|
| Thin Ring | Central axis (perpendicular to the plane) | I = MR² |
| Thin Ring | Diameter (in-plane axis) | ![]() |
| Thin Ring | Parallel axis | I = ICM + Md² |

Q1. Why is the moment of inertia of a ring larger than a disc?
The moment of inertia of a ring is larger because all the mass of the ring lies at the maximum distance from the axis, increasing resistance to rotation.
Q2. A thin ring of mass 3 kg and radius 0.4 m rotates about its central axis. Find its moment of inertia.
Given: Mass = 3 kg, Radius = 0.4 m
I = MR²
= 3 × (0.4)²
= 3 × 0.16
= 0.48 kg·m²
Q3. A thin ring has a mass of 2 kg and a radius of 0.5 m. Find its moment of inertia about the central axis perpendicular to its plane.
Given: Mass = 2 kg, Radius = 0.5 m
I = MR²
= 2 × (0.5)²
= 0.5 kg·m²
JEE Main marks vs rank vs percentile
JEE Advanced Eligibility Criteria
JEE Advanced Chemistry Syllabus
JEE Advanced Registration Dates
Our expert will get in touch with you soon
