Defining the moment of inertia
Consider a rigid body rotating about a fixed axis with angular velocity \(\omega\). Every particle of the body moves in a circle about the axis. A particle of mass \(m_i\) at perpendicular distance \(r_i\) from the axis has linear speed \(v_i = r_i\omega\), so its kinetic energy is \(\tfrac{1}{2}m_i v_i^2 = \tfrac{1}{2}m_i r_i^2 \omega^2\). Because \(\omega\) is the same for every particle, the total kinetic energy of the body is
\[ K = \tfrac{1}{2}\left(\sum_{i} m_i r_i^2\right)\omega^2 . \]NCERT defines the bracketed sum as the moment of inertia \(I\):
\[ \boxed{\,I = \sum_{i} m_i r_i^2\,} \qquad\text{so that}\qquad K = \tfrac{1}{2}I\omega^2 . \]For a continuous body the sum becomes an integral over mass elements:
\[ I = \int r^2\, dm , \]where \(r\) is the perpendicular distance of the element \(dm\) from the axis. From the definition the dimensions of \(I\) are \(\mathrm{ML^2}\) and its SI unit is the kilogram metre squared, \(\mathrm{kg\,m^2}\). The quantity \(I\) is independent of how fast the body spins; it is a characteristic of the body together with the axis about which it rotates.
The rotational analogue of mass
Set the rotational kinetic energy \(K = \tfrac{1}{2}I\omega^2\) beside the familiar translational form \(K = \tfrac{1}{2}mv^2\). The two are identical in structure, with \(\omega\) replacing \(v\) and \(I\) replacing \(m\). NCERT draws the conclusion explicitly: the moment of inertia is the rotational analogue of mass.
The physical meaning follows the analogy. Mass measures a body's resistance to a change in its state of linear motion — its translational inertia. In the same way, the moment of inertia about a given axis measures the body's resistance to a change in its rotational motion about that axis — its rotational inertia. A flywheel is built with most of its mass in the rim precisely to give it a large \(I\); because of that large rotational inertia it resists sudden changes in angular speed and smooths out jerky motion in an engine.
Once \(I\) plays the role of mass, the rotational form of Newton's second law is \(\tau = I\alpha\) — built out in dynamics of rotational motion.
Why mass distribution and axis matter
The single most important property of \(I\) — and the source of most NEET questions on this topic — is that it is not a fixed quantity. NCERT states it directly: unlike the mass of a body, the moment of inertia depends on the distribution of mass about the axis of rotation and on the orientation and position of that axis. Because each mass element is weighted by the square of its distance, mass placed far from the axis contributes far more than the same mass placed close to it.
Two equal masses about the same axis. Quadrupling the distance multiplies the contribution to \(I\) by sixteen, because \(I\) weights mass by distance squared. This is why a ring (all mass at \(R\)) out-weighs a disc, and a hollow body out-weighs a solid one.
Two consequences run through every problem. First, the same body has different moments of inertia about different axes — a uniform rod gives \(ML^2/12\) about its centre but \(ML^2/3\) about one end. Second, comparing two bodies of equal mass and equal radius, the one with mass pushed farther out has the larger \(I\): ring beats disc, hollow sphere beats solid sphere, hollow cylinder beats solid cylinder. Whenever you state a moment of inertia, you must state the axis along with it.
The standard MoI table
NCERT Table 6.1 lists the moments of inertia of regular bodies about specified axes; the derivations belong to higher classes, but the results are canonical and must be memorised exactly. The factor grid below gives each body, its axis, and its \(I\) in terms of mass \(M\) and the linear dimension \(R\) or \(L\). The hollow-sphere and rod-about-end entries are added from the NIOS supplement (§7.3.1) because they recur in NEET.
Read down the column of values and the distribution rule becomes visible. For a fixed mass and radius the moment of inertia falls as mass is drawn inward: hollow cylinder \(MR^2 \to\) solid cylinder \(MR^2/2\); hollow sphere \(\tfrac{2}{3}MR^2 \to\) solid sphere \(\tfrac{2}{5}MR^2\); ring \(MR^2 \to\) disc \(MR^2/2\). Note also that the ring and the hollow cylinder share the value \(MR^2\) about the symmetry axis — both are thin shells of mass at radius \(R\) — and the disc and solid cylinder both give \(MR^2/2\).
Radius of gyration
Every entry in the table can be written in one common form. NCERT notes that in all cases \(I = Mk^2\), where \(k\) has the dimension of length and is called the radius of gyration:
\[ I = Mk^2 \qquad\Longrightarrow\qquad k = \sqrt{\dfrac{I}{M}} . \]The radius of gyration is the distance from the axis at which the whole mass of the body could be concentrated as a single point to give the same moment of inertia the body actually has. It is a geometric property of the body and the axis — it does not depend on the mass. Reading the values straight off the table:
| Body and axis | Moment of inertia \(I\) | Radius of gyration \(k = \sqrt{I/M}\) |
|---|---|---|
| Thin rod, ⊥ at midpoint | \(ML^2/12\) | \(L/\sqrt{12}\) |
| Disc, about a diameter | \(MR^2/4\) | \(R/2\) |
| Disc, ⊥ at centre | \(MR^2/2\) | \(R/\sqrt{2}\) |
| Ring, ⊥ at centre | \(MR^2\) | \(R\) |
| Solid sphere, diameter | \(2MR^2/5\) | \(R\sqrt{2/5}\) |
| Hollow sphere, diameter | \(2MR^2/3\) | \(R\sqrt{2/3}\) |
NCERT gives two of these explicitly: \(k = L/\sqrt{12}\) for a rod about its midpoint and \(k = R/2\) for a disc about its diameter. Because \(k\) is built from \(I\), it carries the same distribution logic — a body with mass farther from the axis has a larger radius of gyration. The radius of gyration is a convenient single number for comparing how "spread out" two bodies are relative to their axes.
Worked example 1 — moment of inertia of a point-mass system
Four particles, each of mass \(m\), sit at the corners of a square of side \(L\). Find the moment of inertia of the system about an axis through the centre of the square, perpendicular to its plane. Then add a further mass \(m\) at each of one pair of opposite corners and find the new value.
Distance of each corner from the centre. The diagonal of the square is \(L\sqrt{2}\), so each corner lies at \(r = L/\sqrt{2}\) from the centre. The axis is perpendicular to the plane through that centre, so this \(r\) is the perpendicular distance for every particle.
Apply \(I = \sum m_i r_i^2\). All four masses are equal and equidistant: \(I = 4\,m\,r^2 = 4m\left(\dfrac{L}{\sqrt2}\right)^2 = 4m\cdot\dfrac{L^2}{2} = 2mL^2.\)
After adding two more masses. Now two corners carry \(2m\) and two carry \(m\), but every corner is still at the same \(r = L/\sqrt2\). The total mass-weight of corners is \(2m+2m+m+m = 6m\), so \(I' = 6m\,r^2 = 6m\cdot\dfrac{L^2}{2} = 3mL^2.\)
Reading. Adding mass changed \(I\) from \(2mL^2\) to \(3mL^2\) — a direct illustration that the moment of inertia tracks the distribution of mass about the axis. The summation definition handles any arrangement of point masses without a formula.
Worked example 2 — using the radius of gyration
A solid sphere and a thin hollow sphere have the same mass \(M\) and the same radius \(R\). Find the radius of gyration of each about a diameter, and the ratio of the two.
Solid sphere. \(I_\text{solid} = \tfrac{2}{5}MR^2\). From \(I = Mk^2\), \(k_\text{solid} = \sqrt{I_\text{solid}/M} = \sqrt{\tfrac{2}{5}R^2} = R\sqrt{\tfrac{2}{5}}.\)
Hollow sphere. \(I_\text{hollow} = \tfrac{2}{3}MR^2\), so \(k_\text{hollow} = \sqrt{\tfrac{2}{3}R^2} = R\sqrt{\tfrac{2}{3}}.\)
Ratio. \(\dfrac{k_\text{solid}}{k_\text{hollow}} = \sqrt{\dfrac{2/5}{2/3}} = \sqrt{\dfrac{3}{5}} = \sqrt{\dfrac{3}{5}}\approx 0.77.\) Equivalently \(k_\text{solid}:k_\text{hollow} = \sqrt3 : \sqrt5\). The hollow sphere has the larger radius of gyration, consistent with its mass sitting entirely at the surface.
Worked example 3 — comparing two bodies
A solid sphere (A), a thin disc (B) and a circular ring (C) each have mass \(M\) and radius \(R\) and spin about their own symmetry axes with the same angular speed \(\omega\). Rank the work needed to bring each to rest.
Moments of inertia about the symmetry axis. Solid sphere \(I_A = \tfrac{2}{5}MR^2 = 0.40\,MR^2\); disc (⊥ at centre) \(I_B = \tfrac{1}{2}MR^2 = 0.50\,MR^2\); ring (⊥ at centre) \(I_C = MR^2 = 1.00\,MR^2\).
Work–energy connection. Bringing a body from angular speed \(\omega\) to rest removes its rotational kinetic energy, so the work required is \(W = \tfrac{1}{2}I\omega^2\). With \(\omega\) common to all three, \(W \propto I\).
Ranking. Since \(I_C > I_B > I_A\), the work satisfies \(W_C > W_B > W_A\). The ring, with all its mass at the rim, stores the most rotational energy for the same spin and is hardest to stop — a clean consequence of mass distribution about the axis.
Moment of inertia in one breath
- \(I = \sum m_i r_i^2\) (point masses) or \(I = \int r^2\,dm\) (continuous body); units \(\mathrm{kg\,m^2}\), dimensions \(\mathrm{ML^2}\).
- \(K = \tfrac{1}{2}I\omega^2\): the moment of inertia is the rotational analogue of mass.
- \(I\) is not fixed — it depends on mass distribution and on the orientation and position of the axis. Always state the axis.
- Standard table: ring \(MR^2\); disc \(MR^2/2\); rod \(ML^2/12\) (centre) or \(ML^2/3\) (end); solid sphere \(\tfrac{2}{5}MR^2\); hollow sphere \(\tfrac{2}{3}MR^2\); solid cylinder \(\tfrac{1}{2}MR^2\); hollow cylinder \(MR^2\).
- For equal \(M\) and \(R\), hollow beats solid — mass farther out raises \(I\).
- Radius of gyration: \(I = Mk^2\), so \(k = \sqrt{I/M}\); \(k = R/2\) for a disc about its diameter, \(k = L/\sqrt{12}\) for a rod about its midpoint.