Moment of inertia — how to calculate for different geometries and axes

Question

How do we calculate the moment of inertia (MOI) of different bodies about different axes? Given a uniform disc of mass $M$ and radius $R$ , find its MOI about (a) an axis through the centre perpendicular to the plane, (b) a diameter, and (c) a tangent perpendicular to the plane.

(CBSE 11 + JEE Main)

Solution — Step by Step

This is the standard result:

I_{centre} = \frac{MR^2}{2}

This is derived by integration: $I = \int r^2 \, dm$ , where each thin ring of radius $r$ and width $dr$ contributes $dm \cdot r^2$ .

For a planar body, if $I_z$ is the MOI about an axis perpendicular to the plane:

I_z = I_x + I_y

By symmetry, $I_x = I_y = I_{diameter}$ . So:

\frac{MR^2}{2} = 2 \times I_{diameter}

I_{diameter} = \mathbf{\frac{MR^2}{4}}

A tangent perpendicular to the plane is parallel to the central axis, shifted by distance $R$ :

I_{tangent} = I_{centre} + MR^2 = \frac{MR^2}{2} + MR^2 = \mathbf{\frac{3MR^2}{2}}

Body	Axis	MOI
Thin rod	Centre, perpendicular	$\frac{ML^2}{12}$
Thin rod	End, perpendicular	$\frac{ML^2}{3}$
Ring	Centre, perpendicular	$MR^2$
Disc	Centre, perpendicular	$\frac{MR^2}{2}$
Disc	Diameter	$\frac{MR^2}{4}$
Solid sphere	Diameter	$\frac{2MR^2}{5}$
Hollow sphere	Diameter	$\frac{2MR^2}{3}$

flowchart TD
    A["Need MOI about a given axis"] --> B{"Is it a standard axis?"}
    B -- Yes --> C["Use the standard formula from table"]
    B -- No --> D{"Is the new axis parallel to a known axis?"}
    D -- Yes --> E["PARALLEL AXIS THEOREM: I = I_CM + Md²"]
    D -- No --> F{"Is the body planar/flat?"}
    F -- Yes --> G["PERPENDICULAR AXIS THEOREM: I_z = I_x + I_y"]
    F -- No --> H["Use integration: I = ∫r² dm"]
    E --> I["Calculate"]
    G --> I
    H --> I

Why This Works

Moment of inertia measures how “spread out” the mass is from the rotation axis. The farther the mass, the harder it is to rotate — that is why $I = \int r^2 \, dm$ has $r^2$ (not $r$ ).

The parallel axis theorem ( $I = I_{CM} + Md^2$ ) adds the extra contribution when the axis shifts away from the centre of mass. The perpendicular axis theorem works only for flat bodies because only in 2D does the geometry allow $z$ -axis MOI to split neatly into $x$ and $y$ components.

Alternative Method

For JEE, memorise the $k^2/R^2$ ratios instead of full formulas. Here $k$ is the radius of gyration ( $I = Mk^2$ ). Ring: $k^2/R^2 = 1$ . Disc: $1/2$ . Solid sphere: $2/5$ . Hollow sphere: $2/3$ . These ratios directly determine which body rolls fastest down an incline (smaller ratio = faster).

Common Mistake

The parallel axis theorem can ONLY be used to shift from a CM axis to a parallel non-CM axis, or vice versa. You cannot shift directly from one non-CM axis to another non-CM axis. For example, to find MOI of a rod about a point at $L/4$ from one end: first find MOI about CM ( $ML^2/12$ ), then shift by $L/4$ using parallel axis theorem. Do NOT try to shift from the end ( $ML^2/3$ ) by $3L/4$ — that gives the wrong answer because neither axis passes through CM.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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