JEE Mains Physics · System of Particles and Rotational Motion

Parallel and Perpendicular Axes and Composite Bodies

The parallel-axis theorem moves an axis away from the centre of mass, I = I_cm + Md²; the perpendicular-axis theorem links three axes of a flat body; a system's moment of inertia is the sum of its parts about the same axis.

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Why this matters

Twenty-nine PYQs, sixteen of them asking for a number, and seven from 2026: the largest page in the chapter. Twelve move one body's axis with the parallel- or perpendicular-axis theorem, often to a tangent. Seventeen add the moments of inertia of several parts, or subtract a piece that has been cut out.

Concept 1 of 2: Parallel-axis and perpendicular-axis theorems

The standard table gives I about axes through the centre of mass. To use any other axis parallel to one of those, add Md²: the whole mass, moved a distance d, adds its own point-mass term. For a flat body, the moment of inertia about the axis normal to it equals the sum about two perpendicular axes lying in its plane.

Definition

Parallel axis: $I = I_{cm} + Md^{2}$ , where d is the distance between the new axis and a PARALLEL axis through the centre of mass.
In radii of gyration: $k^{2} = k_{cm}^{2} + d^{2}$ .
Perpendicular axis, flat bodies only: $I_z = I_x + I_y$ , with x and y in the plane and z normal to it, all through one point. A disc: $I_{\text{diameter}} = \tfrac{1}{2}I_{\text{normal}} = \tfrac{1}{4}MR^{2}$ .
Tangents: solid sphere $\tfrac{7}{5}MR^{2}$ ; hollow sphere $\tfrac{5}{3}MR^{2}$ ; disc, tangent in its plane $\tfrac{5}{4}MR^{2}$ ; disc, normal axis at the rim $\tfrac{3}{2}MR^{2}$ ; ring, tangent in its plane $\tfrac{3}{2}MR^{2}$ ; ring, normal axis at the rim $2MR^{2}$ .
Square plate of side l about a normal axis through a corner: $\tfrac{1}{6}Ml^{2} + M\tfrac{l^{2}}{2} = \tfrac{2}{3}Ml^{2}$ .

Axis theorems

I = I_{cm} + Md^{2} \qquad I_z = I_x + I_y \qquad k^{2} = k_{cm}^{2} + d^{2}

Worked example

A uniform disc has mass 2 kg and radius 0.3 m. Find its moment of inertia about a tangent lying in its plane.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 1 · Q23Moderate

Example 1 · System of Particles and Rotational Motion · Parallel and Perpendicular Axes and Composite Bodies

A solid sphere of radius 10 cm is rotating about an axis which is at a distance 15 cm from its center. The radius of gyration about this axis is

\sqrt{n}\text{ }cm

. The value of

n

The parallel-axis theorem starts at the centre of mass

I = I_cm + Md² holds only when one of the two axes passes through the centre of mass. To move between two other axes, go back to the centre-of-mass axis first, then out again.

The perpendicular-axis theorem is for flat bodies

I_z = I_x + I_y works for a disc, a ring or a plate. It fails for a sphere or a cylinder: a sphere's three diameters all give 2MR²/5, not 4MR²/5 for one of them.

A diameter given in place of a radius

A ring 'of diameter r' has radius r/2. Its tangent in the plane gives (3/2)M(r/2)² = 3Mr²/8, not 3Mr²/2.

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Concept 2 of 2: Composite bodies and bodies with a piece removed

Moments of inertia about the SAME axis simply add. So break a system into parts you know, find each part's I about the given axis (moving the axis with the parallel-axis theorem where needed), and add. A hole is a part with negative mass: take the full body and subtract the piece that is missing.

Definition

$I_{\text{system}} = \sum I_{\text{part}}$ , every part about the same axis.
Point masses: $\sum mr^{2}$ , with r the perpendicular distance from the axis. A mass on the axis adds nothing.
Two spheres (m, R) on a light rod, centres a distance d from the axis: $2\left(\tfrac{2}{5}mR^{2} + md^{2}\right)$ .
A cylinder whose own axis is normal to the rotation axis, centre a distance d away: $M\left(\tfrac{R^{2}}{4} + \tfrac{L^{2}}{12}\right) + Md^{2}$ .
A cut-out: $I_{\text{rest}} = I_{\text{full}} - \left(I_{\text{hole, own centre}} + m_{\text{hole}}d^{2}\right)$ , with the hole's mass in proportion to its area.
A disc with a hole of radius R/2 touching its rim: $\tfrac{1}{2}MR^{2} - \left[\tfrac{1}{2}\tfrac{M}{4}\tfrac{R^{2}}{4} + \tfrac{M}{4}\tfrac{R^{2}}{4}\right] = \tfrac{13}{32}MR^{2}$ .

Adding parts and removing a hole

I_{\text{system}} = \sum I_{\text{part}} \qquad I_{\text{rest}} = I_{\text{full}} - \left(I_{\text{hole}} + m_{\text{hole}}d^{2}\right)

Worked example

Four particles of 2 kg each sit at the corners of a square of side 1 m. Find the moment of inertia (a) about an axis through the centre, normal to the square, and (b) about one diagonal.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 1 · Q8Moderate

Example 2 · System of Particles and Rotational Motion · Parallel and Perpendicular Axes and Composite Bodies

The moment of inertia of a square loop made of four uniform solid cylinders, each having radius R and length

L(R < L)

about an axis passing through the mid points of opposite sides, is (Take the mass of the entire loop as M ) :

Parts about different axes

Each part's I must be about the system's axis before they are added. Adding each sphere's 2mR²/5 about its own centre leaves out the md² that usually dominates.

The removed piece has its own moment of inertia

Subtracting only m_hole·d² treats the hole as a point. Its own (1/2)m_hole·r² about its centre must go too.

Distance from the axis, not from the origin

For a point mass, r is the perpendicular distance to the axis. A mass on the axis contributes nothing, however far it is from the origin.

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Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Parallel-axis and perpendicular-axis theorems
Axis theorems
$I = I_{cm} + Md^{2} \qquad I_z = I_x + I_y \qquad k^{2} = k_{cm}^{2} + d^{2}$
Composite bodies and bodies with a piece removed
Adding parts and removing a hole
$I_{\text{system}} = \sum I_{\text{part}} \qquad I_{\text{rest}} = I_{\text{full}} - \left(I_{\text{hole}} + m_{\text{hole}}d^{2}\right)$

Watch out for (6)

The parallel-axis theorem starts at the centre of mass
→ Parallel-axis and perpendicular-axis theorems
The perpendicular-axis theorem is for flat bodies
→ Parallel-axis and perpendicular-axis theorems
A diameter given in place of a radius
→ Parallel-axis and perpendicular-axis theorems
Parts about different axes
→ Composite bodies and bodies with a piece removed
The removed piece has its own moment of inertia
→ Composite bodies and bodies with a piece removed
Distance from the axis, not from the origin
→ Composite bodies and bodies with a piece removed

Test yourself on System of Particles and Rotational Motion

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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