Moment of inertia and radius of gyration
Moment of inertia quantifies how mass is distributed with respect to an axis of rotation, and the radius of gyration is a measure related to this distribution.
Why this shows up in the exam
You need to calculate and compare moments of inertia for various shapes and axes, which is a common NEET question.
How NEET tests this
Learn the idea
Moment of inertia I tells how hard it is to spin a body about a chosen axis; the radius of gyration k is the single distance that would give the same I if all the mass were at that distance (I = M k²). The key insight is to first pick the simplest axis (usually through the centre) whose I is known, then shift it with the parallel‑axis theorem.
🧠 Memory hook: A disc’s diameter gives a quarter; add a whole radius squared (the shift) → five‑quarters MR²
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- I = Σ m r² for particles; for continuous bodies I = ∫ r² dm
- Parallel‑axis theorem: I = I_cm + M d², where d is the perpendicular distance between axes
- For a thin uniform disc: I about a central diameter = (1/4) M R²; I about a central perpendicular axis = (1/2) M R²
- Radius of gyration k = √(I/M)
- Perpendicular‑axis theorem (planar bodies): I_z = I_x + I_y
- Moment of inertia is additive for separate parts of a system
How to approach it
- 1Identify the axis required in the question
- 2Find a standard I for the same body about a convenient axis (usually through the centre) from NCERT tables
- 3If the required axis is parallel but displaced, apply the parallel‑axis theorem using the distance between the axes
- 4Simplify to obtain I; if asked for k, use k = √(I/M)
Worked example — watch it click
The moment of inertia of a disc of mass M and radius R about an axis, which is tangential to the circumference of disc and parallel to its diameter :
- A)(3/2) MR²
- B)(2/5) MR²
- ✅(5/4) MR²
- D)(4/4) MR²
The concept behind this problem
The problem forces you to recognise that the tangent axis is parallel to a diameter, so you must start from the known I about a central diameter (¼ MR²) and then shift it outward with the parallel‑axis theorem.
Step by step
- 1For a disc about its center: I_cm = (1/2)MR².
- 2Using parallel axis theorem for tangential axis: I = I_cm + Md² where d = R.
- 3Thus I = (1/2)MR² + MR² = (3/2)MR².
- 4However, checking option (d) shows (4/4)MR² = MR², which is incorrect.
- 5Option (c) (5/4)MR² doesn't match standard formula.
- 6Re-examining: tangent parallel to diameter means d = R, so I = MR²/2 + MR² = 3MR²/2.
- 7None match exactly.
Watch out
Most students use I_cm = ½ MR² (perpendicular axis) instead of the correct ¼ MR² for a diameter, leading to the wrong 3/2 MR² result.
Common slip-ups that cost marks
- •Mixing up the perpendicular central axis (½ MR²) with the central diameter (¼ MR²) for a disc
- •Using the perpendicular‑axis theorem for a three‑dimensional body instead of a planar one
- •Forgetting that d in the parallel‑axis theorem is the shortest perpendicular distance between the two axes
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Practise it
These are real questions from past NEET papers that test this exact idea.
A light rod of length l has two masses m₁ and m₂ attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is :
Push further
More challenging1 harder question built from the past papers above — a step up in difficulty, with distractors designed so you can't get there by elimination. Written and checked by our reviewers, not from a real paper.
A solid sphere of mass M and radius R has a moment of inertia I₀ about its diameter. What is its moment of inertia about a tangent to the sphere?
More from Motion of System of Particles and Rigid Body
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Center of mass: definition and calculation
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Rotational kinematics and dynamics
Rotational kinematics describes the motion of rotating bodies, while dynamics relates torque, angular acceleration, and rotational kinetic energy.
Centre of Mass of Discrete Particles
For discrete particles, the centre-of-mass position is the vector sum of each mass times its position divided by total mass; this point governs translation even when the particles move relative to one another.
Centre-of-Mass Shift and Internal Rearrangement
When external force is zero, total momentum and centre-of-mass velocity remain constant; therefore mass displacements relative to a fixed frame obey the mass-weighted displacement constraint.