What factors determine the moment of inertia of a body?

The moment of inertia depends on: (1) Total mass of the body, (2) Distribution of mass relative to the axis of rotation, (3) Shape and size of the body, and (4) The orientation and position of the axis of rotation. Unlike mass, it is not an intrinsic constant of the body.

Can the perpendicular axes theorem be applied to a sphere or cylinder?

No! The perpendicular axes theorem (I_z = I_x + I_y) is strictly restricted to thin planar (two-dimensional) laminas lying entirely in the xy-plane. It cannot be applied to three-dimensional objects like solid spheres, solid cylinders, or cones.

What is the radius of gyration?

The radius of gyration k is the effective distance from the axis of rotation at which, if the whole mass of the body were concentrated, its moment of inertia would remain unchanged: I = M * k^2, so k = sqrt(I / M).

§ 6.5NCERT Class 11 · Physics · Chapter 6

Moment of Inertia, Theorems of Inertia & Radius of Gyration

QCC Notes · Class 11 Physics NCERT-aligned Updated 26 Sep 2026 9 min read

In linear motion, mass M measures an object's resistance to acceleration (inertia). In rotational motion, resistance to angular acceleration depends not merely on the quantity of matter, but crucially on how far that mass is spread from the axis of rotation. This rotational analogue of mass is the Moment of Inertia (I).

1. Definition and Physical Meaning of Moment of Inertia

For a system of n discrete particles with masses m₁, m₂, ..., m_n located at perpendicular distances r₁, r₂, ..., r_n from the rotation axis:

I = ∑ m_i r_i² = m₁r₁² + m₂r₂² + ... + m_nr_n²

For continuous rigid bodies, the summation becomes an integral:

I = ∫ r² dm (where r is the perpendicular distance of mass element dm from the axis)

Units & Dimensions: SI unit is kg·m². Dimensional formula: [ML²T⁰]. Moment of inertia is a tensor (or treated as a scalar for a fixed axis).

Radius of Gyration (k)

The radius of gyration k of a body about a given axis is defined as the distance from the axis at which the entire mass M of the body could be concentrated without altering its moment of inertia:

I = M k² ⇒ k = √(I / M) = √((r₁² + r₂² + ... + r_n²) / n) (Root-mean-square distance)

2. Fundamental Theorems of Moment of Inertia

A. Theorem of Perpendicular Axes (Planar Lamina Only)

For a thin, flat laminar body lying in the xy-plane, the moment of inertia about an axis perpendicular to its plane (z-axis) equals the sum of its moments of inertia about two mutually perpendicular axes in its plane intersecting at the point where the z-axis passes through:

I_z = I_x + I_y

STRICT DOMAIN OF VALIDITY: The perpendicular axes theorem is valid ONLY for 2D flat planar bodies (e.g., thin discs, rings, triangular plates). Applying it to a solid sphere, cylinder, or block is an immediate blunder penalized by NTA!

B. Theorem of Parallel Axes (Universal for All Bodies)

The moment of inertia I of any rigid body of mass M about any arbitrary axis is equal to its moment of inertia I_cm about a parallel axis passing through its centre of mass, plus the product of the mass and the square of the perpendicular distance d between the two axes:

I = I_cm + M d²

Crucial Rule: One of the two parallel axes MUST pass through the Centre of Mass! You cannot jump directly between two arbitrary parallel axes without routing through I_cm.

3. Master Formula Table: Standard Geometries

Rigid Body	Axis of Rotation	Moment of Inertia (I)	k² / R²
Thin Rod (Length L)	Perpendicular through center	(1/12) M L²	L² / 12
Thin Rod (Length L)	Perpendicular through one end	(1/3) M L²	L² / 3
Circular Ring (Radius R)	Central perpendicular axis	M R²	1
Circular Ring (Radius R)	About any diameter (perpendicular theorem)	(1/2) M R²	1 / 2
Circular Ring (Radius R)	Tangent in its plane (parallel theorem)	(3/2) M R²	3 / 2
Circular Disc (Radius R)	Central perpendicular axis	(1/2) M R²	1 / 2
Circular Disc (Radius R)	About any diameter (perpendicular theorem)	(1/4) M R²	1 / 4
Circular Disc (Radius R)	Tangent perpendicular to plane	(3/2) M R²	3 / 2
Circular Disc (Radius R)	Tangent in its plane (parallel theorem)	(5/4) M R²	5 / 4
Hollow Cylinder (Radius R)	Central longitudinal axis	M R²	1
Solid Cylinder (Radius R)	Central longitudinal axis	(1/2) M R²	1 / 2
Hollow Sphere (Thin Shell, R)	About any diameter	(2/3) M R²	2 / 3
Hollow Sphere (Thin Shell, R)	Tangent to surface	(5/3) M R²	5 / 3
Solid Sphere (Radius R)	About any diameter	(2/5) M R²	2 / 5
Solid Sphere (Radius R)	Tangent to surface	(7/5) M R²	7 / 5

4. Visualising Parallel vs Perpendicular Axis Theorems

Figure 6.5 — Visual Proof: Finding Moment of Inertia for a Disc

  [PERPENDICULAR AXES THEOREM (2D Disc)]       [PARALLEL AXIS THEOREM (Any Body)]
                ^ z-axis (I_z = 1/2 MR²)             Axis through CM     Parallel Axis
                |                                          |                   |
            +---+---+                                      |       <---d--->   |
           /    |    \                                     |                   |
    ------O-----*-----O-----> x-axis (Diameter)          [ CM ]                |
         /     / \     \                                   |                   |
        +-----*---*-----+                                  |                   |
             /     \                                       |                   |
            v y-axis (Diameter)                          I_cm                I = I_cm + Md²
        I_z = I_x + I_y = 2 I_dia
        ===> I_dia = (1/4) MR²
      

Exam Read-out: Notice that for a disc, symmetry ensures I_x = I_y = I_diameter. Since I_z = ½ MR², Perpendicular theorem instantly gives I_dia = ¼ MR²!

Step-by-Step Solved Numerical Examples

Solved Example 1: Moment of Inertia of a Disc with a Cutout Hole

Problem: A uniform circular disc has mass M and radius R. A circular hole of radius R/2 is cut such that its edge passes through the center of the original disc. Find the moment of inertia of the remaining portion about an axis passing through the center of the original disc and perpendicular to its plane.

I_remaining = (13 / 32) M R²

Step 1: Determine Masses:
Area of full disc = πR² (Mass = M).
Area of removed circular hole = π(R/2)² = πR²/4.
Mass of removed portion: m = M / 4.

Step 2: Moment of Inertia of the Full Original Disc:
I_total = ½ M R².

Step 3: Moment of Inertia of Removed Disc About Original Center:
Center of removed hole is at distance d = R/2 from original center.
MI of removed hole about its own CM: I_cm,hole = ½ m (R/2)² = ½ (M/4) (R²/4) = M R² / 32.
By Parallel Axes Theorem, MI of removed hole about original center:
I_hole,O = I_cm,hole + m d² = (M R² / 32) + (M/4) (R/2)²
= (M R² / 32) + (M R² / 16) = 3 M R² / 32.

Step 4: Subtract Hole Inertia from Full Disc:
I_rem = I_total − I_hole,O = (½ M R²) − (3/32 M R²)
= (16/32 M R²) − (3/32 M R²) = (13 / 32) M R².

Solved Example 2: Radius of Gyration of a Solid Sphere About Tangent

Problem: Find the radius of gyration of a uniform solid sphere of radius R about an axis tangent to its surface.

k = √(7/5) R ≈ 1.183 R

Step 1: Moment of Inertia About Diameter:
I_cm = (2/5) M R².

Step 2: Apply Parallel Axis Theorem for Tangent:
Distance from diameter to tangent is d = R.
I_tangent = I_cm + M R² = (2/5) M R² + M R² = (7/5) M R².

Step 3: Find Radius of Gyration:
I = M k² ⇒ (7/5) M R² = M k²
k = √(7/5) R ≈ 1.183 R.

5. Solved NEET & JEE Practice Questions

Q1. The ratio of the radii of gyration of a circular ring and a circular disc of the same radius about their respective central transverse axes is:

A. 1 : √2
B. √2 : 1
C. 2 : 1
D. 1 : 2

Answer: B

Explanation: For a ring: I_ring = M R² ⇒ k_ring = R.
For a disc: I_disc = ½ M R² ⇒ k_disc = R / √2.
Ratio k_ring : k_disc = R : (R / √2) = √2 : 1.

Q2. Four point masses, each of mass m, are situated at the four corners of a light square frame of side a. The moment of inertia of the system about an axis passing through one corner and perpendicular to the square is:

A. m a²
B. 2 m a²
C. 3 m a²
D. 4 m a²

Answer: D

Explanation: Let the axis pass through corner (0,0).
Mass 1 at (0,0): r₁ = 0 ⇒ m(0)² = 0.
Mass 2 at (a,0): r₂ = a ⇒ m a².
Mass 3 at (0,a): r₃ = a ⇒ m a².
Mass 4 at opposite diagonal (a,a): r₄ = √(a² + a²) = a√2 ⇒ m(a√2)² = 2 m a².
Total I = 0 + m a² + m a² + 2 m a² = 4 m a².

Q3. The moment of inertia of a uniform thin rod of mass M and length L about an axis passing through a point at distance L/4 from one end and perpendicular to the rod is:

A. (7/48) M L²
B. (1/12) M L²
C. (1/48) M L²
D. (13/48) M L²

Answer: A

Explanation: Center of mass is at L/2. The axis is at L/4 from an end, so distance from CM is d = (L/2) − (L/4) = L/4.
By parallel axis theorem: I = I_cm + M d² = (1/12) M L² + M (L/4)² = (1/12) M L² + (1/16) M L² = (4 + 3)/48 M L² = (7/48) M L².

Q4. A thin wire of length L and uniform linear mass density ρ is bent into a circular loop. The moment of inertia of this loop about its diameter is:

A. (ρ L³) / (4π²)
B. (ρ L³) / (8π²)
C. (ρ L³) / (16π²)
D. (ρ L³) / (2π²)

Answer: B

Explanation: Total mass M = ρ L. Circumference 2πR = L ⇒ R = L / (2π).
Moment of inertia of a circular ring about its diameter is I_dia = ½ M R².
Substituting M and R: I_dia = ½ (ρ L) [ L / (2π) ]² = ½ (ρ L) [ L² / (4π²) ] = (ρ L³) / (8π²).

6. Frequently Asked Questions (FAQs)

Why does a hollow cylinder have a larger moment of inertia than a solid cylinder of the same mass and external radius?

In a hollow cylinder, all of its mass M is concentrated at the outermost radius R (I = MR²). In a solid cylinder, mass is distributed uniformly from r = 0 all the way to R, so the average distance of mass elements from the axis is significantly smaller (I = 1/2 MR²). Therefore, the hollow cylinder possesses twice the rotational inertia.

Can moment of inertia ever be negative?

No. Moment of inertia is defined as I = ∑ m_i r_i² (or ∫ r² dm). Because mass m is strictly positive and the square of perpendicular distance r² is non-negative, the moment of inertia is always a strictly positive quantity for any real physical mass distribution.

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