9. Mechanics — Statics

Moment of inertia

Moment of inertia is the scalar measure of a rigid body's rotational inertia, representing how its mass is distributed relative to a rotational axis.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Parallel Axes Theorem (Steiner's Theorem)

The moment of inertia of a body about any axis is equal to the sum of the moment of inertia about a parallel axis through its center of mass and the product of the total mass of the body and the square of the perpendicular distance between the two axes: $$I = I_{cm} + M h^2$$

📐 Geometric Interpretation

It bridges any arbitrary axis in space to the unique central axis, visually decomposing rotational inertia into an intrinsic part ($I_{cm}$) and an extrinsic spatial shift ($M h^2$).

⚙️ Algebraic & Structural Meaning

Provides a linear-quadratic decomposition that simplifies complex rotational dynamics into standard center-of-mass calculations.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Rigid Body Continuum / Discrete Mass Distribution	`Domain of integration / summation`	Strictly Mandatory / Necessary	If the body deforms, mass distribution changes dynamically, invalidating static spatial integration.
Parallel Axis Alignment	`Spatial orientation vector`	Strictly Mandatory	If axes are non-parallel, the $M h^2$ scalar shift term fails; product of inertia tensors must be invoked.
Perpendicular Axis Condition	`Planar lamina condition ($z = 0$)`	Essential for Perpendicular Axes Theorem	If the body is three-dimensional, $I_z = I_x + I_y$ does not hold without planar mass confinement.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Uniform Rod about End

$$I = \frac{1}{3} M L^2$$

Rule: Direct memory result for a rod of mass $M$ and length $L$ about an axis perpendicular to its length through one end.

Circular Ring vs. Disc (Diametral Axis)

$$I_{\text{ring}} = \frac{1}{2} M R^2, \quad I_{\text{disc}} = \frac{1}{4} M R^2$$

Rule: Diametral moment of inertia is always half of the polar moment of inertia for thin symmetrical planar bodies.

Application: Frequently asked in comparative matching questions in KVS/NVS PGT.

Solid Sphere about Tangent

$$I_{\text{tangent}} = \frac{2}{5} M R^2 + M R^2 = \frac{7}{5} M R^2$$

Rule: Apply Parallel Axes Theorem directly to the standard diametral sphere result ($I_{cm} = \frac{2}{5}MR^2$).

Application: Standard recurring formula in DSSSB PGT Mathematics.

Rectangular Lamina about Center

$$I = \frac{M(a^2 + b^2)}{12}$$

Rule: Sum of perpendicular dimensions squared times mass over 12 for an axis perpendicular to the plane through the center.

Application: Tested in LT Grade mechanics sections.

Hollow Cylinder (Shell) vs. Solid Cylinder

$$I_{\text{shell}} = M R^2, \quad I_{\text{solid}} = \frac{1}{2} M R^2$$

Rule: Cylindrical mass concentrated at radius $R$ gives $MR^2$; uniform distribution halves it.

Application: Quick elimination tool in multiple-choice questions.

Radius of Gyration ($k$) Quick Relation

$$k = \sqrt{\frac{I}{M}}$$

Rule: If moment of inertia is given as a multiple of $MR^2$, immediately drop $M$ and take the square root of the coefficient multiplied by $R$.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Confusing Axis Orientation in Parallel Axes Theorem

Common Pitfall: Applying $I = I_{cm} + Mh^2$ when the axis is not passing through the center of mass or when axes are not parallel.

Counterexample: Using $h$ as the distance from an arbitrary random point instead of the center of mass.

Why it fails: The theorem mandates that one of the two parallel axes *must* pass through the center of mass.

🛡️ Exam Defense: Always verify that $I_{cm}$ is explicitly the moment of inertia about the center of mass before adding $M h^2$.

⚠️ COMMON ERROR #02

Misapplying Perpendicular Axes Theorem to 3D Bodies

Common Pitfall: Using $I_z = I_x + I_y$ for solid spheres or cylinders.

Counterexample: Solid sphere where $I_z = \frac{2}{5}MR^2$, but $I_x + I_y = \frac{2}{5}MR^2 + \frac{2}{5}MR^2 = \frac{4}{5}MR^2 \neq I_z$.

Why it fails: The perpendicular axes theorem is strictly valid only for two-dimensional planar laminas lying in the $xy$-plane.

🛡️ Exam Defense: Check if the object is flat (lamina) before applying $I_z = I_x + I_y$.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Parallel Axes Theorem	Any two parallel axes (one through CM)	$I = I_{cm} + Mh^2$	Shifting axis parallel to itself	Axis given at distance $h$ from center
Perpendicular Axes Theorem	Planar body ($z=0$ lamina)	$I_z = I_x + I_y$	Sum of orthogonal in-plane inertias gives out-of-plane inertia	Mention of 'perpendicular to plane' or lamina
Radius of Gyration	Mass distribution metric	$k = \sqrt{I/M}$	Effective distance where entire mass is concentrated	Asks for 'radius of gyration' or $k$
Moment of Inertia (Rod)	Uniform linear density	$I = \frac{1}{3}ML^2$ (end), $\frac{1}{12}ML^2$ (center)	Inertia of 1D line segment about axis	Keywords: uniform rod, wire, thin bar

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the moment of inertia of a uniform circular disc of mass $M$ and radius $R$ about a tangent in its own plane.

(A) $\frac{1}{2}MR^2$(B) $\frac{3}{2}MR^2$(C) $\frac{5}{4}MR^2$(D) $\frac{3}{4}MR^2$

📚 Method A: Formal Textbook Derivation

Step 1: Moment of inertia of disc about a diametral axis is $I_{cm} = \frac{1}{4}MR^2$. Step 2: Distance to parallel tangent in its plane is $h = R$. Step 3: Apply Parallel Axes Theorem: $I = I_{cm} + MR^2 = \frac{1}{4}MR^2 + MR^2 = \frac{5}{4}MR^2$.

⚡ Method B: 25-Second Exam Shortcut

Recall diametral $I = 0.25 MR^2$, add $1.0 MR^2$ for the parallel shift to get $1.25 MR^2 = \frac{5}{4}MR^2$.

💡 Tip: Note whether the tangent is *in its own plane* or *perpendicular to its plane* (which would give $\frac{3}{2}MR^2$).

CANONICAL PROBLEM 02 Dual-Method Breakdown

If the radius of gyration of a circular ring of radius $R$ about an axis passing through its center and perpendicular to its plane is $k$, what is $k$?

(A) $R$(B) $\frac{R}{\sqrt{2}}$(C) $2R$(D) $\frac{R}{2}$

📚 Method A: Formal Textbook Derivation

Step 1: For a circular ring, $I = MR^2$. Step 2: By definition of radius of gyration, $I = Mk^2$. Step 3: Equating gives $Mk^2 = MR^2 \implies k = R$.

⚡ Method B: 25-Second Exam Shortcut

For a ring about polar axis, all mass is at distance $R$, hence radius of gyration is identically $R$.

💡 Tip: Do not confuse ring ($k=R$) with disc ($k=R/\sqrt{2}$).

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about an axis through its center and perpendicular to its length is given by $$I = \dots$$

A $\frac{1}{3}ML^2$

B $\frac{1}{6}ML^2$

C $\frac{1}{12}ML^2$

D $\frac{1}{4}ML^2$

Derivation & Analysis:

Integrating $\int_{-L/2}^{L/2} x^2 \left(\frac{M}{L}\right) dx$ yields $\frac{1}{12}ML^2$.

MCQ 2: What is the moment of inertia of a solid sphere of mass $M$ and radius $R$ about any of its diameters?

A $\frac{2}{5}MR^2$

B $\frac{2}{3}MR^2$

C $\frac{1}{2}MR^2$

D $\frac{7}{5}MR^2$

Derivation & Analysis:

Standard integration for a solid sphere gives $I = \frac{2}{5}MR^2$ about any diametral axis.

MCQ 3: If $I_x$ and $I_y$ are moments of inertia of a plane lamina about mutually perpendicular axes $ox$ and $oy$ in its plane, the moment of inertia about an axis perpendicular to the plane and passing through their intersection is $$I_z = \dots$$

A $I_x - I_y$

B $\frac{I_x + I_y}{2}$

C $I_x + I_y$

D $\sqrt{I_x I_y}$

Derivation & Analysis:

By the Perpendicular Axes Theorem, $I_z = I_x + I_y$ for any planar lamina.

MCQ 4: The radius of gyration of a solid cylinder of radius $R$ about its own geometric axis is $$k = \dots$$

A $R$

B $\frac{R}{\sqrt{2}}$

C $\frac{R}{2}$

D $\frac{R}{\sqrt{3}}$

Derivation & Analysis:

For a solid cylinder, $I = \frac{1}{2}MR^2 = Mk^2 \implies k = \frac{R}{\sqrt{2}}$.

MCQ 5: Parallel axes theorem is expressed as $I = I_{cm} + Mh^2$. Here $h$ represents:

A Diameter of the body

B Radius of gyration

C Perpendicular distance between the two parallel axes

D Length of the rod

Derivation & Analysis:

By definition, $h$ is the strictly perpendicular distance between the axis passing through the CM and the parallel axis in question.

MCQ 6: The moment of inertia of a thin circular ring of mass $M$ and radius $R$ about a tangent perpendicular to its plane is:

A $2MR^2$

B $\frac{3}{2}MR^2$

C $MR^2$

D $\frac{1}{2}MR^2$

Derivation & Analysis:

Using Parallel Axes Theorem with $I_{cm} = MR^2$ (about diameter perpendicular to plane axis? No, about polar axis $I_{cm} = MR^2$, parallel axis at distance $R$ gives $I = MR^2 + MR^2 = 2MR^2$).

MCQ 7: Dimensional formula for moment of inertia is:

A $[MLT^{-2}]$

B $[ML^2 T^{-2}]$

C $[ML^2]$

D $[ML^{-2}]$

Derivation & Analysis:

Since $I = \sum m_i r_i^2$, its dimensional formula is $[M][L]^2 = [ML^2]$.

MCQ 8: Assertion (A): The moment of inertia of a body depends on the axis of rotation. Reason (R): Moment of inertia is a scalar quantity, not a tensor.

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B Both (A) and (R) are true but (R) is not the correct explanation of (A)

C (A) is true but (R) is false

D (A) is false but (R) is true

Derivation & Analysis:

Assertion is true because MOI changes with axis orientation. Reason is false because MOI is strictly a tensor of rank 2 (though represented as a scalar for a fixed axis).

MCQ 9: What is the moment of inertia of a thin spherical shell of mass $M$ and radius $R$ about its diameter?

A $\frac{2}{3}MR^2$

B $\frac{2}{5}MR^2$

C $MR^2$

D $\frac{4}{5}MR^2$

Derivation & Analysis:

For a thin spherical shell, $I = \frac{2}{3}MR^2$ about any diametral axis.

Q10

MCQ 10: If two bodies of same mass and material are in the form of a solid sphere and a thin spherical shell, which has greater moment of inertia about their diameters?

A Solid sphere

B Thin spherical shell

C Both have equal inertia

D Depends on velocity

Derivation & Analysis:

Shell has $I = \frac{2}{3}MR^2 \approx 0.67 MR^2$, whereas solid sphere has $I = \frac{2}{5}MR^2 = 0.4 MR^2$. Thus, shell has greater moment of inertia.

Common questions

Strictly speaking, moment of inertia is neither a simple scalar nor a vector; it is a second-rank symmetric tensor. However, for a fixed axis of rotation, it behaves and can be treated as a scalar quantity.

Candidates frequently confuse the formulas for solid vs. hollow bodies and misapply the perpendicular axes theorem to three-dimensional solids instead of strictly two-dimensional laminas.

Memorize standard standard results ($I_{cm}$ and radii of gyration for rods, rings, discs, spheres, and cylinders) and master the quick application of Parallel and Perpendicular Axes Theorems without re-deriving via integration.

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