9. Mechanics — Statics

Centre of gravity

The centre of gravity of a body or a system of particles is the unique point through which the resultant of the weights of all the constituent particles acts, regardless of the orientation of the body.

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

📖 Topic progress

○ Mark as complete

Definition and statement

Principle of Moments for Centre of Gravity

The moment of the weight of a system of particles about any point or axis is equal to the sum of the moments of the weights of the individual particles about the same point or axis, acting at the coordinates of the centre of gravity $(\bar{x}, \bar{y}, \bar{z})$ given by $W\bar{x} = \sum w_i x_i$, $W\bar{y} = \sum w_i y_i$, and $W\bar{z} = \sum w_i z_i$, where $W = \sum w_i$.

📐 Geometric Interpretation

Geometrically, the centre of gravity represents the weighted average position of mass distribution. For bodies possessing axes or planes of symmetry, the centre of gravity must lie on every axis or plane of symmetry.

⚙️ Algebraic & Structural Meaning

Algebraically, it reduces a continuous distribution of parallel forces (gravity) into a single resultant force vector acting at a single coordinate point, simplifying equilibrium equations to $\sum F = 0$ and $\sum \tau = 0$.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Uniform Gravitational Field	`Entire system spatial domain`	Strictly Mandatory / Necessary	If gravity varies significantly across the body (non-uniform field), the centre of gravity does not coincide with the centre of mass.
Continuity of Mass Distribution	`Volume $V$ or Area $A$ or Arc $L$`	Essential for integration	Discontinuous density functions require summation or Riemann-Stieltjes integrals rather than standard definite integrals.
Rigidity of the Body	`Configuration space`	Mandatory for statics	Deformable bodies alter their shape and mass distribution under load, shifting the centre of gravity dynamically.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Composite Bodies Formula

$$ \bar{x} = \frac{\sum A_i x_i}{\sum A_i} \quad \text{or} \quad \bar{x} = \frac{\sum V_i x_i}{\sum V_i} $$

Rule: Treat composite standard shapes (cones, hemispheres, cylinders) with signed areas/volumes (subtracting cavities).

Application: Use this result in the stated type of problem.

Centre of Gravity of Standard Arcs and Sectors

$$ \bar{y} = \frac{a \sin \alpha}{\alpha} \text{ (Circular Arc)}, \quad \bar{y} = \frac{2a \sin \alpha}{3 \alpha} \text{ (Circular Sector)} $$

Rule: Direct substitution of semi-vertical angle $\alpha$ for circular segments and sectors.

Application: Direct formula recall questions in DSSSB and KVS PGT.

Hemisphere and Cone Centroid Heights

$$ \bar{y} = \frac{3}{8}a \text{ (Solid Hemisphere)}, \quad \bar{y} = \frac{3}{4}a \text{ (Hollow Hemisphere)}, \quad \bar{y} = \frac{1}{4}h \text{ (Solid Cone)} $$

Rule: Memorize standard distance fractions from base or vertex to eliminate integration steps.

Symmetry Exploitation

$$ \text{If } f(-x, y) = f(x, y) \implies \bar{x} = 0 $$

Rule: Instantly set coordinates lying on axes of symmetry to zero without calculation.

Application: Saves critical time in multi-part mechanics problems.

Pappus-Guldinus Theorem for Volumes/Surfaces

$$ V = 2\pi \bar{y} A, \quad S = 2\pi \bar{y} L $$

Rule: Relates surface area and volume of revolution directly to the path length of the centre of gravity.

Triangular lamina / Tetrahedron Rule

$$ \bar{x} = \frac{x_1 + x_2 + x_3}{3}, \quad \bar{y} = \frac{y_1 + y_2 + y_3}{3} $$

Rule: The centroid of a triangle or tetrahedron is the arithmetic mean of its vertex coordinates.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Centre of Mass vs. Centre of Gravity Confusion

Common Pitfall: Assuming centre of gravity and centre of mass are always at different points regardless of the gravitational field.

Counterexample: Uniform gravitational field $g = \text{constant}$

Why it fails: They coincide only when the gravitational acceleration vector $\vec{g}$ is uniform in magnitude and direction across the entire body.

🛡️ Exam Defense: In standard textbook problems, assume uniform field unless specified otherwise, but note the conceptual distinction.

⚠️ COMMON ERROR #02

Sign Error in Cavity / Removed Portions

Common Pitfall: Adding the volume/mass of a removed cutout portion instead of subtracting it in composite bodies.

Counterexample: A sphere with a spherical cavity

Why it fails: A cavity represents negative mass/weight/volume, requiring negative sign in summation: $\sum V_i x_i - \sum V_{\text{cavity}} x_{\text{cavity}}$.

🛡️ Exam Defense: Always treat cutouts with a negative weight factor.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Centre of Mass	Independent of external gravity field	$\vec{R} = \frac{\sum m_i \vec{r}_i}{\sum m_i}$	Average position of mass distribution	Involves masses $m_i$
Centre of Gravity	Uniform parallel gravitational field	$\bar{x} = \frac{\sum w_i x_i}{\sum w_i}$	Point of application of resultant weight	Involves weights $w_i = m_i g$
Centroid	Uniform geometric density	$\bar{x} = \frac{\int x \, dA}{\int dA}$	Geometric center of figure	Involves lengths, areas, or volumes
Pappus Theorem	Plane curve / area revolved about axis	$V = 2\pi \bar{y} A$	Volume generated by revolving area	Keywords: 'surface of revolution', 'volume', 'revolved'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the centre of gravity of a uniform solid hemisphere of radius $a$ from its base.

(A) $\frac{3}{8}a$(B) $\frac{3}{5}a$(C) $\frac{1}{2}a$(D) $\frac{2}{3}a$

📚 Method A: Formal Textbook Derivation

Using integration in spherical polar or cylindrical coordinates: $\bar{y} = \frac{\int_0^a y \cdot \pi x^2 dy}{\int_0^a \pi x^2 dy}$, where $x^2 = a^2 - y^2$. Evaluating gives $\frac{3}{8}a$.

⚡ Method B: 25-Second Exam Shortcut

Standard result memorization for solid hemisphere centroid is $\frac{3}{8}a$ from base.

💡 Tip: Do not derive during exam; recall solid hemisphere is $\frac{3}{8}a$ and hollow hemisphere is $\frac{1}{2}a$ from center (or $\frac{1}{2}a$ from base).

CANONICAL PROBLEM 02 Dual-Method Breakdown

A uniform circular sector of radius $a$ has semi-vertical angle $\alpha$. Its centre of gravity from the vertex is given by:

(A) $\frac{2a \sin \alpha}{3 \alpha}$(B) $\frac{a \sin \alpha}{\alpha}$(C) $\frac{3a \sin \alpha}{4 \alpha}$(D) $\frac{4a \sin \alpha}{3 \alpha}$

📚 Method A: Formal Textbook Derivation

Integrate polar moments over the sector domain: $\bar{r} = \frac{\int_{-\alpha}^{\alpha} \int_0^a r^2 \cos \theta \, dr \, d\theta}{\int_{-\alpha}^{\alpha} \int_0^a r \, dr \, d\theta} = \frac{2a \sin \alpha}{3\alpha}$.

⚡ Method B: 25-Second Exam Shortcut

Direct standard formula for circular sector distance from vertex.

💡 Tip: Pay attention whether distance is asked from the vertex or from the centre of the arc.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: The centre of gravity of three particles of equal mass placed at the vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is:

A $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$

B $\left(\frac{x_1+x_2+x_3}{2}, \frac{y_1+y_2+y_3}{2}\right)$

C $(x_1+x_2+x_3, y_1+y_2+y_3)$

D $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

Derivation & Analysis:

For equal weights $w_1 = w_2 = w_3 = w$, $\bar{x} = \frac{w(x_1+x_2+x_3)}{3w} = \frac{x_1+x_2+x_3}{3}$.

MCQ 2: The distance of the centre of gravity of a uniform solid cone of height $h$ from its vertex is:

A $\frac{1}{4}h$

B $\frac{1}{2}h$

C $\frac{3}{4}h$

D $\frac{2}{3}h$

Derivation & Analysis:

From the base, the CG of a solid cone is at height $\frac{1}{4}h$. Therefore, from the vertex, it is $h - \frac{1}{4}h = \frac{3}{4}h$.

MCQ 3: If a portion of radius $b$ is removed from a uniform circular lamina of radius $a$, the distance of the CG of the remaining part from the center is:

A $\frac{b^2}{a-b}$

B $\frac{a^2 b^2}{a^2 - b^2}$

C $\frac{b^3}{a^2 - b^2}$

D $\frac{a^3}{a^2 - b^2}$

Derivation & Analysis:

Using composite body formula with negative area: $\bar{x} = \frac{A_1 x_1 - A_2 x_2}{A_1 - A_2} = \frac{0 - (\pi b^2)(d)}{\pi a^2 - \pi b^2}$, yielding $\frac{b^2 d}{a^2 - b^2}$. For tangent circles, $d = b$, so $\frac{b^3}{a^2-b^2}$.

MCQ 4: The centre of gravity of a uniform circular arc of radius $a$ subtending an angle $2\alpha$ at the centre is at a distance from the centre of the arc equal to:

A $\frac{a \sin \alpha}{\alpha}$

B $\frac{a \cos \alpha}{\alpha}$

C $\frac{2a \sin \alpha}{\alpha}$

D $\frac{a \sin \alpha}{2\alpha}$

Derivation & Analysis:

By standard derivation for a circular arc, $\bar{y} = \frac{\int_{-\alpha}^{\alpha} a \cos \theta \cdot a d\theta}{\int_{-\alpha}^{\alpha} a d\theta} = \frac{a \sin \alpha}{\alpha}$.

MCQ 5: Centre of gravity and centre of mass coincide when:

A The body is non-homogeneous

B The gravitational field is uniform

C The body is in motion

D The body has no axes of symmetry

Derivation & Analysis:

In a uniform gravitational field ($g$ is constant vector), $w_i = m_i g$, hence the factors $g$ cancel out in the CG formula, making CG identical to CM.

MCQ 6: The centre of gravity of a triangular lamina coincides with its:

A Circumcentre

B Incentre

C Centroid

D Orthocentre

Derivation & Analysis:

For a uniform triangular lamina, integration yields the centroid (intersection of medians).

MCQ 7: What is the height of the centre of gravity of a hollow hemisphere of radius $a$ from its base?

A $\frac{1}{2}a$

B $\frac{1}{4}a$

C $\frac{3}{8}a$

D $\frac{2}{3}a$

Derivation & Analysis:

For a hollow hemisphere, $\bar{y} = \frac{1}{2}a$ from the centre (which is also the distance from the base plane).

MCQ 8: Assertion (A): The centre of gravity of a body may lie outside the material of the body. Reason (R): CG is a point representing the mean position of weight distribution.

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

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

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

Derivation & Analysis:

For hollow bodies like rings, circular wires, or hollow spheres, the CG lies at the geometric center which is outside the solid material.

MCQ 9: If the density of a rod of length $l$ varies as $\rho(x) = kx$ from one end $x = 0$, its centre of gravity is at:

A $\frac{l}{2}$

B $\frac{2l}{3}$

C $\frac{3l}{4}$

D $\frac{l}{3}$

Derivation & Analysis:

$\bar{x} = \frac{\int_0^l x (kx) dx}{\int_0^l (kx) dx} = \frac{[x^3/3]_0^l}{[x^2/2]_0^l} = \frac{l^3/3}{l^2/2} = \frac{2}{3}l$.

Q10

MCQ 10: The centroid of the region bounded by $y = x^2$ and $y = 4$ in the upper half plane has $\bar{x}$ equal to:

A 1

B 0

C 2

D 4

Derivation & Analysis:

Due to symmetry about the y-axis ($x = -x$), the x-coordinate of the centroid $\bar{x} = 0$.

Common questions

They coincide whenever the gravitational field is uniform across the entire body. In a non-uniform gravitational field (such as an enormous structure spanning planetary scales), they diverge.

Failing to assign negative signs to removed parts (cavities or cutouts). Every removed volume or area must be subtracted in both numerator and denominator.

Create a formula chart for hemispheres, cones, arcs, and sectors (solid vs. hollow, distance from vertex vs. base) and revise them daily to solve questions.

← Previous Topic

Work, power and energy

Next Topic →

Common catenary