Centre of gravity

Balance a ruler on one fingertip and something remarkable happens: at exactly one point along its length, the whole ruler — every gram of wood or plastic in it — stops trying to rotate and simply rests. Move your finger a centimetre either way and the ruler tips and falls. That one magic point is the centre of gravity: the single location where the entire weight of a body can be imagined to act, as if the object were replaced by a single heavy dot. The idea is not just a classroom curiosity. It is why a tightrope walker carries a long pole angled low, why a loaded truck with a high centre of gravity is more prone to rolling over on a sharp bend than a low sports car, why the Leaning Tower of Pisa has not yet toppled even after centuries of tilting, and why a hanging picture frame always settles with a particular edge down no matter how you first hang it. In every one of these situations, one question decides everything: where is the point through which gravity effectively pulls, and does the vertical line through that point still fall inside the base the object is standing on? This article builds the centre of gravity from the ground up. We connect it tightly to the centre of mass you already know from the chapter's opening sections, show why for every object you will meet in Class 11 the two points are numerically identical, walk through the plumb-line experiment used to locate the centre of gravity of an irregular lamina, and turn the idea into a working test for whether an object stands or topples. Two worked examples — a composite L-shaped lamina and a tilting rectangular block — carry the numbers all the way to a final answer, exactly as a board or competitive exam would demand.

Key takeaways

The centre of gravity (CG) is the single point where the total weight of a body can be considered to act; equivalently, it is the point about which the net gravitational torque on the body is zero.
For every object in the CBSE Class 11 syllabus, $g$ is effectively constant over the body's size, so the centre of gravity coincides exactly with the centre of mass.
Locating the CG of a composite shape uses the same weighted-average formula as centre of mass: $X_{cg} = \dfrac{\sum m_i x_i}{\sum m_i}$ , treating each regular piece as a point mass at its own centre.
The CG of an irregular flat lamina is found experimentally by the plumb-line method: suspend the lamina freely from two different points in turn; the CG is the intersection of the two vertical lines traced.
A body is stable as long as the vertical line through its CG falls inside its base of support; if tilted so this line falls outside the base, gravity's torque about the base edge topples it.
A lower CG and a wider base of support both increase stability against toppling — this is why race cars are built low and wide, and why a cone balances on its flat base but not its point.
Key idea: the centre of gravity is the point where a body's total weight acts, it coincides with the centre of mass whenever g is uniform over the body, and an object stays standing only as long as the vertical line through this point falls within its base of support.

The core insight

The centre of gravity is the single point where a body's entire weight can be taken to act — under the uniform $g$ of every Class 11 problem it coincides exactly with the centre of mass, and a body topples the instant the vertical line through this point drifts outside its base of support.

Where it fits in the syllabus

This topic sits inside the CBSE Class 11 chapter "System of Particles and Rotational Motion," immediately alongside centre of mass and the moment/torque ideas used for rigid-body equilibrium. The prescribed syllabus builds centre of mass formally through $\vec R_{cm} = \dfrac{m_1\vec r_1+m_2\vec r_2}{m_1+m_2}$ and its extension to many particles and rigid bodies; centre of gravity is the natural gravity-facing twin of that same idea, and CBSE expects you to know why the two points merge into one under everyday conditions, how to find the CG of an irregular lamina experimentally, and how to use it to reason about equilibrium and stability of rigid bodies. Nothing here goes beyond algebra and simple trigonometry, and no calculus or integration is used to derive any mass-distribution result, consistent with the board's keep-out on deriving quantities like moment of inertia from first principles.

The physical picture

Every particle of a body — every tiny piece of the L-shaped lamina, every atom of the leaning tower — has its own small weight $m_i g$ pulling straight down. A real body is not one particle; it is billions of them, each pulled down independently. Physics replaces this cloud of tiny downward forces with one single equivalent force, the total weight $Mg$ , acting through one single point. That point is the centre of gravity, and it is defined by requiring that this one equivalent force produces exactly the same total torque about any axis as the sum of all the individual particle weights actually does.

\vec R_{cg} = \frac{\sum_i m_i g_i\,\vec r_i}{\sum_i m_i g_i}

where: $\vec r_i$ is the position of the $i$ -th particle from some chosen origin, $m_i$ is its mass, and $g_i$ is the local acceleration due to gravity at that particle's location. Read physically, this says the CG is the weight-weighted average position of every particle in the body — particles with larger weight pull the average position more strongly toward themselves, exactly the way a heavier child sitting far from the pivot of a see-saw shifts the balance point toward their side. Now compare this with the centre of mass you already know:

\vec R_{cm} = \frac{\sum_i m_i\,\vec r_i}{\sum_i m_i}

where: the only difference is that each particle's contribution is weighted by its mass $m_i$ alone, with no $g_i$ attached. If $g$ is the same number for every particle in the body — true for any object small enough that Earth's gravity does not measurably change from its top to its bottom, which is every single object in this syllabus — then $g_i = g$ is a common factor in both the numerator and denominator of the $\vec R_{cg}$ formula, and it cancels exactly:

\vec R_{cg} = \frac{\sum_i m_i g\,\vec r_i}{\sum_i m_i g} = \frac{g\sum_i m_i \vec r_i}{g\sum_i m_i} = \frac{\sum_i m_i \vec r_i}{\sum_i m_i} = \vec R_{cm}

where: this cancellation is the entire reason the two concepts behave as one in Class 11 — the centre of gravity is not a different point requiring a separate calculation; it is the centre of mass viewed through the lens of "where does the weight act." Only for astronomically large bodies, where $g$ genuinely varies from one side of the object to the other (imagine a body so large that its near side to a planet feels noticeably stronger gravity than its far side), would $\vec R_{cg}$ and $\vec R_{cm}$ actually separate — a situation entirely outside the Class 11 syllabus.

Figure

A ruler balanced on a fingertip with the centre of gravity marked at the balance point and small downward weight arrows shown along its length

Locating the CG in practice

For a regular uniform body — a rod, a rectangular sheet, a disc, a sphere — symmetry alone places the CG at the geometric centre, because for every particle on one side there is a matching particle on the exact opposite side, and their contributions cancel around the centre. For a composite object built from several uniform pieces, treat each piece as a single point mass located at its own known centre, and apply the same weighted average:

X_{cg} = \frac{m_1x_1+m_2x_2+\cdots}{m_1+m_2+\cdots}, \qquad Y_{cg} = \frac{m_1y_1+m_2y_2+\cdots}{m_1+m_2+\cdots}

where: $(x_i, y_i)$ is the known centre of the $i$ -th piece and $m_i$ its mass; because $g$ is uniform, this is simultaneously the centre of mass formula and the centre of gravity formula — no separate calculation is needed. For a genuinely irregular lamina, where no formula applies, CBSE prescribes an experimental method: the plumb-line method. The lamina is punched with a small hole near one edge and hung freely from a nail through that hole, so it can swing until it settles. Since the lamina hangs in equilibrium, its CG must lie directly below the point of suspension — if it were off to one side, gravity's torque about the nail would keep rotating the lamina until the CG swings underneath. A plumb line (a string with a small weight) is hung from the same nail, and the vertical line it traces is marked on the lamina with a pencil. The lamina is then taken down, a second hole is punched near a different edge, and the whole process — hang freely, mark the vertical plumb line — is repeated. The two marked lines are straight lines that both must pass through the CG, so their intersection point is the centre of gravity of the lamina. A single suspension only tells you a line the CG lies on, not the point itself; that is why two trials are essential.

Stability and the base of support

Once the CG is located, it controls whether an object stands or falls. Drop an imaginary vertical line straight down from the CG. If that line lands inside the object's base of support — the region enclosed by its points of contact with the ground — gravity's torque about any point on the edge of the base pulls the object back toward standing upright, and it is stable. If the object is tilted far enough that the vertical line through the CG falls outside the base, gravity's torque about the nearest edge now works to rotate the object further away from upright, and it topples. This single geometric test explains why a lower CG (mass packed close to the ground) and a wider base (a bigger region the vertical line has to escape) both make an object harder to tip over — the reasoning behind the low, wide stance of a racing car or a table lamp, and the tall, narrow instability of a top-heavy bookshelf.

Derivation AThe critical toppling angle of a standing block

The goal is to find, using only the geometry of a rigid block and the definition of the centre of gravity, the exact tilt angle at which a uniform block balanced on one bottom edge is on the verge of toppling — the angle beyond which its own weight, acting through its CG, produces a torque about the pivoting edge that flips the block over rather than restoring it to standing. This uses nothing beyond the CG coinciding with the geometric centre for a uniform block and simple right-triangle trigonometry; no calculus or moment-of-inertia integration is needed, matching the Class 11 keep-out on deriving mass-distribution results from first principles.

Step 1. Set up the geometry before any tilt. A uniform rectangular block of height $H$ and base width $W$ stands flat on the floor. Since the block is uniform, its centre of gravity sits at its exact geometric centre: a height $H/2$ above the base, and a horizontal distance $W/2$ from either bottom edge. While the block sits flat, this vertical line through the CG obviously falls inside the base, midway between the two bottom edges, so the block is stable.

Step 2. Tilt the block about one bottom edge and track the CG. Imagine slowly rotating the block about its front-bottom edge (the pivot) by an angle $\theta$ measured from the vertical. Because the block is rigid, the CG stays fixed at height $H/2$ and horizontal distance $W/2$ from the centre of the base — but as the whole block rotates by $\theta$ , so does the straight-line segment joining the pivot edge to the CG, and with it the direction of the vertical line dropped from the CG relative to the base.

Step 3. Find the critical angle by requiring the CG's vertical line to pass exactly over the pivot edge. At the tipping point, the vertical dropped from the CG lands exactly on the pivot edge — any further tilt and it falls outside the base, beyond the pivot, and the block topples. Because the CG is offset from the pivot edge by a horizontal distance $W/2$ (measured along the base, from the centre of the base to the edge) and a vertical distance $H/2$ (the height of the CG above the base) when the block sits flat, tilting the block by angle $\theta$ from vertical rotates this offset with it. The critical condition — vertical line through CG lands exactly on the pivot — is reached when the tilt angle from vertical equals the angle the CG-to-pivot line already makes with the vertical in the untilted block, since tilting the block by that same angle swings the CG's vertical projection exactly onto the pivot.

Step 4. Write the critical angle using the right triangle formed by the pivot edge, the base centre, and the CG. In the untilted block, this triangle has a horizontal leg $W/2$ (base centre to pivot edge) and a vertical leg $H/2$ (base to CG height), so the angle this diagonal makes with the vertical is

\theta_{max} = \tan^{-1}\!\left(\frac{W/2}{H/2}\right) = \tan^{-1}\!\left(\frac{W}{H}\right)

where: $W$ is the full base width and $H$ is the full height of the uniform block, and $\theta_{max}$ is measured from the vertical. Tilting the block beyond $\theta_{max}$ swings the CG's vertical line past the pivot edge, outside the base, and the block topples; tilting less than $\theta_{max}$ leaves the CG's vertical line still over the base, and gravity's torque restores the block to standing.

Step 5. Read off the physical trend. A block that is wide (large $W$ ) and short (small $H$ ) has a large $\theta_{max}$ and is hard to topple; a block that is narrow (small $W$ ) and tall (large $H$ ) has a small $\theta_{max}$ and topples with only a slight tilt. This is exactly the "low and wide is stable, tall and narrow is unstable" rule stated qualitatively earlier, now pinned down as a single formula.

Worked examples

Board level - centre of gravity of a composite L-shaped lamina

A uniform L-shaped lamina is made of two uniform rectangular pieces: piece 1 is $4\ \text{m}\times2\ \text{m}$ with mass $8\ \text{kg}$ , centred at $(2, 1)$ ; piece 2 is $2\ \text{m}\times3\ \text{m}$ with mass $6\ \text{kg}$ , centred at $(1, 3.5)$ (all coordinates in metres, with the pieces joined so these centres are correct for the combined shape). Find the centre of gravity of the whole lamina.

Step 1. Recognise that since $g$ is uniform over the lamina, the centre of gravity coincides with the centre of mass, so the ordinary weighted-average formula applies directly, treating each rectangular piece as a point mass located at its own known centre.

Step 2. Compute the $x$ -coordinate of the CG.

X_{cg} = \frac{m_1x_1+m_2x_2}{m_1+m_2} = \frac{8(2)+6(1)}{8+6} = \frac{16+6}{14} = \frac{22}{14} \approx 1.57\ \text{m}

Step 3. Compute the $y$ -coordinate of the CG.

Y_{cg} = \frac{m_1y_1+m_2y_2}{m_1+m_2} = \frac{8(1)+6(3.5)}{14} = \frac{8+21}{14} = \frac{29}{14} \approx 2.07\ \text{m}

Step 4. Combine into the final position. The centre of gravity of the composite lamina is at $(1.57\ \text{m},\ 2.07\ \text{m})$ , pulled toward piece 1's centre in $x$ (since piece 1 is heavier) and toward piece 2's centre in $y$ (since piece 2's centre lies much higher).

Competitive level - critical toppling angle of a standing block

A uniform rectangular block of height $H = 0.6\ \text{m}$ and base width $W = 0.4\ \text{m}$ stands on a horizontal floor, and is slowly tilted about one bottom edge. Find the maximum angle from the vertical to which the block can be tilted before it topples, and find the minimum base width the block would need (keeping the same height) to survive being tilted to $45°$ without toppling.

Step 1. Apply the toppling-angle result derived above directly. With $W = 0.4\ \text{m}$ and $H = 0.6\ \text{m}$ ,

\theta_{max} = \tan^{-1}\!\left(\frac{W}{H}\right) = \tan^{-1}\!\left(\frac{0.4}{0.6}\right) = \tan^{-1}(0.667) \approx 33.7°

So the block topples once tilted beyond about $33.7°$ from the vertical; within this angle gravity's torque restores it to standing.

Step 2. Set up the reverse question: for the block to just survive a $45°$ tilt, the critical angle itself must equal $45°$ , so solve $\tan\theta_{max} = W'/H$ for the new width $W'$ with $\theta_{max} = 45°$ and $H$ unchanged at $0.6\ \text{m}$ .

Step 3. Solve for $W'$ .

W' = H\tan(45°) = 0.6 \times 1 = 0.6\ \text{m}

Step 4. State both final answers together. The original $0.4\ \text{m}\times0.6\ \text{m}$ block topples beyond $\approx 33.7°$ from vertical; to push that critical angle out to $45°$ without changing its height, the base would need to be widened to at least $0.6\ \text{m}$ — one and a half times its original width, because $\tan 45° = 1$ demands the CG's horizontal and vertical offsets from the pivot be exactly equal. Timing note: this pair of angle calculations (one direct, one reversed) should take under 3 minutes total in a competitive exam — the whole problem reduces to remembering that $\tan\theta_{max} = W/H$ for a uniform block tilted about a bottom edge, so no fresh derivation is needed once you recall the ratio.

Common traps

Treating centre of gravity and centre of mass as two separate quantities requiring two separate calculations — for every Class 11 problem, with uniform $g$ , they are numerically the identical point, and computing one gives you the other for free.
Believing the plumb-line method locates the CG from a single suspension — one trial only gives a line the CG lies on; a second suspension from a genuinely different point is required, and the CG is the intersection of the two lines.
Forgetting that stability depends on the width of the base of support, not simply the height of the centre of gravity — a tall object with a sufficiently wide base can be more stable than a short object with a narrow base, as the $\tan\theta_{max} = W/H$ result makes precise.
Assuming the CG of an irregular or composite object sits at its visual or geometric midpoint rather than computing the correct mass-weighted position — a heavier piece pulls the CG toward itself, just as in the L-shaped lamina example.
Mixing up which leg of the toppling triangle is $W/2$ and which is $H/2$ , or forgetting the factor of one-half when a formula is stated in terms of full width and height rather than half-width and half-height.

Common trapCentre of gravity is not automatically the same as centre of mass in general — it only coincides for uniform g, which happens to be every situation the syllabus tests

Students often state "CG equals CM" as an unconditional identity, but it is really a consequence of $g$ being the same number for every particle of the body. The formulas $\vec R_{cg} = \sum m_ig_i\vec r_i/\sum m_ig_i$ and $\vec R_{cm} = \sum m_i\vec r_i/\sum m_i$ are genuinely different definitions; they merge into one only because the common factor $g$ cancels when $g_i = g$ for every particle. For any object you will meet in this course this is always true, so treating them as identical is completely safe here — but understand why, since an exam can test the reasoning by asking what would happen for an astronomically large body where $g$ varies across its size.

You should be able to…

State the definition of centre of gravity and explain, using the cancellation of $g$ , exactly why it coincides with the centre of mass for every object in this syllabus.
Compute the centre of gravity of a composite object built from simple uniform pieces using the weighted-average formula.
Describe, step by step, the plumb-line experimental method for locating the CG of an irregular lamina, and explain why two suspension trials are required.
Apply the base-of-support test for stability, and derive or recall the critical toppling angle $\theta_{max} = \tan^{-1}(W/H)$ for a uniform block.
Explain qualitatively why a lower centre of gravity and a wider base both improve stability against toppling.

Ask the tutor: want to work through the CG of a differently shaped composite lamina, or see the toppling-angle idea applied to a tilting bookshelf or a loaded truck — ask the AI tutor to walk through it with you.

If CG and centre of mass are always the same in this chapter, why does the syllabus bother teaching both terms?

Because they answer physically different questions even when they land on the same point. Centre of mass answers "where is the mass concentrated, for the purposes of describing motion" — it is the point that moves as if all external force acted there, useful even with zero gravity, like a spinning wrench tumbling in space. Centre of gravity answers "where does gravity effectively pull" — it is the point relevant to weight, torque due to gravity, and stability against toppling. The two questions happen to have the same answer whenever $g$ is uniform, which is every case here, but keeping the concepts distinct matters once you meet situations (astrophysics, later courses) where a body is large enough that $g$ genuinely varies across it.

Why does a taller object not automatically tip over more easily than a shorter one?

Height alone does not decide stability — the ratio of base width to height does, through $\tan\theta_{max} = W/H$ . A tall object with a wide enough base, like a well-designed tower, can have a larger critical tipping angle than a short object with a very narrow base, like a pencil balanced on its tip. What actually matters is how far the vertical line through the CG has to travel before it escapes the base — a wide base gives that line more room to move as the object tilts, regardless of how tall the object is.

Does the plumb-line method work for any solid object, or only flat laminas?

The method as commonly taught in Class 11 is specifically for a flat, roughly two-dimensional lamina, because the CG of a lamina lies in its own plane, and two suspension trials pin down a single point within that plane. For a genuinely three-dimensional solid, the same physical idea extends (hang it freely from different points and track the vertical lines in three dimensions), but the syllabus keeps the demonstration to the two-dimensional lamina case, which is what you should describe if asked to explain the experiment.

In the toppling formula, why is it W over H and not the other way round?

Because $\theta$ is measured from the vertical, and $\tan\theta$ compares the side opposite the angle to the side adjacent to it in the right triangle formed by the pivot edge, the base centre, and the CG. The horizontal leg ( $W/2$ , opposite the angle from vertical) sits over the adjacent vertical leg ( $H/2$ ), giving $\tan\theta_{max} = (W/2)/(H/2) = W/H$ . If you instead measured the angle from the horizontal floor, the ratio would flip to $H/W$ — so always check which reference line an angle is measured from before writing down a tangent ratio.