Centre of gravity

Tilt a loaded truck a little on a slope and it stays upright; tilt it a bit more and it suddenly rolls over. Nothing about the road changed in that instant — what changed is where an invisible line from the truck's centre of gravity (C.G.) falls relative to its wheels. The centre of gravity is the single point where a body's entire weight can be taken to act, whatever way the body is turned, and it is the key to two everyday questions: where does an irregular object balance, and when does it topple? This article builds the ICSE Class 10 picture from that one idea: locating the C.G. for regular shapes, reading off stable, unstable and neutral equilibrium from what the C.G. does on a small displacement, and — one step past the syllabus statement — turning "wider base, lower C.G. is more stable" into an actual toppling angle.

The core insight

The centre of gravity is the fixed point through which a body's total weight acts, and a body topples the instant the vertical line through its C.G. moves outside its base of support.

The core idea

What the C.G. is. Every particle of a body is pulled down by gravity, so the body carries countless tiny weight forces, one on every particle. The C.G. is the single point through which the resultant of all these parallel forces — the total weight — can be considered to act, no matter how the body is oriented.

Where it sits for regular shapes. For a uniform rod, the C.G. is at the midpoint. For a disc or ring, it is at the centre of the circle. For a square, rectangle or parallelogram, it is where the diagonals cross. For a triangular lamina, it is the centroid, where the three medians meet. For a sphere or cube, it is the geometric centre. None of this requires the C.G. to sit inside material at all — the C.G. of a ring or a hollow bangle is at the centre of the hole, in empty space, because the missing material there is exactly compensated by the symmetric ring of mass around it.

Locating it for an irregular body. Suspend an irregular lamina freely from a point and it settles with its C.G. directly below the point of suspension — otherwise the weight would create a net moment and it would keep rotating. Mark that vertical line, suspend the lamina from a second point, and mark a second line; the C.G. is where the two lines cross. Balancing an object on a knife-edge finds the same vertical line directly, because balance requires zero moment of the weight about the edge:

τ=W×d=0⇒d=0\tau = W \times d = 0 \quad \Rightarrow \quad d = 0

where: τ\tau is the moment (torque) of the weight about the pivot, WW is the weight, and dd is the perpendicular distance from the pivot to the vertical line through the C.G. — zero only when the pivot lies exactly under the C.G.

Stability and the base of support. A body stands as long as the vertical line through its C.G. falls inside its base of support, and it topples the moment that line moves outside the base. Displace a body slightly and watch what its C.G. does: if it rises and the vertical line stays inside the base, gravity pulls the body back — stable equilibrium. If it falls, gravity carries the body further away — unstable equilibrium. If the C.G. neither rises nor falls (a ball on a flat table), it is neutral equilibrium. A wide base and a low C.G. both make it harder for that vertical line to escape the base — why race cars and loaded trucks are built low and wide.

Figure
A block on a base being tilted about its front edge, showing the centre of gravity, the height h above the base, the half-width a of the base, and the tilt angle theta between the vertical and the line from the pivot to the centre of gravity
Tilting a block about its front edge upright, stable tilted toward the critical angle C.G. line inside base h edge edge a theta pivot edge C.G. line approaching the pivot edge Once the C.G. line crosses the pivot edge, gravity's moment flips from restoring to overturning.
Derivation AThe critical tilt angle at which a block topples

The goal is to turn the qualitative rule "a body topples once the vertical line through its C.G. leaves the base" into a single formula for the exact tilt angle at which that happens, for a block of base half-width aa resting with its C.G. centred at height hh above the base, using only the moment of the weight about the pivot edge — no calculus, just geometry and the condition for zero restoring moment.

Step 1. Set up the geometry. Before any tilt, the C.G. sits at horizontal distance aa from the front edge of the base (the pivot when the block is tipped) and height hh above the base. The line joining the pivot to the C.G. makes a fixed angle ϕ\phi with the vertical, where tan⁡ϕ=a/h\tan\phi = a/h — fixed because the pivot and the C.G. tilt together as one rigid body.

Step 2. Tilt the block about the pivot through angle θ\theta. The pivot-to-C.G. line rotates with it, making angle (ϕ−θ)(\phi - \theta) with the vertical. As long as θ<ϕ\theta < \phi, the C.G. is still on the same side of the pivot, and its weight creates a restoring moment that tips the block back onto its full base.

Step 3. Find the critical angle. The moment of the weight about the pivot is proportional to the horizontal distance between the pivot and the vertical line through the C.G. — which shrinks to zero exactly when θ=ϕ\theta = \phi. Beyond this, the C.G. has swung past the pivot, its vertical line falls outside the base, and the moment reverses, adding to the tilt instead of opposing it.

Step 4. State the result. The critical tilt angle θc\theta_c at which the block topples is therefore

tan⁡θc=ah\tan\theta_c = \frac{a}{h}

where: aa is the base half-width (centre of base to pivot edge), hh is the C.G. height above the base, and θc\theta_c is the tilt beyond which the body no longer returns. This confirms the qualitative rule algebraically: a larger aa or smaller hh increases θc\theta_c, so the body can tilt further before it topples — why widening the base or lowering the C.G. improves stability.

Worked example

A toy of mass 0.5 kg has a base of half-width a=4 cma = 4\ \text{cm} and its C.G. is centred at height h=3 cmh = 3\ \text{cm} above the base. Find the critical tilt angle at which the toy topples, and check whether it stays upright when tilted by 30°.

Step 1. Identify the quantities. Here a=4 cma = 4\ \text{cm} and h=3 cmh = 3\ \text{cm}, so tan⁡θc=a/h=4/3=1.333\tan\theta_c = a/h = 4/3 = 1.333.

Step 2. Solve for the critical angle. θc=tan⁡−1(1.333)=53.1∘\theta_c = \tan^{-1}(1.333) = 53.1^\circ. Up to this tilt, the vertical line through the C.G. still falls within the toy's base, so the weight produces a restoring moment.

Step 3. Compare with the given tilt of 30°. Since 30∘<53.1∘30^\circ < 53.1^\circ, the vertical line through the C.G. is still well inside the base, the restoring moment is still acting, and the toy rotates back to its original position rather than toppling.

Answer: θc=53.1∘\theta_c = 53.1^\circ; at a 30° tilt the toy is still inside its stable range and returns upright — matching the low, wide-based design that keeps a toy in stable equilibrium even under a fairly large push.

Common traps

  • Assuming the C.G. must lie inside the material of the object — it can lie in empty space, as in a ring, a hollow bangle, or an L-shaped bracket.
  • Treating centre of gravity and geometric centre as always the same point — they coincide only for a uniform, regularly shaped body; a non-uniform body's C.G. shifts toward its heavier side.
  • Believing a taller object is automatically less stable than a shorter one — stability depends on the ratio of base half-width to C.G. height (a/ha/h), not on height alone; a tall object with a very wide base can be more stable than a short, narrow one.
  • Forgetting that the critical angle is measured from the pivot edge, not from the object's own centre — the formula tan⁡θc=a/h\tan\theta_c = a/h uses distances from that edge to the C.G.

Ask the tutor: want to work out the C.G. of an L-shaped sheet of metal, or see why Formula 1 cars are built almost flat to the ground — ask the AI tutor for a step-by-step walkthrough.

Why does the C.G. of a hollow ring lie at its centre, where there is no material at all?

The C.G. marks the point through which the resultant of all the tiny weight forces on the ring can be taken to act — it does not have to sit on any actual particle of the body. A ring is symmetric all the way around, so every small piece of mass on one side is balanced by an equal piece diametrically opposite. The resultant of all these balanced pairs works out to be exactly at the centre of the circle, even though that point sits in the empty hole at the middle of the ring.

Is the critical tilt angle formula, tan of theta c equals a over h, something ICSE expects us to derive in an exam?

The ICSE syllabus statement asks you to define centre of gravity, locate it for regular shapes, and connect the C.G. and the base to stable, unstable and neutral equilibrium in words — it does not name this angle formula. It is included here because it turns "wider base, lower C.G. is more stable" into a concrete, checkable number. If your exam only asks for the qualitative explanation, the reasoning in Step 3 above — the restoring moment vanishing once the C.G.'s vertical line reaches the pivot edge — is exactly the explanation to write.

How is finding the C.G. by suspending a lamina different from finding it with a knife-edge?

Both methods use the same principle — a body settles so that its C.G. lies directly below (or above) the support, because that is the only position where the weight creates zero moment about the support. Suspending a lamina from a point lets it hang freely, so it rotates until the C.G. is directly below the suspension point; repeating from a second point gives a second line, and the C.G. is where the two lines intersect. A knife-edge instead supports the body from below, so it balances flat only when the edge lies directly under the C.G. — useful for a rod or a flat sheet you want to keep horizontal rather than hanging.

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