Center of Mass & Center of Gravity

Pick any axis. A small piece of mass mᵢ at position rᵢ from it contributes a gravitational torque rᵢ × mᵢg, and the body's total is the sum over every piece. If g is the same vector at every piece, it factors out: Σ rᵢ × mᵢg = (Σ mᵢrᵢ) × g = r_cm × Mg. The distributed weight exerts exactly the torque of one force Mg applied at the center of mass, about every axis at once. Test it on a 3.0 m uniform beam of 12 kg carrying a 6.0 kg motor 2.5 m from its left end. About the left end the pieces give 12 × 9.81 × 1.5 + 6.0 × 9.81 × 2.5 = 323.7 N m. The center of mass is at (12 × 1.5 + 6.0 × 2.5)/18 = 1.833 m, and 176.6 N × 1.833 m gives the same 323.7 N m; about the right end both routes give 206.0 N m. Put the axis through the CM itself and the gravitational torque is zero, which is why a body supported at that one point balances in any orientation.

Real objects are rarely one simple shape, but they are usually a few. Choose an origin, split the body into pieces whose centers you know by symmetry, list each piece's mass and center, and take the mass-weighted average. A hole is a piece of negative mass: the object is the solid shape minus the material that is not there. Take a steel plate 0.80 m × 0.40 m that had a mass of 16 kg before a 0.20 m × 0.20 m square was cut out of it, centered 0.20 m from its left edge on the long midline. The areal density is 16 kg ÷ 0.32 m² = 50 kg m⁻², so the cut removed 2.0 kg. Then x_cm = (16 × 0.40 − 2.0 × 0.20)/(16 − 2.0) = 0.429 m from the left edge, 2.9 cm to the right of the plate's midpoint and away from the hole, while y_cm stays on the midline because the hole keeps that symmetry. The denominator is the 14 kg that remains, and the removed mass sits at the center of the removed square.

Before integrating, look for symmetry. The center of mass lies on every plane or axis about which the mass, not just the outline, is symmetric, so a uniform rod, disc or sphere needs no calculation. When density varies along one direction, replace the sum by an integral: slice the body into elements dm = λ(x) dx, so x_cm = ∫ x λ(x) dx / ∫ λ(x) dx. A beam of length L whose mass per unit length grows linearly, λ = λ₀(1 + x/L), has ∫ λ dx = 3λ₀L/2 and ∫ x λ dx = 5λ₀L²/6, so x_cm = 5L/9. For L = 2.7 m that is 1.50 m from the light end, 0.15 m beyond the midpoint, and λ₀ has cancelled: the balance point depends on how the mass is distributed, not on how much there is. A single support at 1.50 m carries this beam level; moved to the midpoint, it leaves an unbalanced torque that drops the heavy end.

A body hung from a frictionless pivot comes to rest with its center of gravity directly below the pivot. Off that vertical, Mg has a lever arm about the pivot and the body swings. The gravitational torque vanishes at one more place, with the CG directly above the pivot, but there any nudge lowers the CG and the body keeps turning; only the lowest position is a resting one. Hang a flat card from one point and draw the vertical through the pivot with a plumb line; hang it from a second point and draw again. The CG is where the lines cross, and a third suspension is the check, because its line must pass through the same point. The method finds the CG even where there is no material: for a ring, a horseshoe or a boomerang the lines cross in empty space. Precision depends on geometry. Two lines of width w crossing at angle α overlap in a rhombus with sides w/sin α, so lines meeting at 30° leave an uncertainty patch twice as long as lines meeting at 90°. Choose suspension points far apart.

The factoring step assumed one g for every piece. Drop that and the torque-weighted point becomes r_cg = Σ mᵢgᵢrᵢ / Σ mᵢgᵢ, pulled toward the pieces that are pulled harder. Near Earth g ∝ 1/r², so it falls by a fraction of about 2h/R over a height h: over a 100 m tower that is 2 × 100 m ÷ (6.37 × 10⁶ m) = 3.1 × 10⁻⁵, or 0.003%. For a uniform 100 m column the g-weighted point sits H²/(6R) = 0.26 mm below the midpoint, far below anything a structural engineer measures, so every problem in this course sets CG = CM and should say so. The distinction becomes real only when a body spans a noticeable fraction of the distance to the center of the attracting mass. Then the weighting depends on how the body is oriented, no fixed point carries the weight for every orientation, and gravity exerts a torque about the center of mass itself: the gravity-gradient torque that keeps a long orbiting tether pointing at Earth.

A body resting on a base is in equilibrium while the vertical through its CG meets the floor inside the base of support, the region enclosed by its contact points. Tilt it about one edge and the CG rises and swings toward that edge; once it passes directly above the edge, the torque of Mg reverses and the body topples. That happens at tan θ_c = d/h. Read as energy, U = Mg y_cm: equilibrium is stable where any small tilt raises the CG, unstable where it lowers it, and neutral where it stays level, as for a ball on a flat floor. An 80 kg wardrobe 0.70 m wide with its CG 0.90 m up tips at θ_c = tan⁻¹(0.35/0.90) = 21.3°, and toppling it from upright takes Mg(√(0.35² + 0.90²) − 0.90) = 51.5 J. Put 40 kg in its bottom drawer, 0.20 m up, and h falls to 0.667 m: θ_c rises to 27.7° and the barrier almost doubles, to 101.6 J. A load that slides as the wardrobe tilts is another matter: it carries the CG toward the edge, and the calculation must follow it.