Concept of Centre of Gravity

Every physical object is made up of countless tiny mass elements, each experiencing a gravitational pull toward the Earth. Rather than tracking the force on every individual particle, physicists use a single special point called the centre of gravity (CG) — the unique location where the entire weight of the object can be considered to act as a single resultant force. This concept dramatically simplifies the analysis of equilibrium, rotation, and stability in rigid bodies. The centre of gravity is closely related to, but subtly distinct from, the centre of mass (CM). The centre of mass is the mass-weighted average position of all particles in a body, whereas the centre of gravity is the weight-weighted average position. In a uniform gravitational field — which is an excellent approximation for objects small compared to Earth's radius — the two points coincide exactly. For very large objects (such as a skyscraper or a mountain) where gravitational field strength $g$ varies slightly with height, the two points can differ by a measurable amount. Understanding the centre of gravity is essential for engineers designing stable structures, gymnasts and dancers controlling balance, and aerospace engineers ensuring rockets fly without tumbling. A body is in rotational equilibrium when its centre of gravity lies directly above (or below) its base of support, a principle that governs everything from stacking blocks to the posture of the human spine.

Key takeaways

Weight of Individual Particles
Total Weight and Torque
Locating the Centre of Gravity
Uniform Gravitational Field: CG = CM
Experimental Determination by Suspension

Weight of Individual Particles

Divide the rigid body into $n$ small particles with masses $m_1, m_2, \ldots, m_n$ located at positions $x_1, x_2, \ldots, x_n$ (measured along one axis for simplicity). Each particle experiences a gravitational force $w_i = m_i g_i$ , where $g_i$ is the local gravitational field strength at the position of particle $i$ .

Total Weight and Torque

The total weight of the body is $W = \sum_{i=1}^{n} w_i = \sum_{i=1}^{n} m_i g_i$ . The total torque about the origin due to all these weight forces is $\tau_{\text{total}} = \sum_{i=1}^{n} w_i x_i = \sum_{i=1}^{n} m_i g_i x_i$ .

Locating the Centre of Gravity

The centre of gravity $x_{\text{CG}}$ is the single point where the total weight $W$ must act to produce the same torque as all individual weights combined. Setting $W \cdot x_{\text{CG}} = \sum_{i=1}^{n} m_i g_i x_i$ and solving: x_{\text{CG}} = \frac{\sum_{i=1}^{n} m_i g_i x_i}{\sum_{i=1}^{n} m_i g_i}. In three dimensions this generalises to vector form: $\vec{r}_{\text{CG}} = \frac{\sum_{i} m_i g_i \vec{r}_i}{\sum_{i} m_i g_i}$ .

Uniform Gravitational Field: CG = CM

When $g$ is constant throughout the body (i.e., $g_i = g$ for all $i$ ), the factor $g$ cancels from numerator and denominator: x_{\text{CG}} = \frac{g\sum_{i} m_i x_i}{g\sum_{i} m_i} = \frac{\sum_{i} m_i x_i}{M} = x_{\text{CM}}. This is why, for everyday objects, we use centre of mass and centre of gravity interchangeably.

Experimental Determination by Suspension

For an irregularly shaped lamina (flat plate), the CG can be found experimentally. Suspend the object freely from one point — it will hang so that the CG lies directly below the pivot (since this is the only position of rotational equilibrium). Draw a vertical plumb line through the pivot point onto the object. Repeat from a second pivot point. The intersection of the two plumb lines is the centre of gravity.

The Key Equation

$$\vec{r}_{\text{CG}} = \frac{\sum_{i=1}^{n} m_i g_i \,\vec{r}_i}{\sum_{i=1}^{n} m_i g_i}

- **\vec{r}_{\text{CG}}** — Position of centre of gravity (m): The vector position of the single point where net gravitational torque acts. - **m_i** — Mass of particle i (kg): The mass of the i-th small element of the rigid body. - **g_i** — Local gravitational field strength at particle i (N kg⁻¹): The magnitude of gravitational acceleration at the location of particle i; varies with height above Earth. - **\vec{r}_i** — Position vector of particle i (m): The displacement of the i-th particle from the chosen reference origin. - **\sum_{i=1}^{n} m_i g_i** — Total weight W (N): The scalar sum of the gravitational forces on all particles, equal to the total weight of the body. ## Worked Example **Problem.** A non-uniform horizontal rod of total length 4.0 m lies along the x-axis. It consists of two sections: a light left half (mass $m_1 = 2.0$ kg, centre at $x_1 = 0.5$ m from the left end) and a heavier right half (mass $m_2 = 6.0$ kg, centre at $x_2 = 3.0$ m from the left end). Assuming uniform gravity, find the position of the centre of gravity measured from the left end.

x_{\text{CG}} = 2.375 \text{ m}

**Answer.** The centre of gravity is 2.375 m from the left end of the rod.

Figure

Diagram

Origins · 1 minArchimedes of Syracuse and how this was discovered

Archimedes of Syracuse

-287–-212 · Greek (Sicilian)

Archimedes is credited with the first rigorous mathematical treatment of the centre of gravity. In his works 'On the Equilibrium of Planes' and 'The Method', he established that the centre of gravity of a triangle lies at the intersection of its medians, and proved results about centres of gravity of parabolic segments. His lever law — that a system is in equilibrium when the torques about the fulcrum balance — is the direct physical foundation from which the concept of centre of gravity is derived. Archimedes allegedly claimed 'Give me a lever long enough and a fulcrum on which to place it, and I shall move the world.' His confidence was rooted in his own deep understanding of torques and the centre of gravity — if the Earth's CG could be positioned correctly relative to a pivot, an arbitrarily small force applied far enough from the fulcrum could indeed move it.

Centre of Mass — The centre of mass is mathematically identical to the centre of gravity in a uniform gravitational field. Understanding the distinction and relationship between these two concepts is fundamental to rigid body dynamics.
Torque and Rotational Equilibrium — The centre of gravity is the point about which the net gravitational torque on a body is zero. The condition for rotational equilibrium — that net torque equals zero — directly determines when and where the CG must lie relative to the support.
Stability and Toppling Conditions — Whether a body topples or remains stable depends entirely on whether the vertical line through the CG falls within the base of support. A lower and more central CG leads to greater stability.
Newton's Laws of Motion — Newton's second law for a system of particles states that the net external force equals total mass times the acceleration of the centre of mass. Gravity acting through the CG is the key external force in most rigid body problems.