Physics · System Of Particles And Rotational Motion · NEET
No. It is a mass-weighted average, not a simple average. Plain average would be (x₁ + x₂ + ... + xₙ)/n, which ignores mass. The correct formula is X = Σmᵢxᵢ / Σmᵢ, so each position is multiplied by its own mass first. Only when all masses are equal does the mass cancel and the CoM become the simple average (the centroid).
Because heavier mass gets a bigger weight in the average. In a two-particle case X = (m₁x₁ + m₂x₂)/(m₁+m₂). If m₂ is larger, the term m₂x₂ dominates and pulls X towards x₂. Physically, to balance the system on a pivot the pivot must be nearer the heavy end, exactly like a see-saw where the lighter child sits farther out.
Use one formula per axis. X = Σmᵢxᵢ / M and Y = Σmᵢyᵢ / M (and Z if in 3D). Treat x and y completely separately: add up mass times x-coordinate for X, then mass times y-coordinate for Y. For 3 equal masses the CoM is the centroid of the triangle, but with unequal masses you must do the weighted sums.
No. It can lie in empty space where no particle exists. For example, the CoM of a ring or a horseshoe is outside the material. For a system of separate particles the CoM is a geometric point in space that need not coincide with any particle. It is just the balance point of the mass distribution.
It is a compact way to write all coordinate equations at once. The position vector rᵢ carries the x, y and z coordinates of particle i together, so R = Σmᵢrᵢ / M automatically gives X, Y and Z in one line. For NEET numericals you usually break it back into components, but the vector form is handy when momentum (Σmᵢvᵢ = MV) is involved.
Two particles of mass 5 kg and 10 kg are attached to the ends of a rigid massless rod of length 1 m. The distance of the centre of mass from the 5 kg particle is nearly:
Two objects of mass 10 kg and 20 kg are connected to the ends of a rigid massless rod of length 10 m. The distance of the centre of mass from the 10 kg mass is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
For positions along one axis, X = (m₁x₁ + m₂x₂ + ... + mₙxₙ) / M, where M = m₁ + m₂ + ... + mₙ is the total mass. In 2D and 3D you also compute Y = Σmᵢyᵢ / M and Z = Σmᵢzᵢ / M. In vector form R = Σmᵢrᵢ / M.
When all masses are equal, the mass cancels out and the CoM becomes the simple average of the positions. For two equal masses it is the midpoint; for three equal masses it is the centroid of the triangle they form.
Yes. The CoM is a geometric balance point, not a physical particle, so it can lie in empty space. Classic examples are a ring, a horseshoe, or an L-shaped plate whose CoM falls in the gap where there is no material.
The whole system moves as if all its mass were at the CoM and all external forces acted there. This lets you treat a complicated system as a single point, which is the basis for momentum conservation, projectile-explosion problems, and rotational motion questions that appear every year.
The centroid is a purely geometric average of positions. The centre of mass is a mass-weighted average. They coincide only when the mass is uniformly distributed (equal masses or a body of uniform density). With unequal masses, the CoM shifts towards the heavier region while the centroid stays fixed.
