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SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 97 small, we can treat the body as a continuous Let us consider a thin rod, whose width and distribution of mass. We subdivide the body into breath (in case the cross section of the rod is n small elements of mass; ∆m , ∆m ... ∆m ; the rectangular) or radius (in case the cross section i element ∆m is taken to be located about the of the rod is cylindrical) is much smaller than i its length. Taking the origin to be at the point (xi yi zi). The coordinates of the centre of mass are then approximately given by geometric centre of the rod and x-axis to be along the length of the rod, we can say that on X = ∑ ( m ix i,Y = ∑ ( m iy i,Z = ∑ ( m izi account of reflection symmetry, for every ∑ ∆m i ∑ ∆m i ∑ ∆ mi element dm of the rod at x, there is an element of the same mass dm located at –x (Fig. 6.8). As we make n bigger and bigger and each ∆m smaller and smaller, these expressions become exact. In that case, we denote the sums The net contribution of every such pair to over i by integrals. Thus, the integral and hence the integral itself is zero. From Eq. (6.6), the point for which the ∑ ∆m →idm ∫ M, integral itself is zero, is the centre of mass. Thus, the centre of mass of a homogenous thin ∑ (∆m ixi→ x∫dm, rod coincides with its geometric centre. This can (∆m )y → ydm, beunderstoodonthebasisofreflectionsymmetry. ∑ i i ∫ The same symmetry argument will apply to homogeneous rings, discs, spheres, or even and ∑ (∆m iz i z∫m thick rods of circular or rectangular cross Here M is the total mass of the body. The section. For all such bodies you will realise that coordinates of the centre of mass now are for every element dm at a point (x, y, z) one can 1 1 1 always take an element of the same mass at X= M ∫x dm,Y = M ∫ydm and Z =M ∫z dm (6.5a) the point (–x, –y, –z). (In other words, the origin is a point of reflection symmetry for these The vector expression equivalent to these three scalar expressions is bodies.) As a result, the integrals in Eq. (6.5 a) all are zero. This means that for all the above R = 1 rdm (6.5b) bodies, their centre of mass coincides with their M ∫ geometric centre. If we choose, the centre of mass as the origin of our coordinate system, Example 6.1 Find the centre of mass of R = 0 three particles at the vertices of an equilateral triangle. The masses of the i.e.,rdm = 0 particles are 100g, 150g, and 200g ∫ respectively. Each side of the equilateral or ∫x dm = ∫dm = z d∫ = 0 (6.6) triangle is 0.5m long. Often we have to calculate the centre of mass of Answer homogeneous bodies of regular shapes like rings, discs, spheres, rods etc. (By a homogeneous body we mean a body with uniformly distributed mass.) By using symmetry consideration, we can easily show that the centres of mass of these bodies lie at their geometric centres. Fig. 6.8 Determining the CM of a thin rod. Fig. 6.9 Reprint 2026-27