System of Particles and Rotational MotionCentre of Mass
In villages and at pilgrimages people still carry loads on a kaanwar: a long pole across the shoulder with a bucket hanging from each end. Where must the pole rest on the shoulder so that it stays level? Sina has two buckets on a pole 120 cm long. Drag the point where the pole rests on her head, and balance all three pairs of buckets.
Take the pole as the x-axis, with the masses m1 and m2 at distances x1 and x2 from an origin O. The balance point is the centre of mass C of the two particles, at a distance X from O given by
X is a mass-weighted mean of x1 and x2: the heavier particle counts for more, so C lies nearer to it. For two equal masses, m1 = m2 = m, it is the ordinary mean, (x1 + x2)/2, exactly midway between them.
The same rule works for any number of particles on a line. For n particles of masses m1, m2, … mn at x1, x2, … xn,
The Greek capital sigma, Σ, means “add up over all the particles”. The denominator Σmi = M is the total mass of the system.
Particles in a Plane
Now let three particles lie in a plane, not on one line. Choose x- and y-axes in that plane, with the particles of masses m1, m2, m3 at (x1, y1), (x2, y2) and (x3, y3). Each coordinate of the centre of mass C is a mass-weighted mean on its own:
Three steel balls of 1 kg, 2 kg and 3 kg lie on a grid. Drag them so that their centre of mass lands on each of the three flags in turn.
The centre of mass C is always pulled hardest by the 3 kg ball. If the three masses were equal, m, the factor m would cancel and X = (x1 + x2 + x3)/3, Y = (y1 + y2 + y3)/3: for three equal masses the centre of mass is the centroid of their triangle.
For n particles spread out in space, not necessarily in one plane, the centre of mass is at (X, Y, Z), where
With position vectors ri = xiî + yiĵ + zik̂ for the particles and R = Xî + Yĵ + Zk̂ for the centre of mass, the three equations become one:
If we put the origin at the centre of mass, then R = 0, so Σmiri = 0 about the centre of mass. Remember this: it comes back when we find the centre of gravity.
From Particles to a Solid Body
A metre stick or a flywheel is a system of particles too, but with so many atoms, packed so closely, that adding them one by one is impossible. Instead we treat the body as a continuous distribution of mass. Cut it into n small pieces of masses Δm1, Δm2, … Δmn, the i-th piece at (xi, yi, zi). Then, approximately,
and the same for Y and Z. The thinner the slices, the better the answer. The bars show the mass of each slice.
Slice a uniform rod and a tapered rod (thick at one end) into more and more pieces.
Number of slices n =
As n grows and each Δm shrinks, the sums become integrals: ΣΔmi → ∫dm = M, and Σ(Δmi)xi → ∫x dm. So for a continuous body
and with the origin at the centre of mass,
For a uniform body we rarely need to integrate at all; symmetry does the work. Take a thin uniform rod with the origin at its geometric centre and the x-axis along it (Fig. 6.8). For every small piece dm at a distance x on one side there is an identical piece dm at −x on the other. Each pair adds x dm + (−x) dm =
The same reflection argument works for uniform rings, discs, spheres and thick rods: for every piece at (x, y, z) there is an equal piece at (−x, −y, −z). So for all these bodies the centre of mass is at the geometric centre. Notice that for a ring this is a point where there is no material at all.
Example 6.1
Example 6.1 In the school lab Riya places three small steel balls of mass 100 g, 150 g and 200 g at the corners O, A and B of an equilateral triangle of side 0.5 m chalked on the bench. Find their centre of mass.
Answer Take O as the origin and OA along the x-axis. Then O = (0, 0), A = (0.5, 0) and B = (0.25, 0.25√3), all in metres.
so X = 5/18 m =
so Y = 1/(3√3) m =
Fig. 6.9 — the three balls, their centre of mass C and the centroid G of the triangle.
C is not the geometric centre G of the triangle. The masses are unequal, so the heaviest ball, at B, pulls C towards itself; C coincides with G only when all three masses are equal.
Example 6.2
Example 6.2 Find the centre of mass of a uniform triangular plate (a lamina) LMN.
Answer Cut the triangle into thin strips parallel to one side. Each strip is a thin uniform rod, so its centre of mass is at its midpoint. Slice the triangle three ways (Fig. 6.10).
The midpoints of all the strips parallel to MN lie on the median LP, so the centre of mass of the whole triangle lies on LP. By the same argument it lies on MQ and on NR. It must therefore be the point where the three medians meet: the centroid G of the triangle.
Example 6.3
Example 6.3 A uniform L-shaped lamina, of mass 3 kg, is made of three squares of side 1 m. Find its centre of mass.
Answer Take the axes along the outer edges of the L, with O at the corner. The lamina is uniform, so each square has a mass of 1 kg, and the centre of mass of each square is at its own centre: C1 = (½, ½), C2 = (³⁄₂, ½) and C3 = (½, ³⁄₂). Replace each square by a 1 kg particle at its centre:
so X = Y = 5/6 m =
The centre of mass lies on the line OD, the diagonal through the inner corner D. We could have guessed that without any working: the L is symmetric about OD. And if the three squares had different masses? Replace each by a particle of its own mass at its centre and use the same formulas. Tap the squares to make them heavier.
The line OD is a line of symmetry only while the two mirror-image squares, C2 and C3, have equal masses. That is when the centre of mass C stays on it.
Speed Round: Where Is the Centre of Mass?
Mass-weighted means on a line, and the symmetric bodies. Get five right to move on.