Physics · Ch 4 — Laws of Motion

Mathematical Understanding of Centre of Mass

4.13.1

Mathematical Understanding of Centre of Mass

For a system of n discrete particles of masses $m_1, m_2, \ldots, m_n$ (total mass $M=\sum_{i=1}^n m_i$ ) with position vectors $\vec{r_1}, \vec{r_2}, \ldots, \vec{r_n}$ from some chosen origin O, the position vector of the CENTRE OF MASS from the same origin is the mass-weighted average, $\vec{r}=\frac{\sum_{i=1}^n m_i\vec{r_i}}{M}$ . If the origin is itself chosen to BE the centre of mass, then $\sum m_i\vec{r_i}=0$ -- i.e. the centre of mass is precisely the point about which the sum of the 'moments of mass' ( $m_i\vec{r_i}$ , directly analogous to moment of force) of all the particles is zero. In Cartesian component form, $x=\frac{\sum m_ix_i}{M}$ , and similarly for y and z.

For a CONTINUOUS mass distribution, the sums become integrals: $\vec{r}=\frac{\int \vec{r}\,dm}{M}$ , with $x=\frac{\int x\,dm}{M}$ , $y=\frac{\int y\,dm}{M}$ , $z=\frac{\int z\,dm}{M}$ , where $M=\int dm$ is the total mass. Working out these integrals for common uniform, symmetric shapes gives standard results, tabulated in Table 4.1: for two point masses, the c.m. divides the joining line in inverse proportion to the masses (closer to the heavier one); for any uniform, geometrically symmetric object, the c.m. is at the geometrical centre; and specific formulas exist for shapes such as an isosceles triangular plate ( $y_c=H/3$ from the base), a right-angled triangular plate, thin semicircular rings and discs, hemispherical shells, solid hemispheres, and hollow/solid right circular cones.

Worked illustration (letter 'E' shape, Example 4.12): a letter E made of ten identical unit squares (mass m each, concentrated at each square's centre) has its centre of mass located, using the symmetry of the shape to combine groups of squares into effective point masses, at 0.7 cm from the spine, closer to the side with the heavier effective mass (the two outer arms) -- illustrating the general two-step technique of first combining sub-groups by symmetry, then combining the resulting effective masses.

Worked illustration (three hollow spheres at a triangle's vertices, Example 4.13): thin hollow spheres of radii 1, 2 and 3 cm (masses proportional to $r^2$ , so in ratio 1:4:9) sit at the vertices of an equilateral triangle of side 10 cm; placing the origin at the heaviest mass simplifies the arithmetic, and applying $x_c=\frac{\sum m_ix_i}{M}$ , $y_c=\frac{\sum m_iy_i}{M}$ locates the combined centre of mass at approximately (45/14, $10\sqrt3/28$ ) cm from that vertex. …

Figure 4.11Fig 4.11: Centre of mass for n particles

What this figure shows. A diagram showing a fixed origin O and n discrete point masses m1, m2, ..., mn scattered at arbitrary positions around it (drawn as small dots or circles at various locations), each connected to the origin O by its own position-vector arrow (r1, r2, ..., rn, drawn as straight arrows from O to each mass). An additional, distinctly marked point representing the system's overall centre of mass is shown somewhere among the scattered masses (a location that is the mass-weighted average of all the individual position vectors), with its own position vector r drawn from the same origin O to this centre-of-mass point, i.e. …

Table 4.1Table 4.1: Centre of mass coordinates for uniform symmetric objects

Uniform Symmetric Object | Coordinates of centre of mass (c.m.)

System of two point masses | c.m. divides the joining distance in inverse proportion of the masses (closer to the heavier mass)

Any geometrically symmetric object of uniform density | Centre of mass at the geometrical centre of the object

Isosceles triangular plate (height H from base) | xc = 0, yc = H/3 (measured from the base, along the axis of symmetry)

Right-angled triangular plate (legs along axes, right angle at origin, other vertex at (p,q)) | xc = p/3, yc = q/3

Thin semicircular ring of radius R | xc = 0, yc = 2R/pi

Thin semicircular disc of radius R | xc = 0, yc = 4R/(3 pi)

Hemispherical shell of radius R | xc = 0, yc = R/2

Solid hemisphere of radius R | xc = 0, yc = 3R/8 …

Misc Ex.12Centre of mass of a letter 'E' cut from uniform cardboard

Worked out. A letter 'E' shape is built from ten identical small squares, each of mass m concentrated at its own centre (points labelled 1 to 10); using symmetry, the three squares forming the top arm (1,2,3) combine to an effective mass 3m at point 2's coordinates (1,2), and the bottom arm (8,9,10) similarly combines to 3m at (1,-2); these two combine (by symmetry, ignoring y) to 7m (including the middle-arm square 6) at x=1, while the vertical spine squares (4,5,7) combine by symmetry to 3m at the origin x=0; taking the weighted average of these two effective masses (3m at x=0 and 7m at x=1) gives the c.m. at x = 0.7 cm from the …

Misc Ex.13Centre of mass of three hollow spheres at the vertices of a triangle

Worked out. Three thin-walled uniform hollow spheres of radii 1 cm, 2 cm and 3 cm have their centres located at the three vertices A, B, C of an equilateral triangle of side 10 cm; since the mass of a thin hollow sphere is proportional to its surface area (proportional to radius squared), the masses are in ratio 1:4:9 (mA=m, mB=4m, mC=9m); choosing the origin at C (the heaviest mass) and B on the positive x-axis, the weighted-average formulas give the system's centre of mass at approximately xc = 45/14 cm and yc = (10 sqrt3 …

Misc Ex.14Centre of mass of a disc with a hole cut out of it

Worked out. A disc of radius 2r has a circular hole of radius r cut from it, with the hole's centre a distance r from the original disc's centre; Method I treats the original full disc (mass 4m, c.m. at the centre O) as the combination of the (already removed) small disc (mass m, c.m. at distance r from O) plus the remaining annular-like piece (mass 3m), and solves the weighted-average equation for the remaining piece's centre of mass, getting it at a distance r/3 from O on the side opposite to the hole; Method II re-derives the identical result by treating the missing disc as a 'negative …

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