Centre of Mass

For an extended body or a collection of particles, the centre of mass (CM) is a single point that behaves as if all the mass were concentrated there and all external forces applied there. This simplifies analysis enormously — a wobbling wrench thrown across a room has a CM that traces a clean parabola.

Concept

For a system of discrete particles of masses $m_i$ at positions $\vec{r}_i$, the position of the CM is

$\vec{R}_{\text{CM}} = \frac{\sum_i m_i \vec{r}_i}{\sum_i m_i} = \frac{1}{M}\sum_i m_i \vec{r}_i$

where $M = \sum m_i$.

For a continuous body with mass density $\rho(\vec{r})$,

$\vec{R}_{\text{CM}} = \frac{1}{M}\int \vec{r}\,dm = \frac{1}{M}\int \vec{r}\,\rho(\vec{r})\,dV$

Key properties:

• CM depends on mass distribution, not on motion.
• For symmetric, uniform bodies, CM lies at the geometric centre.
• CM may lie outside the body (e.g., a ring, a horseshoe).

CM of standard uniform bodies:

• Uniform rod: midpoint.
• Triangular lamina: centroid (1/3 of the way from base to vertex).
• Uniform disc / ring: geometric centre.
• Hollow / solid sphere: geometric centre.
• Hollow / solid hemisphere: along the symmetry axis at $3R/8$ (solid) or $R/2$ (hollow) from the flat face.
• Uniform cone (solid): along axis at $h/4$ from base.

Derivation

CM of a system of two particles. Particles of masses $m_1$ and $m_2$ at positions $x_1$ and $x_2$ on the $x$-axis:

$X_{\text{CM}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$

If we take $x_1 = 0$ and $x_2 = d$, then $X_{\text{CM}} = \frac{m_2 d}{m_1+m_2}$ — closer to the heavier mass.

CM of a uniform rod of length $L$. Place the rod along the $x$-axis from 0 to $L$, with linear mass density $\lambda = M/L$:

$X_{\text{CM}} = \frac{1}{M}\int_0^L x\,\lambda\,dx = \frac{\lambda}{M}\cdot\frac{L^2}{2} = \frac{L}{2}$

The CM is at the midpoint, as expected.

CM of a solid hemisphere of radius $R$ (along symmetry axis, measuring from flat face):

$z_{\text{CM}} = \frac{1}{M}\int z\,dm$

Using thin disc elements of radius $\sqrt{R^2 - z^2}$ and density $\rho = M/(\tfrac{2}{3}\pi R^3)$, one gets $z_{\text{CM}} = 3R/8$.

Worked Example

Three point masses are placed at the corners of an equilateral triangle of side $a$: $m$, $2m$, $3m$. Find the position of the CM.

Solution:

Place the masses: $m$ at $(0,0)$, $2m$ at $(a, 0)$, $3m$ at $(a/2, a\sqrt{3}/2)$.

Total mass $M = 6m$.

$X_{\text{CM}} = \frac{m(0) + 2m(a) + 3m(a/2)}{6m} = \frac{2a + 1.5a}{6} = \frac{3.5a}{6} = \frac{7a}{12}$

$Y_{\text{CM}} = \frac{0 + 0 + 3m \cdot a\sqrt{3}/2}{6m} = \frac{3 a\sqrt{3}/2}{6} = \frac{a\sqrt{3}}{4}$

So $\vec{R}_{\text{CM}} = (7a/12, \, a\sqrt{3}/4)$, closer to the heaviest mass at the top.

Common Confusions

• CM is not always inside the body. A ring's CM is at its centre, where no mass exists.
• CM is not always at the geometric centre. Only when the body is uniform and symmetric. A barbell with heavier weights on one side has its CM shifted.
• CM is a property of the mass distribution, not of forces or motion.
• Position vector of CM depends on the chosen origin — but the point itself doesn't.

Key Takeaways

• $\vec{R}_{\text{CM}} = \frac{1}{M}\sum m_i \vec{r}_i$ (discrete) or $\frac{1}{M}\int\vec{r}\,dm$ (continuous).
• For symmetric uniform bodies, CM is at the geometric centre.
• CM of two particles is on the line joining them, closer to the heavier.
• CM may lie outside the body (ring, hemisphere).
• The CM is the "balance point" — gravity acting at CM gives the same net torque as gravity distributed.