Definition of the centre of mass

When a system is made up of several objects, there is a single point that summarises its overall position: the centre of mass (CM). It is not just any point, but a weighted average of the positions of the bodies, in which the weight of each is its mass. A heavier body “pulls” the centre of mass towards itself more than a lighter one.

Principle — Centre of mass

For a system of $n$ bodies, the centre of mass has position $\ev{\vec{s}_{\text{CM}} = \frac{m_1\,\vec{s}_1 + m_2\,\vec{s}_2 + \cdots + m_n\,\vec{s}_n}{m_1 + m_2 + \cdots + m_n} = \frac{\sum_i m_i\,\vec{s}_i}{M_{\text{tot}}}}$ It is a weighted average of the positions, with the masses as weights.

The calculation is vectorial: the formula is applied separately to each component ($x$, $y$, possibly $z$). The denominator is always the total mass of the system $M_{\text{tot}} = \sum_i m_i$.

Key formula

$\vec{s}_\text{CM} = \frac{\sum m_i\,\vec{s}_i}{M_\text{tot}}$ $\vec{v}_\text{CM} = \frac{\vec{p}_\text{tot}}{M_\text{tot}}$ $E_\text{pot,g} = M_\text{tot}\,g\,h_\text{CM}$

Example — Three masses in the plane

$m_1 = 4\;\text{kg}$ at $(3,1)$, $m_2 = 1\;\text{kg}$ at $(1,0)$, $m_3 = 2\;\text{kg}$ at $(2,-2)$. The total mass is $7\;\text{kg}$: $\vec{s}_{\text{CM}} = \frac{4(3,1) + 1(1,0) + 2(2,-2)}{7} = \frac{(17,\; 0)}{7} = \left(\tfrac{17}{7},\; 0\right) \approx (2{,}43;\; 0)$ The $y$ component vanishes because the upward contribution ($m_1$) and the downward one ($m_3$) balance each other.

The figure shows how larger masses pull the CM towards themselves: the gold point (CM) is closer to the $4\;\text{kg}$ mass than to the other two.

The centre of mass of three point masses is the weighted average of the positions. Larger masses “pull” the CM towards themselves.

Links

Topics: Centro di massa Concepts: Centro di massa Skills: Calcolo del centro di massa

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