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We have a system of $n$ particles, and the particle $i$ has a point mass $m_i$. The center of mass is then given by:

$X = \frac{\sum_i^nm_ix_i}{\sum_i^nm_i}$ $Y = \frac{\sum_i^nm_iy_i}{\sum_i^nm_i}$ $Z = \frac{\sum_i^nm_iz_i}{\sum_i^nm_i}$

Now imagine that we want to add a specific amount of extra mass $M_a$, to the existing particle system, in order to guide the center of gravity to a new position, X', Y', Z'. Taking into account that the number of particles can be a large number (e.g 1M), is there an algorithm (iterative or not) that I can use to find how to distribute the mass to the existing particles? The whole amount of $M_a$ must be used, and the initial mass $m_i$ of the $i$-th particle cannot be changed.

I'm thinking of starting with each particle having an equal amount of extra mass, e.g $M_a / n$, and then, through some iterative process (which I still have to figure out), guide the center of mass to the desired position.

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    Sounds like @joriki's answer takes care o$f$ it then.2012-03-07

3 Answers 3

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For the purposes of determining the centre of mass, you can treat the entire mass $M=\sum_im_i$ of the existing particles as concentrated in the centre of mass $\vec R=(X,Y,Z)$. Now you want to add mass $M_a$ to move the centre of mass to \vec R'=(X',Y',Z'). This requires the additional mass $M_a$ to be added such that its centre of mass is at $\vec R_a$ with

\frac{M\vec R+M_a\vec R_a}{M+M_a}=\vec R'\;,

that is,

\vec R_a=\frac{(M+M_a)\vec R'-M\vec R}{M_a}=\vec R'+\frac M{M_a}\left(\vec R'-\vec R\right)\;.

This point lies beyond \vec R' on the line from $\vec R$ to \vec R'. You can distribute the additional mass such that its centre of mass is at $\vec R_a$ if and only if this point lies in the convex hull of the positions $\vec r_i$ of the point masses. If so, then by definition $\vec R_a$ can be expressed as a convex combination $\vec R_a=\sum_i\mu_i\vec r_i$ with $\mu_i\ge0$ and $\sum_i\mu_i=1$, and you get the desired mass distribution by placing mass $\mu_iM_a$ at $\vec r_i$.

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    @John: You're welcome.2012-03-07
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Intuitively, there can be no guaranteed solution to the problem, given that there are no bounds to where the new X'Y'Z' is placed. Even a trivial example could show this, if you are constrained by only manipulating the mass of the existing particles and by the total amount of mass available to add.

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Let $M_i = {\sum_i^nm_i}.$ Place $M_a$ at position ($X^a$, $Y^a$, $Z^a$) which is collinear with points ($X$, $Y$, $Z$) and ($X'$, Y', $Z'$) and also satisfies the equation $M_i*d_i = M_a*d_a.$ Here $d_i$ is the distance between points ($X$, $Y$, $Z$) & ($X'$, Y', $Z'$). Similarly $d_a$ is the distance between ($X'$, Y', $Z'$) & ($X^a$, $Y^a$, $Z^a$).

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    The mass must be distributed to the particle system and not added to a single point, which may not be part of the system.2012-03-05