201902200820 Derivation 1.2

Prove that the magnitude $R$ of the position vector for the center of mass from an arbitrary origin is given by the equation

$M^2R^2=M \displaystyle{\sum_i}m_ir_i^2-\displaystyle{\frac{1}{2}\sum_{i\neq j}m_im_jr_{ij}^2}$ .

Solution.

This solution is not mine. It was found on the Internet some years ago, to whose author(s) I lost references.

$\displaystyle{M\mathbf{R}}=\sum m_i \mathbf{r}_i$ .

Taking squares on both sides,

$\displaystyle{M^2\mathbf{R}^2}=\sum_{i,j}m_im_j\mathbf{r}_i\cdot \mathbf{r}_j$ .

Notice $\mathbf{r}_{ij}=\mathbf{r}_i-\mathbf{r}_j$ , squaring it,

$r_{ij}^2=r_i^2-2\mathbf{r}_i\cdot \mathbf{r_j}+r_j^2$ ,

arranging,

$\displaystyle{\mathbf{r}_i}\cdot \mathbf{r}_j=\frac{1}{2}(r_i^2+r_j^2-r_{ij}^2)$ .

Plugging this,

$\begin{aligned} \displaystyle{M^2R^2} & =\frac{1}{2}\sum_{i,j}m_im_jr_i^2+\frac{1}{2}\sum_{i,j}m_im_jr_j^2-\frac{1}{2}\sum_{i,j}m_im_jr_{ij}^2\\ & = \displaystyle{\frac{1}{2}M\sum_im_ir_i^2}+\frac{1}{2}M\sum_jm_jr_j^2-\frac{1}{2}\sum_{i,j}m_im_jr_{ij}^2\\ & = \displaystyle{M\sum_im_ir_i^2}-\frac{1}{2}\sum_{i\neq j}m_im_jr_{ij}^2 \\ \end{aligned}$

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