(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Prove the magnitude R of the position vector [tex]\vec{R}[/tex] for the center of mass from an arbitrary point of origin is given by the equation

M[tex]^{2}[/tex]R[tex]^{2}[/tex] = M[tex]\sum{m_{i}r^{2}_{i}[/tex] - [tex]\frac{1}{2}[/tex][tex]\sum{m_{i}m_{j}r_{ij}^{2}[/tex]

2. Relevant equations

[tex]\vec{R}[/tex] = [tex]\frac{1}{M}[/tex] [tex]\sum{m_{i}\vec{r}_{i}[/tex]

M=[tex]\sum{m_{i}[/tex]

[tex]\vec{r}_{ij}[/tex] = [tex]\vec{r}_i[/tex] - [tex]\vec{r}_j[/tex]

3. The attempt at a solution

OK, the simplest thing to start with is to simply dot [tex]\vec{R}[/tex] with itself to get the magnitude R[tex]^{2}[/tex] and then bring over M[tex]^{2}[/tex] to the LHS. However, I'm not sure how to handle the dotting/squaring with the indices.

Would it be

M[tex]^{2}[/tex]R[tex]^{2}[/tex] = [tex]\sum{m_{i}\vec{r}_{i}[/tex] [tex]\bullet[/tex] [tex]\sum{m_{i}\vec{r}_{i}[/tex]

OR

M[tex]^{2}[/tex]R[tex]^{2}[/tex] = [tex]\sum{m_{i}\vec{r}_{i}[/tex] [tex]\bullet[/tex] [tex]\sum{m_{j}\vec{r}_{j}[/tex] ?

Can I then simplify that by bringing terms out of the summation, or the dot product inside?

My recollection of the summation rules is a bit fuzzy. Actually, I'm not sure I've used the summation notation along with vector operations before.

I've tried writing the squared magnitudes in the desired answer as dot products, too, which allowed me to do a lot of rewriting but I can't connect the givens to the answer yet.

I.e.,

r[tex]^{2}_{i}[/tex] = [tex]\vec{r}_{i}[/tex] [tex]\bullet\vec{r}_{i}[/tex]

r[tex]^{2}_{ij}[/tex] = [tex]\vec{r}_{ij}[/tex] [tex]\bullet\vec{r}_{ij}[/tex] = [tex](\vec{r}_{i}-\vec{r}_{j})[/tex] [tex]\bullet (\vec{r}_{i}-\vec{r}_{j})[/tex] = [tex]\vec{r}_{i}[/tex] [tex]\bullet\vec{r}_{i}[/tex] - 2[tex]\vec{r}_{i}[/tex] [tex]\bullet\vec{r}_{j}[/tex] + [tex]\vec{r}_{j}[/tex] [tex]\bullet\vec{r}_{j}[/tex]

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# Homework Help: Center of mass prob. with sigma notation, vectors

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