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Homework Help: Steps btw 2 equ. goes wrong

  1. Oct 31, 2004 #1


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    My classical mechanics textbook (Symon) gives two equations. So I try to do the calculations that lead from one to the other but there's something that doesn't work. Tell me what I'm doing wrong please.

    We start from

    [tex]\sum_{k=1}^N (\mathbf{r}_k-\mathbf{r}_Q)\times \mathbf{F}^i_k - m_k(\mathbf{r}_k-\mathbf{r}_Q)\times \mathbf{\ddot{r}}_k[/tex]

    and want to get to

    [tex]\sum_{k=1}^N (\mathbf{r}_k-\mathbf{r}_Q)\times \mathbf{F}^i_k - M(\mathbf{R}-\mathbf{r}_Q)\times \mathbf{\ddot{r}}_k[/tex]


    [tex]M\mathbf{R} = \sum_{k=1}^N m_k \mathbf{r}_k[/tex]

    I'm not sure if the [itex]-M(\mathbf{R}-\mathbf{r}_Q)\times \mathbf{\ddot{r}}_k[/itex] term is included in the sum or not but in either case, it doesn't make sense to me.

    Here are the steps I made:

    [tex]=\sum_{k=1}^N (\mathbf{r}_k-\mathbf{r}_Q)\times \mathbf{F}^i_k -\sum_{k=1}^N m_k(\mathbf{r}_k-\mathbf{r}_Q)\times \mathbf{\ddot{r}}_k[/tex]

    Now let's work on the second sum; we should get [itex]M(\mathbf{R}-\mathbf{r}_Q)\times \mathbf{\ddot{r}}_k[/tex] (either with or without the sum sign before)

    [tex]\sum_{k=1}^N m_k(\mathbf{r}_k-\mathbf{r}_Q)\times \mathbf{\ddot{r}}_k
    = \sum_{k=1}^N m_k\mathbf{r}_k \times \mathbf{\ddot{r}}_k -\mathbf{r}_Q \times \mathbf{\ddot{r}}_k
    = \sum_{k=1}^N m_k\mathbf{r}_k \times \mathbf{\ddot{r}}_k -\sum_{k=1}^N \mathbf{r}_Q \times \mathbf{\ddot{r}}_k
    =\left[ \sum_{k=1}^N m_k\mathbf{r}_k - \sum_{k=1}^N \mathbf{r}_Q\right] \times \mathbf{\ddot{r}}_k[/tex]
    [tex] = M\left[ \mathbf{R} - \sum_{k=1}^N \mathbf{r}_Q \right] \times \mathbf{\ddot{r}}_k[/tex]

    And this is when it doesn't make sense because [itex]\sum_{k=1}^N \mathbf{r}_Q = N \mathbf{r}_Q[/itex] and there is no N in the final equation from the book. This is if the second term is NOT included in the sum. If it IS, then it doesn't make more sense because it would mean the second term is

    [tex]\sum_{k=1}^N M( \mathbf{R} - \mathbf{r}_Q ) \times \mathbf{\ddot{r}}_k
    = M\left[ \sum_{k=1}^N\mathbf{R} - \sum_{k=1}^N \mathbf{r}_Q \right] \times \mathbf{\ddot{r}}_k[/tex]

    which isn't what we have either!
  2. jcsd
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