Why Do My Classical Mechanics Equations Not Align?

[quasar987]

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

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

and want to get to

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

given

$$M\mathbf{R} = \sum_{k=1}^N m_k \mathbf{r}_k$$

I'm not sure if the $-M(\mathbf{R}-\mathbf{r}_Q)\times \mathbf{\ddot{r}}_k$ 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:

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

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)<br /> <br /> $$\sum_{k=1}^N m_k(\mathbf{r}_k-\mathbf{r}_Q)\times \mathbf{\ddot{r}}_k <br /> = \sum_{k=1}^N m_k\mathbf{r}_k \times \mathbf{\ddot{r}}_k -\mathbf{r}_Q \times \mathbf{\ddot{r}}_k <br /> = \sum_{k=1}^N m_k\mathbf{r}_k \times \mathbf{\ddot{r}}_k -\sum_{k=1}^N \mathbf{r}_Q \times \mathbf{\ddot{r}}_k<br /> =\left[ \sum_{k=1}^N m_k\mathbf{r}_k - \sum_{k=1}^N \mathbf{r}_Q\right] \times \mathbf{\ddot{r}}_k$$<br /> $$= M\left[ \mathbf{R} - \sum_{k=1}^N \mathbf{r}_Q \right] \times \mathbf{\ddot{r}}_k$$<br /> <br /> And this is when it doesn't make sense because $\sum_{k=1}^N \mathbf{r}_Q = N \mathbf{r}_Q$ 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 <br /> <br /> $$\sum_{k=1}^N M( \mathbf{R} - \mathbf{r}_Q ) \times \mathbf{\ddot{r}}_k <br /> = M\left[ \sum_{k=1}^N\mathbf{R} - \sum_{k=1}^N \mathbf{r}_Q \right] \times \mathbf{\ddot{r}}_k$$<br /> <br /> which isn't what we have either![/itex]

 

[member 553083]

Where am I wrong? It looks like I'm missing something very simple but I can't figure it out. Is there a rule I'm not aware of about sums and cross products? The answer is that you are forgetting to use the given information about the total mass M. In the first equation, the second term should be written as: \sum_{k=1}^N m_k(\mathbf{r}_k-\mathbf{r}_Q)\times \mathbf{\ddot{r}}_k = M(\mathbf{R}-\mathbf{r}_Q)\times \mathbf{\ddot{r}}_k since M\mathbf{R} = \sum_{k=1}^N m_k \mathbf{r}_k

 

[wais]

Based on the equations you provided, it looks like you are trying to derive the equations of motion for a system of particles, where \mathbf{r}_k is the position vector of the kth particle, \mathbf{F}^i_k is the sum of the external forces acting on the kth particle, and \mathbf{\ddot{r}}_k is the acceleration of the kth particle. The first equation you provided is the sum of the torque exerted on the system by the external forces, while the second equation is the sum of the torque exerted on the system by the particles themselves.

The issue with your derivation is that you are not taking into account the fact that the position vector \mathbf{r}_k is measured from the center of mass of the system, \mathbf{R}, and not from some arbitrary point \mathbf{r}_Q. This is why there is a difference between the two equations and why the \mathbf{r}_Q term should not be included in the second sum.

To properly derive the second equation, you can use the fact that the center of mass is defined as \mathbf{R} = \frac{1}{M}\sum_{k=1}^N m_k \mathbf{r}_k, where M is the total mass of the system. Plugging this into the second sum, we get:

\sum_{k=1}^N m_k(\mathbf{r}_k-\mathbf{R})\times \mathbf{\ddot{r}}_k
= \sum_{k=1}^N m_k\mathbf{r}_k \times \mathbf{\ddot{r}}_k -\sum_{k=1}^N m_k\mathbf{R} \times \mathbf{\ddot{r}}_k
= \sum_{k=1}^N m_k\mathbf{r}_k \times \mathbf{\ddot{r}}_k - M\mathbf{R} \times \mathbf{\ddot{r}}_k
= \sum_{k=1}^N m_k\mathbf{r}_k \times \mathbf{\ddot{r}}_k - M\left(\frac{1}{M}\sum_{k=1}^N m