# Using work on particle system

If $\vec{a}$ and $\vec{v}$ is the acceleration and velocity of a particle system, and the external forces $\vec{F}_i$ are acting on the system, then:

$$\int_{\vec{r}_1}^{\vec{r}_2} \sum_{i}\vec{F_i} \cdot d\vec{r} = \frac{m}{2}\left(\vec{v}_2^2-\vec{v}_1^2\right)$$

(Where $\vec{v}_1$ is the velocity at position $\vec{r}_1$, etc.)

To the question, my book asks about the velocity of the boxes when they have moved a certain distance (in this case $1\text{ m}$): You can get that velocity in several ways. I solved it by just examining one of the blocks, but the problem with this approach is that you need to find an expression for the tension force, which involves solving a system of equations. Nevertheless, it worked.

However, in a PDF, they simply solved the problem using:

$$(20\text{ kg})(9.81\text{ m/s}^2)(1 \text{ m}) = \frac{1}{2}(24\text{ kg})v^2$$

This wasn't very well-motivated, but I assume that they're letting both boxes be the particle system, which means that the tension-forces are internal. This gives that $\sum_{i}\vec{F_i} = (20\text{ kg})(9.81\text{ m/s}^2)$, thus giving the equation above.

Now to my question: Judging by the initial equation I posted, it seems to me that this only is right if they assume that the velocity of the CM has the same magnitude as the speed of each individual box. I basically began examining this, assuming that $\vec{v} \equiv \dot{\vec{r}}_{CM}$:

$$\vec{r}_{CM} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2}$$ $$\dot{\vec{r}}_{CM} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2}{m_1 + m_2}$$ $$v = \left| \dot{\vec{r}}_{CM} \right| = \left|\frac{m_1 \vec{v}_1 + m_2 \vec{v}_2}{m_1 + m_2}\right|$$

To me, it doesn't seem like this formula simplifies to $v = |\vec{v}_1| = |\vec{v}_2|$ (even if you let $|\vec{v}_1| = |\vec{v}_2|$) which is why I'm rather confused by the whole thing.

Questions:

1. If the above formula doesn't simplify correctly, why does assuming that the center of mass moves at the same speed as the individual boxes work? What have I done wrong to reach my "contradicting" conclusion?
2. If the above formula does simplify correctly (in case I just don't see how), are there any general guidelines as to when you can make these simplifications?

The answer to the problem does not use the centre of mass it equates the addition of the kinetic energy of each mass to the loas of gravitational potential energy of one of the masses.

Why would you expect the speed of the centre of mass of the system to equal the speeds of the individual particle?
Pulley with equal masses each side moving in opposite directions at constant speed or equal magnitude acceleration has the centre of mass not moving.

To find the velocity of the centre of mass in this case the following addition has to be made which will certainly not give the speed of the centre of mass as the speed of the individual masses. • I don't expect the speed of the CM to be equal to the sum of the individual speeds, but I thought the answer in the PDF assumed that it was, which is why I asked the question. I do understand your first paragraph, and it makes intuitive sense now, but I am still puzzled about this: the picture you added shows that CM has a speed which is not equal to the individual speeds, but this: prntscr.com/e5bd3d is the L.H.S. in my first equation, and then, it should follow that this speed: prntscr.com/e5be0r is the same as that of the CM. But they're asking for the individual speeds. – Max Feb 6 '17 at 18:57
• I did think that I understand what you were thinking and am sorry if I gave you another impression. I thought that you were concerned that the speed of the centre of mass of the system of two masses was not equal to the speed of each of the individual masses. To me the equation in words is loss in gravitational potential energy of the 20 kg mass is equal to the gain in kinetic energy of the 4 kg mass plus the gain in kinetic energy of the 20 kg mass. – Farcher Feb 6 '17 at 19:06
• Would that not imply however that the center of mass is equal to the speeds of the individual masses? Even though it shouldn't be. The R.H.S. has a $\vec{v}^2 = |\vec{v}|^2$, which should be referring to the CM, so would expressing that in terms of $|\vec{v}_1|$ and $|\vec{v}_2|$ yield a R.H.S. that is $\frac{1}{2}m\vec{v}_{1,2}^2$? To me it seems impossible, which is what I cannot get my head around. Your interpretation of the equation makes more sense, but to me it seems to mathematically be saying another thing. – Max Feb 6 '17 at 19:14
• If the answer had been written as $(20\text{ kg})(9.81\text{ m/s}^2)(1 \text{ m}) = \frac{1}{2}(4\text{ kg})v^2+\frac {1}{2}(20\text{ kg})v^2$ would you have thought about centre of mass? – Farcher Feb 6 '17 at 19:20
• The vertical drop of the centre of mass is not 1 m. – Farcher Feb 6 '17 at 19:37