### IIT JEE, Centre of Mass, and mugging

During the time I was preparing for IIT-JEE, I was confused about very many things. I spent a lot of time trying to unravel concepts rather than solve problems and memorise formulae. Subsequently, I screwed up my entrance exams, and got into a rather well-known institute purely on the basis of the English/General Aptitude section.

One concept which was a major pain in the posterior was “centre of mass”. When you throw a body up in the air, say a rod, then it will rotate about a special point, which will lie at the geometrical centre in the case of a uniform rod. More importantly, we could treat the centre of mass of an object as containing the entire mass of the body.

What?!! So if you cut out the geometrical centre and weight the rest of the rod, will it not weigh anything??

“No”, the teacher said. “This is just for calculation purposes.” And without any further explanations, he’d stomp off towards his overpriced car.

I tried to interpret the centre of mass as some “special point” that is ‘gifted’ the mass of the body when the body is in motion. Dissatisfied, I finally construed an explanation last year about why the concept of the centre of mass is so useful. To the best of my knowledge, I haven’t read this explanation in any book or website. A part of this explanation, in a rather inaccessible form, can be found in Resnick-Halliday.

Take a solid body, and let external force $\sum{\overline{F}_{ext}}$ act on it. Take any particle $p_i$. This particle exerts forces on other particles, and other particles exert forces on it. The resultant force on $p_i$, due to Newton’s Second Axiom, is $m_i\overline{a}_i$. Let us determine $\sum\limits_{i=1}^n{m_i\overline{a}_i}$. This is equal to $\sum{\overline{F}_{ext}}+\sum\limits_{i=1}^n{\sum\limits_{j=1,j\neq i}^n{\overline{F}_{ij}}}$, where $\overline{F}_{ij}$ is the force $p_i$ applies on $p_j$. By Newton’s Third Law, $\sum\limits_{i=1}^n{\sum\limits_{j=1,j\neq i}^n{\overline{F}_{ij}}}=0$. Hence, we have $\sum{\overline{F}_{ext}}=\sum\limits_{i=1}^n{m_i\overline{a}_i}$.

Now what? We want to find a point such that under the given circumstances, it has the same acceleration as a body of mass $\sum\limits_{i=1}^n{m_i}$ would have when $\sum{\overline{F}_{ext}}$ is applied on it. There are two things to note here:

1. We don’t KNOW the point yet. We want to DETERMINE this point.
2. This seems like an arbitrary condition. We could also have wanted to determine the point whose acceleration is $(\sum{\overline{F}_{ext}}/\sum\limits_{i=1}^n{m_i})^2$. However, it is not that arbitrary. We will always be able to calculate the acceleration of this particle based on external forces, which are easily determined, in however complex situations. And although we’ll also be able to easily determine the acceleration of the point whose acceleration is $(\sum{\overline{F}_{ext}}/\sum\limits_{i=1}^n{m_i})^2$, the former condition is more relevant to Newtonian mechanics.

Let the particle be called $cm$. We were saying $\overline{a}_{cm}=\sum{\overline{F}_{ext}}/\sum\limits_{i=1}^n{m_i}$. This is equivalent to saying $\overline{a}_{cm}=\sum\limits_{i=1}^n{m_i\overline{a}_{i}}/\sum\limits_{i=1}^n{m_i}$. On integrating twice and applying boundary conditions, we get $\overline{r}_{cm}=\sum\limits_{i=1}^n{m_i\overline{r}_{i}}/\sum\limits_{i=1}^n{m_i}$.

So this is the formula for locating the point whose acceleration depends on the external forces and the total mass of the body. So how does the whole mass of the body get concentrated at this point? It doesn’t. The mass of this particle remains the same!! $\Delta m$, if you like. It is just that $\overline{a}_{cm}=\sum\limits_{i=1}^n{m_i\overline{a}_{i}}/\sum\limits_{i=1}^n{m_i}$ is equivalent to saying $(\sum\limits_{i=1}^n{m_i}).\overline{a}_{cm}=\sum\limits_{i=1}^n{m_i\overline{a}_{i}}=\sum{\overline{F}_{ext}}$. Remember this. The mass of the whole body does NOT get magically transferred to this point.

Similarly,one may determine points such that their acceleration is totally dependant on external forces in the following fashion: $(\sum{\overline{F}_{ext}})^3/\sqrt{\sum\limits_{i=1}^n{m_i}}$. Or any other combination of this form.

If you too are preparing for IIT JEE and think it is the stupidest shit in the world, you’re not alone. Stop solving numericals! Go look outta the window. Explore.