Centre of Mass of a System of Particles

Transcript

0:00Introduction

Hello everyone! One small puzzle first. The Earth and the Moon are about three hundred and eighty four thousand kilometres apart, and the Earth's mass is eighty one times the Moon's. This pair has one special point, its centre of mass. Where is it? Out in space, between the two? Or inside the Earth itself? Hold on to your guess; Example one will settle it. This point opens the whole unit on rotational motion, and JEE questions on it come up often. So, let's begin.

0:33One point that moves simply

First, the idea. Throw a cricket bat spinning into the air. The handle and the blade tumble all over the place, right? So a rigid body, or a group of particles, moves in a complicated way. But one special point of it moves very simply, as if the whole mass were sitting at that point. Okay? And as if all the external forces acted right there. So for our bat, that point moves smoothly, like a small ball thrown on its own. That point is the centre of mass. With me?

1:05Defining the centre of mass

Now, take n particles with masses m one, m two, up to m n; the small numbers below are labels, and the three dots mean and so on. They sit at position vectors r one up to r n; the little arrow on top means a vector, measured from the origin. The centre of mass is at r c m; the small c m below stands for centre of mass. r c m equals m one r one, plus m two r two, up to m n r n, all over the sum of the masses. Each position is weighted by its mass, so a heavier particle pulls the average towards itself. Clear? In short: r c m equals sigma m i r i over capital M, where the big Greek letter sigma means add them all up, for every particle i. And capital M, the sum of all the masses, is the total mass. Got it? In components, do the same for each axis: x c m equals sigma m i x i over M, and the same for y c m and z c m. Okay?

2:10Two particles: closer to the heavier one

Now the simplest system: two particles, m one and m two, a distance d apart. Their centre of mass lies on the line joining them. Right? Let r one and r two be its distances from m one and m two. Then r one equals m two d over m one plus m two, and r two equals m one d over m one plus m two. Look carefully: r one has the other mass, m two, on top. Easy to slip here, so watch it! Multiply r one by m one, and r two by m two: you get the same thing. So m one r one equals m two r two; d is divided in the inverse ratio of the masses. Now, if m one is heavier, which distance is smaller, r one or r two? r one, yes, exactly! Bigger mass, smaller distance. So the centre of mass is always closer to the heavier particle. Remember this for our Earth and Moon puzzle. Clear?

3:11Topper's trick: choose the origin smartly

Now, a topper's trick. The formula needs an origin, and you are free to choose it. So where should you put it? On one of the particles! Its position vector is then the zero vector, the zero with an arrow, so its term simply drops out of the top. But careful: its mass still counts in capital M below. Nice, right? Second, symmetry. For a uniform body, one whose mass is spread evenly, the centre of mass lies on every axis of symmetry, okay? So put the origin, and one axis, along that symmetry line. Then the other coordinate is zero for free, and only one is left to calculate. Got it?

3:52Careful: two common traps

Now, two traps. Take a ring, a thin circle of material. Where is its centre of mass, and is there any material at that point? At the centre, by symmetry. But that centre is empty! So the centre of mass need not lie inside the material. Okay? Ring: the empty centre. Horseshoe: in the gap. If your answer lands in empty space, it can still be right. Clear? Second trap. Shift the origin, and the centre of mass stays at the same physical point, but its coordinates do change. Right? So always state your origin. Coordinates without an origin mean nothing. Got it?

4:33Example 1: Earth-Moon centre of mass, inside the Earth?

Alright, Example one: our puzzle. The Earth's mass is eighty one times the Moon's, and their centres are three point eight four times ten to the five kilometres apart; k m means kilometres. How far from the Earth's centre is the centre of mass of the Earth and Moon system? And is it inside the Earth, of radius six thousand four hundred kilometres? Pause the video, and try it first. Done? Okay, let's check. Use our trick: put the origin at the Earth's centre. r with E written below is the distance from the Earth's centre, and m with M written below is the Moon's mass. The Earth's mass is eighty one m M, so r E equals m M d over eighty one m M plus m M, where d is the distance between the centres, okay? The bottom is eighty two m M, and m M cancels, so r E is just d over eighty two. Now the numbers: three point eight four times ten to the five, over eighty two, about four point six eight times ten to the three kilometres; the wavy equals sign means approximately. That is about four thousand six hundred and eighty kilometres. With me? The Earth's radius is six thousand four hundred kilometres. So is this point inside the Earth, or out in space? Inside! Yes, exactly! Four thousand six hundred and eighty is less than six thousand four hundred; the little angle sign means less than. Puzzle solved: the Earth is so much heavier that the centre of mass lies inside the Earth itself. Okay?

6:18Example 2: hexagon of equal masses, one mass removed

Now a harder one, a JEE pattern question. Six equal masses sit at the corners of a regular hexagon of side a, and one mass is removed. Find the distance of the new centre of mass from the centre of the hexagon. Pause the video and try it. Hint: use symmetry. Done? Okay, let's check. Before removing anything: where is the centre of mass of all six equal masses? At the centre of the hexagon, by symmetry. Correct! So put the origin there. Then m r one plus m r two, up to m r six, equals the zero vector: the six position vectors add to zero. Okay? Now remove the mass at r one. Together with r one, the six made zero, so the other five add up to minus r one. Right? So r c m equals m times the sum of those five, over five m. Oh no, careful! Not six m; only five masses are left. So r c m is minus m r one over five m. The minus sign means: opposite to the removed corner. Clear? Last, the size. The two straight bars mean the magnitude of r one. In a regular hexagon, each corner is a distance a from the centre, equal to the side. So the new centre of mass is a over five from the centre, away from the removed corner. That's it!

7:48Quick revision: formula, two particles, choosing the origin

So, quick revision. One: r c m equals sigma m i r i over capital M, and the same for x, y and z. Two: for two particles, m one r one equals m two r two, closer to the heavier one. Three: choose your origin smartly, and the centre of mass may lie outside the material. Now try the practice problems on this page. All the best!