Motion of the Centre of Mass

Transcript

0:00Introduction

Hello everyone! Okay, picture this. You stand at one end of a small boat on a still lake, and walk to the other end. You move forward, fine. But the boat slides backwards! No paddle, no rope, nobody pushed it. So who moved the boat? Today, one big idea answers that: the motion of the centre of mass. Boats and explosions come up often in JEE, so let's begin.

0:28From position to velocity and acceleration

So, start from what we know. Capital M, the total mass, times r c m, equals the sum of m i r i. The arrow on top means a vector, the small c m below means centre of mass, and the Greek sigma means add over all the particles, labelled by the small i. Now the particles move. So what happens when we differentiate both sides with respect to time? Position becomes velocity, yes! v c m equals the sum of m i v i, over capital M, okay? Now, the top of that fraction, the sum of m i v i, is the total momentum. So capital P total, the total momentum, equals M times v c m. Clear? Differentiate once more: a c m equals the sum of m i a i, over capital M, with me?

1:17Only external forces move the centre of mass

Now multiply by capital M, and M a c m equals the sum of m i a i. By Newton's second law, that is the sum of all the forces on all the particles. But some forces are internal: the particles push and pull on each other. What happens to those when we add up? They cancel, correct! If one particle pulls another, the other pulls back equally, the opposite way. Newton's third law! So internal forces cancel in pairs. Got it? So only the external forces survive. F ext, the total external force, equals M a c m. Newton's second law, for the whole system! Right? So the centre of mass moves like one particle of mass capital M, acted on by the total external force. Internal forces alone can never change its motion. Clear?

2:10No external force: the centre of mass stays put

Now the most useful result. Suppose no external force acts at all. What happens to v c m? It stays constant, yes! F ext equals zero, and the double arrow means implies: v c m is constant. Okay? And if the system starts at rest, the centre of mass stays exactly where it is, whatever the internal forces: walking, explosions, springs. Right? Now a shell that explodes in flight. The explosion forces are internal, so the centre of mass of the pieces carries on along the original parabola. But careful, only until any fragment hits the ground! Then the ground pushes on it, an external force. With me?

2:55Topper's trick: man on a boat

So, our boat puzzle! Still water, no friction, so no horizontal external force acts on you plus the boat, and everything starts at rest. In the picture, the dashed line is the centre of mass, fixed, before and after. So when you walk forward, the boat slides back, by x, the green arrow. You and the boat pushed each other! Okay? Now the toppers' shortcut. Say a person of mass small m walks a distance capital L relative to a boat of mass capital M, with no friction from the water. The boat moves m L over, m plus M, backwards. Right? Now think. Over the ground, who moves more: the light person, or the heavy boat? The person, of course! The person moves M L over, m plus M, forward relative to the ground. The heavier boat moves less. Clear? The same formula works for a block sliding on a wedge, a particle inside a bowl, and similar problems. Got it?

3:58Kinetic energy of a system

One more result, about energy. The kinetic energy capital K of a system equals half M v c m squared, plus K rel. K rel, with the small r e l below for relative, is the kinetic energy measured in the centre of mass frame. Okay? For two particles, K rel equals half mu v rel squared, with v rel their relative speed. Mu, the Greek letter, is the reduced mass: m one times m two, over, m one plus m two, where m one and m two are the two masses. With me?

4:33Example 1: how far the boat moves as the man walks

Now, Example one: our puzzle, with numbers. Here, a sixty kilogram person stands at one end of a boat, three metres long, of mass one hundred and twenty kilograms, at rest on still water. The person walks to the other end. How far does the boat move? Pause the video, and try it first. Done? Okay, let's check. Still water: no horizontal external force, and everything starts at rest. So our shortcut works. Okay? Boat: m L over, m plus M, so sixty times three, over one hundred and eighty. One metre, backwards. Got it? The person walked the whole boat, three metres. So the person moves three metres over the ground, right? No, no, careful! Three metres is relative to the boat. Relative to the shore, the person moves M L over, m plus M: one hundred and twenty times three, over one hundred and eighty, that is two metres forward. Clear? Check: two plus one is three, the boat's length. Person two metres forward, boat one metre back, centre of mass not moving at all. Puzzle solved!

5:47Example 2: shell explodes at the top, where it lands

Now a harder one: an explosion. Here, a shell is fired with a range of one hundred and twenty metres. At the highest point, it explodes into two equal fragments. One fragment falls vertically, starting from rest. Where does the other land? Pause and try. Hint: think about momentum at the top. Done? Okay, let's check. At the highest point the vertical velocity is zero, so the velocity is horizontal. Call it u x, with a small x below. And the top is halfway along the range, at sixty metres. Okay? Gravity is vertical and the explosion forces are internal, so horizontal momentum is conserved. One piece stops dead. What must the other do? Carry all the momentum, exactly! Each piece has mass m, so the shell is two m. Two m u x equals m times zero, plus m v. So v equals two u x: double the speed! Right? How long does it fall? It starts at the top with no vertical velocity, like the shell. So it takes the time of the second half of the flight, when the shell would cover sixty metres at u x. With me? So it lands at one hundred and twenty metres, where the shell would have landed, right? No, no, careful! That is where the centre of mass lands. At double the speed, this piece covers one hundred and twenty metres beyond the top. Sixty plus one hundred and twenty: it lands at one hundred and eighty metres from the launch point. Clear? Now check with the centre of mass. Equal pieces at sixty and one hundred and eighty: the average is one hundred and twenty metres, the original range. Both pieces land together, so the rule holds till the ground. That's it!

7:41Quick revision: F ext = M a cm, boat shift, exploding shell

So, quick revision. F ext equals M a c m, so internal forces cannot change the motion of the centre of mass. No external force, and starting at rest? The centre of mass stays fixed. The boat moves m L over, m plus M, backwards, and an exploding shell's centre of mass stays on the parabola. Now go and try the practice problems on this page. You will enjoy them. All the best!