Angular Momentum of a Particle

Transcript

0:00Introduction

Hello everyone! Let's start with a small puzzle. Picture a ball moving in a perfectly straight line, at constant speed. It is not going round anything. So, does it have angular momentum? Most students say no, nothing is rotating, right? By the end of this video, you will see the honest answer: it depends on the point you measure from! Today: angular momentum of a particle. It comes up often in JEE, so let's begin.

0:29Angular momentum about a point O

So, take a particle of mass m, moving with velocity v, and choose a point O, our reference point. Let r be its position vector, drawn from O, and p, which is m times v, its momentum. The little arrows on top mean these are vectors, okay? The angular momentum about O is L vector. It equals r cross p, where the cross sign means the cross product, and that is m times r cross v. Keep the order: r first, then p. Swap them, and the sign flips. Right? Now, you have done cross products in vectors. So tell me: what is the magnitude of r cross v? r v sine theta! Yes, exactly. Theta, the Greek letter here, is the angle between r and v. So the size of L is m v r sine theta. And m v is just p, so this is also p times r perp, okay? Now, r perp is r with a small upside down T below it, the perpendicular sign. Drop a perpendicular from O on to the line along which the particle moves. That length is r perp, and it equals r sine theta. So, angular momentum is momentum times this perpendicular distance. With me? Now the units. Angular momentum is measured in kilogram metre squared per second; the small minus one up on the s means per second. That is the same as joule second. Its dimensions, in square brackets: M L squared T to the power minus one. Got it? Now, a small quiz. Which famous constant in physics has exactly these dimensions? Planck's constant! Correct. Angular momentum and Planck's constant have the same dimensions, so keep this pair in mind for questions on dimensions. Okay?

2:17Moving in a straight line

Now, back to our puzzle. Look at the figure. The particle moves along the dashed line with constant velocity, the green arrows, through the points A, B and C in turn. O, the orange dot, is our reference point, and d, in blue, is the perpendicular distance from O to the line. The thin grey lines from O are the position vectors, and they keep getting longer. Yet the label on the right says L equals m v d, the same at A, B and C, at all three points. So tell me: r keeps growing, so why does L not grow? Because r perp does not change! Yes, exactly. r gets longer, but the angle theta changes too, and r sine theta, the perpendicular distance, always stays equal to d. So L equals m v d, the same at A, at B, at C, at every point. Puzzle solved: a particle moving in a straight line with constant velocity does have angular momentum about O, and it stays constant. Right? And this works about any point. Just take d as the distance of the line of motion from that point. So, a topper's trick: for straight line motion, skip the cross product and write m v d directly. Clear? Now tell me. Same particle, same velocity, but I move the point O farther away from the line. Is L still the same? No! Now d is bigger, so L is bigger. Same motion, different point, different angular momentum. So L depends on the reference point. Never say just the angular momentum; always say about which point. Got it?

3:55Moving in a circle

Now, the second important case. Take a particle going round a circle of radius r, and find L about the centre. Tell me, what is the angle between r and v here? Ninety degrees! Yes, exactly. The velocity is along the tangent, and the tangent is perpendicular to the radius. So r is perpendicular to v, with the upside down T sign again, and sine theta is one. With me? So, L equals m v r. And v equals r omega, where omega, the Greek letter, is the angular speed. Put that in, and L equals m r squared omega. Clear? Last, the direction: L vector is perpendicular to the plane of the circle. That's it!

4:40Example 1: angular momentum about the origin from r x v

Now, example one, an easy one. Here, a particle of mass zero point five kilogram is at r equals two i hat plus three j hat, in metres. i hat and j hat, with the little caps, are the unit vectors along x and y, okay? Its velocity is four i hat minus j hat, metres per second. Find its angular momentum about the origin. Pause the video, and try it first. Done? Okay, let's check. The reference point is the origin, so r is just the given position. We write down m, r and v. With me? Both vectors lie in the x y plane, so r cross v points along k hat, the unit vector along z. Its k hat component, which can be negative, is x times v y, minus y times v x, where v y and v x, with the small subscripts, are the components of v. Right? So, put in the numbers: x is two, y is three, v x is four, and v y is minus one. What do you get? Careful with the signs. Inside the square brackets: two times minus one is minus two. Three times four is twelve. Minus two minus twelve gives minus fourteen. So r cross v equals minus fourteen k hat. Clear? So the angular momentum is minus fourteen k hat, right? Oh no, careful! We forgot the mass. L equals m times r cross v: zero point five times minus fourteen k hat. That is minus seven k hat, kilogram metre squared per second. The minus sign gives the direction: along minus k hat. Got it?

6:20Example 2: L about (0,0) and about a point on the line

Now a harder one, with a trap. Here, a two kilogram particle moves with constant velocity, three i hat metres per second. At time t equals zero, it passes through the point zero, four, in metres. Find its angular momentum about the origin, and about the point five, four. Pause the video, and try both parts. Done? Let's check. The velocity is along i hat, so the particle moves parallel to the x axis, along the line y equals four metres. So about the origin, the distance of this line is d equals four metres. Okay? Straight line, constant velocity: use our trick, L equals m v d. Two times three times four: twenty four kilogram metre squared per second. The particle passes above the origin, moving to the right, so it turns clockwise about it: minus k hat. Clear? Now, the point five, four. At t equals zero, the particle is at zero, four, which is five metres from this point. So L equals two times three times five: thirty, right? No, no, wait! That is the trap. Look at the point five, four. Its y coordinate is four, so it lies on the line of motion itself! With me? So the line of motion passes right through this point, and r perp is zero. The position vector from this point lies along the velocity, so sine theta is zero. L equals zero, at all times! Same motion: twenty four about the origin, zero about five, four. That is why you always state the reference point. Got it?

7:58Quick revision: L = r x p, straight line and circle

So, quick revision. L is r cross p; its size is m v r sine theta, or p times r perp, in kilogram metre squared per second. Straight line, constant velocity: L is m v d. Circle, about the centre: m r squared omega. And always state the reference point. Now go and try the practice problems on this page. All the best!