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# PHYS 200: Fundamentals of Physics I

## Lecture 2

## - Vectors in Multiple Dimensions

### Overview

In this lecture, Professor Shankar discusses motion in more than one dimension. Vectors are introduced and discussed in multiple dimensions. Vector magnitude and direction are also explained. Null vectors, minus vectors, unit and velocity vectors are discussed along with their properties. Finally, several specific problems are solved to demonstrate how vectors can be added, and problems of projectile motion are expounded.

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html## Fundamentals of Physics I## PHYS 200 - Lecture 2 - Vectors in Multiple Dimensions## Chapter 1. Review of Motion at Constant Acceleration [00:00:00]
v + ½ _{0}tat . You can easily check by taking two derivatives that this particle does have the acceleration ^{2}a. So what are these two other numbers–x and _{0}v and _{0}x-not and v-not? I think we know now what they mean. They tell you the initial location and initial velocity of the object. For example, suppose you use vertical motion and you use y instead of x; and a would be -g; that’s a particle falling down under the affect of gravity. If all you know is the particle is falling under the affect of gravity, that’s not enough to say where the particle is, right? Suppose we all go to a tall building and start throwing things at various times and various speeds. We’re all throwing objects whose acceleration is -g. But the objects are at different locations at a given time. That’s because they could have been released from different heights with different initial velocities.Therefore, it’s not simply enough to say what the acceleration is. You have to give these two numbers. Then once you’ve got those two numbers, they’re no longer free parameters; they’re concrete numbers, maybe 5 and 9. Then for any time at. Again, you should not memorize the formula. I hope you know why it makes sense, right? You want to know how fast the guy is moving. This is the starting speed, that’s the rate at which it is gaining velocity, and that’s how long it’s been gaining it. So you add the two. This also means the following. It is true that if you give me the time, I can tell you the velocity. Conversely, if I knew the velocity of this object, I also know what time it is, provided I knew the initial velocity.Therefore, mathematically at a given time v + 2 _{0}^{2}a (x-x). This is a matter of simple algebra, of taking this and putting it here. But I showed you in the end how we can use calculus to derive that. I got the feeling, when I was talking to some of you guys, that maybe you should brush up on your calculus. You have done it before, but when you say, “I know calculus,” sometimes it means you know it, sometimes it means you know of some fellow who does or you met somebody who knows. That’s not good enough. You really have to know calculus. This is a constant problem. It’s nothing to do with you guys. It is just that you have done it at various times._{0}It’s a problem we have dealt with in the Physics Department year after year. One solution for that is to get a copy of a textbook I wrote called ## Chapter 2. Vector Motion 2D Space: Properties [00:04:45]Okay, so I will now proceed to the actual subject matter for today. I told you, the way I’m going to teach any subject is going to start with the easiest example and lull you into some kind of security and then slowly increase the difficulty. What’s the next difficult thing? The next difficult thing is to consider motion in higher dimensions. How high do you want to go? For most of us, we can go up to three dimensions because we know we live in a three-dimensional world. Everything moves around in 3D. That’ll be enough for this course. In fact, I’m going to use only two dimensions for most of the time because the difference between one dimension and two is very great. Between two and three and four and so on is not very different. There’s only one occasion where it helps to go to three dimensions because there are certain things you can do in 3D you cannot do in less than 3D. But that’s later, so we don’t have to worry about that. We’ll stop at two. If you talk to string theorists, they will tell you there are really, how many dimensions? Do you know? There are actually ten, including space and time. There are nine spatial dimensions. That’s why I call them string guys. Mathematicians have luckily told us how to analyze the mathematics in any number of dimensions. We don’t pay attention to them when they are going beyond three, but now we know we need that. We’re going to do two. Our picture now is going to be some particle that’s traveling in the That’s what we’re going to talk about a little bit, talk a little bit about vectors. You guys have also seen vectors, I’m pretty sure, but it is worth going over some properties and some may be new and some may be old. For the simplest context in which one can motivate a vector and also motivate the rules for dealing with vectors, is when you look at real space, the coordinates But in 2D, the options are not just left and right or north and south, but infinity of possible directions in which I could go those 5 km. What I could do on the first day is to come here. On the second day, could be to go there. These two guys are 5 km long. For that purpose, to describe that displacement, we use a vector. This is called a vector; we are going to give it the name This is Alright, next thing I want to do is to define the vector that plays the role of the number 0. The number 0 has a property. When you add it to any number, it doesn’t make a difference. I want a vector that I want to call the 0 vector. It should have the property that when I add it to anybody, I get the same vector. So you can guess who the 0 vector is. The 0 vector is a vector of no length. If this is Then, I want to think of a vector that I can call You go back to the same You can ask yourself, “If you gave me a particular vector, what do I use for Ay. The angle ^{2}θ that it makes with the x-axis satisfies the condition tan (θ) is Ay over Ax.. What this means–Now here’s the main point. If you give me a pair of numbers, Ax and Ay, that’s as good as giving me this arrow, because I can find the length of the arrow by Pythagoras’ theorem. And I can find the orientation of the arrow by saying the angle θ that satisfies tan (θ) is Ay over Ax. You have the option of either working with the two components of A or with the arrow. In practice, most of the time we work with these two numbers, Ax and Ay. If you are describing a particle with location r, the vector we use typically to locate a particle r, then r is just (I times x) + (J times y), because you all know that’s x and that’s y.I’ve not given you any other example besides the displacement vector, but at the moment, we’ll define a vector to be any object which looks like some multiple of A very important result is that if two vectors are equal, if ## Chapter 3. Choice of Basis Axis and Vector Transformation [00:22:57]Now, when you work with components, In fact, just because the world is round, you can already see if this is the Earth, to somebody that’s the natural direction. If you go to another neighboring country, that’s what they think is naturally Very important point that comes up is the following. What if somebody comes along and has got a new axis? Here is the unit vector Let’s draw a little picture on the side here. This is Here is what I’d say you should do. This is the kind of thing I don’t want to do in classroom. It takes time and I’m probably going to screw up. Imagine taking This should be prime. By the way, when you guys do this, did anybody notice prime? If you noticed it, you got to stop. You cannot let me write anything that’s incorrect. You got to follow it in some detail. I know you couldn’t do the details, but you should at least know that it’s What’s the purpose of the exercise? The purpose of the exercise is to make the following precise remark. You can pick your unit vectors, or what are called basis vectors, any way you like. Pick any two perpendicular directions which may be related to the ones I picked by an angle
Maybe I will add this on as a problem you should do in your homework. Do you realize that this is a pair of simultaneous equations in which you can solve for these two unknowns, if you like, in terms of these two knowns and these coefficients, which are like these numbers, 3, 2, 4, and 6? You can do that, but we like to get the answer the way this gentleman described it, because we like to get an answer more readily than by doing the mundane work. The idea that he had was, if you go from me to you, with a clockwise rotation, you go from you to me by a counterclockwise rotation. Therefore, if I go by Now, here is a very important message, which is the following. When you go from one set of axes to another set of axes, the numbers change. The components of the vector are not the same. You might think it’s all along
Ay = ^{2}Ax′ squared + Ay′ squared. That’s a property of this expression. You can square this guy and add it to the square of this guy and you will find, using the magic of trigonometry, that this is true. I’m going over these points, because they’re very, very important. When we do relativity, we’ll be dealing with vectors in space-time and we’ll find that different observers disagree on what is this and what is this. But they will agree on certain things. Yes?
## Chapter 4. Velocity Vectors: Derivatives of Displacement Vectors [00:38:20]Let’s take a particle that is moving in the In other words, when you move on a line, you wait a small time Why is the derivative of a vector also a vector? Because the difference in the vector between two times is its vector. Dividing by For example, if I say a particle’s location is J times 9t, for every value of time, you can put the numbers in and you can find the velocity by just taking derivatives. The derivative of this guy will be 2^{3}t times I plus what? 27t times ^{2}J. Rule for taking derivatives, if you want to do this mindless application of calculus, you can do it. I and J are constant; ignore them when you take derivatives; x and y are dependent on time. Just take the derivatives and that’s the velocity vector. You can take a derivative of the derivative and you can get the acceleration vector, will be d over ^{2}rdt, and you can also write it as ^{2}dv over dt.Anybody have a question with what I’ve done now or what we’re doing here? Let me summarize what I’m saying. Particle’s moving in a plane. At every instant, it’s got a location given by the vector ## Chapter 5. Derivatives of Vectors: Application to Circular Motion [00:43:40]Now, I want to do one concrete problem where you will see how to use these derivatives. I’m going to write the particular case of
Let’s sneak in one more derivative here, which is to take the derivative of the derivative. That’s a very important result. To do that, let’s take one more derivative. Let’s try to do part of it in the head, so I can just write down the answer. If you take one more derivative, the sine is going to become a cosine and yield another r itself. That’s a very interesting result. It tells you when a particle moves in a circle, it has an acceleration in a negative r direction, namely directed towards the center. What’s the magnitude of that acceleration? I write it without an arrow here. That’s equal to ω times a magnitude of this vector, which is just ^{2}r.But now, I can also write it as r. Here, ready? This is a very important formula. I don’t know how many kids, generation after generation, get in trouble because they do not remember the following fact. I’m going to say it once more with feeling. When a particle moves in a circle, it has an acceleration towards the center of this size, v over ^{2}r. Very simple. It comes from the fact that velocity is a vector and you can change your velocity vector by changing your direction. This particle is constantly changing its direction, but it therefore has an acceleration and the acceleration, we have shown here, is pointing towards the center. The size of that is v over ^{2}r. For example, if a car is going on a racetrack and you’re seeing it from the top, if the speedometer says 60 miles per hour, you might say it’s not accelerating. That’s the layperson’s view. If it’s going in a circle, you will say from now on, that it, indeed, has an acceleration, even though no one’s stepping on the accelerator, of amount v over ^{2}r. That’s a very important thing. That’s what you’re learning in this course. The fact that when you step on your gas you accelerate, everybody knows. The fact that when you go in a circle, you accelerate is what we’re learning here, coming from the fact that velocity is a vector and its change can be due to change in the magnitude or change in direction. This was a problem, but the magnitude of velocity was nailed down and fixed at ωr, but the direction is constantly changing.So I’m going to now do a second class of problems. We will return to this issue later. By the way, one thing I should mention to you. Suppose the particle is not moving in a circle, but does this. Let’s take a circle and just keep a quarter of the circle. During this part, when you’re doing a quarter of a circle, this is supposed to be a quarter of a circle. You have the same acceleration directed towards the center. In other words, you don’t have to be moving actually in a circle to have the acceleration. At any instant, your motion, if it follows any curve, locally can be approximated as being part of some circle. You can take circles of different size and place them against your trajectory and see which one fits. That’s the r is the acceleration directed towards the center of that circle.## Chapter 6. Projectile Motion [00:54:00]One miscellaneous result, which we don’t use very much now, but which I should mention to you is the following. Suppose this is the ground with some origin here. That’s the plane with some origin here. And in the plane there is some other object. What’s the location of the object? Let us say The problem we want to do today, a whole family of problems, looks like this. I want to consider a particle which has definite acceleration h. Car is traveling with some initial speed v in the horizontal direction. This equation is a pair of equations. One along _{0}x and one along y. For the x part, I’m going to write x = x + _{0}v. _{0}tx I’m going to reduce to 0 by choice of my origin being here. The _{0}x coordinate of the car is the 0, because this is my origin.How about the t* from this one and you put it there. That tells you where you land. I don’t think I have to actually do that step. I’m assuming you can fill in the blanks. Solve this equation for t*, put it there and that’s where you land. That’s one class of problems. Here’s the second class.The second class is the most popular application of what I’m doing now. I want you not to memorize every formula the book gives you for this problem. That’s a problem of projectile motion. Projectile motion. You start here and you fire a projectile with some velocity θ. It’s going to go up and it’s going to come down. One question is: Where is it going to land? Another question is: At what angle should you fire your projectile so it will go the furthest? You can find the equation, but you’ve got to think a little bit before you solve everything. It’s good to have an idea of what’s coming. Imagine you got this monster cannon to fire things. It’s got a fixed speed v. How do you want to aim it so you can be most effective? Go as far as you can? There are two schools of thought. One says, aim at your enemy and fire like this. Then it lands on your foot, because assuming the cannon is at zero height, the cannonball certainly comes out towards the enemy, but has no time of flight. Other one says, maximize the time of flight and you point a cannon like this. It goes up, stays in the air for a very long time, but it falls on your head. We know the truth is somewhere between 0 and 90. Now, the naïve guess may be 45, but it turns out the naïve guess is actually correct. I just want to show you how that comes out. I don’t want you to cram this formula. This is the kind of thing you should be able to deduce. Let’s go back to the same thing I wrote earlier. _{0}x = 0 plus– What’s the horizontal velocity? Horizontal velocity is v cos _{0}θ. So x is v cos _{0}θ times t. And y is v sin _{0}θ times t minus ½ gt. Now we know everything about this particle. I told you once you know the free parameters, ^{2}r and _{0}v, you know everything about the future of the object._{0}Let’s ask, what’s the range? Range is that distance. What’s the strategy for range? You see how long you are in the air and the whole time you are in the air, you are traveling horizontally at this speed. Again, let θ - ½ gt*) times t*. I just pulled out a common key. It says you are on the ground on two occasions. One is initially. We are not interested in that. If the time you are interested in is not zero, you’re allowed to cancel it and get the time from here. That time is t* = 2v sin _{0}θ over g. That’s how long you are in the air. This says, if you want to be in the air for a long time, maximize sin θ, so you may think that 90 degrees is the best angle. But that’s not the goal. The goal is to get the biggest range. Go back to x and put your value for t*, which is 2v sin _{0}θ/g. That becomes v times 2 sin _{0}^{2}/gθ cos θ. If you go back to your trigonometry, you should find that is really the formula for sin 2θ. That’s another example where knowing the trigonometry is very helpful. This tells you exactly what you want. It says, make sin 2θ as big as you want. That tells you 2θ is going to be 90 and θ is going to be 45. It’s not from a naïve guess that it’s halfway between 0 and 90. It turns out to be– there’s a fairly complicated balance between the time of flight and the range. Some people memorize this. I would say, don’t do that. After a while, you won’t have room in your head for anything. Just go back as often as possible to this formula and work your way from there. There are more variations, but it’s always the same thing.Here’s another variation. In my days, they would say, “Find out when you want an object to go through that point.” I tell you the initial velocity t. At that time t, the x coordinate must have a certain value. Then go to the x equation and demand that this be equal to the desired x value and find the time. I take the time and put it in the y equation and demand that the y I get, agrees with this y. If you do that, you will find there is one unknown, which is v, and we can solve for _{0}v. This is the most general problem you can have._{0}In the textbook, of course, people make them interesting. Suddenly there’s a mountain. There’s a physicist hiking on this mountain. See, you’re already laughing. It’s not a very credible problem. We don’t hike and when we get stuck, we don’t want food, we want our table of integrals. That’s what we want somebody to send to us. These problems are embellished in many ways to make you all feel involved. For example, instead of a stone dropping, nowadays there’s a monkey that’s falling down. The people in life sciences feel, “Hey, we are represented in this subject.” All those creatures, which are very interesting, in the end, you are told, “There’s a horse inside a railway carriage.” You cannot see the horse, but the horse is moving. You can deduce it because the carriage is moving the other way. There’s a picture of a horse, a picture of the carriage, all glossy stuff. Then it says, “Treat the horse as a point particle.” See? That’s what we learned in the old days. But nowadays, it’s a horse, but in the end, if you’re going to treat it as a point particle, it seems to me to be a real waste. If you’re the Godfather, right? You want to get the contract for Johnny Fontaine, you don’t tell your consiglieri, “Hey Tom, put a point particle on Jack’s bed.” When you want a horse, use a horse. When you want a point particle, if you wake up and find a point particle on your bed, what’s your reaction? I think these extra pictures sometimes they’re helpful, sometimes they just make the book cost a lot more. There are problem in that a horse must be treated when we study rigid bodies. Even there, a horse is not really a rigid body, unless it’s been dead for a long time. The fact is not a point particle is important. I think the reason books are bigger and bigger but still carry the same information is that examples are more and more interesting. Sometimes they serve a purpose. Sometimes they are distracting. I haven’t written a book in this field, so I will not say anything more. Alright, so I think I should stop now. Next week, we’re going to get a clock that you guys can see in the back. I know you’re very anxious to know the time like I am. But then you can just look back or maybe they’ll put one up here. [end of transcript] Back to Top |
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