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  • I would ideally like to have a class with maybe just ten

  • students so that we can have a lot of dialogue and I can find

  • out what you're thinking about, address your questions.

  • That'll make a lot of fun but that's not possible in an

  • introductory course. But I think you should feel

  • free to interrupt or ask questions or discuss any

  • reasonably related issue that I didn't get to.

  • Okay? But that'll make it interesting

  • for everybody because I told you many times the subject is not

  • new to me. It's probably--it's new to you.

  • So what makes it interesting for me is the fact that it's a

  • different class with a different set of people with different

  • questions. And trust me,

  • your ability to surprise me is limitless because even on a

  • subject which I thought is firmly fixed in my mind,

  • some of you guys come up with some point of view or some

  • question that's always of interest to me,

  • of people who've been practicing this business for a

  • long time. So I welcome that,

  • and that'll make it more lively.

  • All right, so I think last time I got off on a rant about all

  • the different pedagogical techniques, some of which I

  • don't endorse fully. I'm going to go back now 300

  • years in time, to one of the greatest laws

  • that we have because--look at the power of this law,

  • right? Here is the equation,

  • and all the mechanical phenomena that we see,

  • around the world, can be understood with this

  • law. And I was starting to give you

  • examples on how to put this law to work, because I think I at

  • least made you realize that simply writing down the law does

  • not give you a good feeling for how you actually use it.

  • So, maybe you have understood it, but I'm going to remind you

  • one more time on how you're supposed to use this law.

  • So, I'm going to take a concrete example.

  • The use of any law of physics is to be able to predict

  • something about the future, given something about the

  • present. So, all problems that we solve

  • can be categorized that way. So I'm going to take a very

  • simple problem, a problem to which I will

  • return in great detail later on; but let me first start

  • with--let me start with this problem, which I will do very

  • quickly now and we'll come back and do it more slowly later.

  • But at least it's a concrete problem.

  • The problem looks like this. So, here is a table and here's

  • a spring and here's the mass m.

  • There's a force constant k.

  • I want to pull it by some amount A,

  • and let it go. So that's the knowledge of the

  • present. The question is,

  • when I let it go, what's this guy going to do?

  • That's the typical physics problem.

  • It can get more and more complex.

  • You can replace the mass by a planet;

  • you can replace the spring by the Sun, which is attracting the

  • planet; you can put many planets,

  • you can make it more and more complicated.

  • But they all boil down to a similar situation.

  • I know some information now and I want to be able to say what'll

  • happen next. So, here I pull the mass,

  • when I want A and I want to know what'll it do next.

  • Remember, when we go back to the laws of Newton,

  • the laws of Newton only tell you this--and we've been talking

  • about this is useful information.

  • The first thing you have to know in order to use the laws of

  • Newton, you have to separately know the left-hand side.

  • You have to know what force is going to act on a body.

  • You cannot simply say, "Oh, I know the force on the

  • body, it is m times a;

  • ma is not a force acting on a body;

  • a is the response to a force;

  • you got to have some other means of finding the force.

  • And in this case, the force on the mass is due to

  • the spring. So, I pull the spring by

  • various amounts and I see what force it exerts.

  • Now, I think you know now in practice how I know what force

  • it exerts, right? I pull it by some amount,

  • attach, say, the one kilogram standard mass.

  • I see what acceleration it experiences, and m times

  • that acceleration, or 1 times the acceleration,

  • is the force the spring exerts. So I pull it by various amounts

  • and I study the spring. And I've learned,

  • by studying the spring, that the force it exerts is

  • some number k, called a force constant,

  • times the amount by which I pull it.

  • If I start off the mass in a position where the spring is

  • neither expanded nor contracted, that's what we like to call

  • x = 0. So I pulled it to x =

  • A. Now, what I'm told is when you

  • pull it to any point x, that's the force the spring

  • exerts. So, this is part of an

  • independent study. People who work in spring

  • physics will study springs and they will find out from you,

  • find out and tell you that any time you buy a spring from me

  • it'll exert this force. And they have done--they

  • figured that out by pulling the spring and attaching it to

  • various entities and seeing what acceleration it produces.

  • There, the masses are taken to be known, because you can always

  • borrow the mass from the Bureau of Standards,

  • or we discussed last time how if you have an unknown mass you

  • can then compare it to this mass and find out what its value is.

  • So every object's mass can be measured.

  • And then the guys making the spring have studied what it does

  • to different masses and figured this out.

  • Now, you come with this mass and you say, what happens when I

  • connect it? Well, I'm assuming the mass of

  • this guy can also be found by comparisons, the way I described

  • to you last time. So we can always find a mass of

  • any object, as we went into in some length.

  • Then, Newton's law says this is equal to ma,

  • but I want to write a as the second derivative of

  • x.

  • So, you now go from a physical law, which is really a

  • postulate. There is no way to derive F

  • = ma. You cannot just think about it

  • and get it. So, whenever I do physics I

  • will sometimes tell you this is a law;

  • that means don't even try to derive it.

  • It just summarizes everything we know in terms of some new

  • terms, but it cannot be deduced. On the other hand,

  • the fate of this mass can now be deduced by applying Newton's

  • law to this equation. Now, this is a new equation,

  • you may not have seen this equation before.

  • For example, if I told you -- forget the

  • left-hand side -- if I told you the right-hand side is 96,

  • I think you guys know how to solve that, right?

  • You have to find a function whose second derivative is 96

  • divided by m, and you all know how to do

  • that; it's t^(2) times a

  • number, and you can fudge the number so it works.

  • This is more complicated. The time derivative of this

  • unknown function is not a given number but the unknown function

  • itself; in other words,

  • x itself is a function of time.

  • This is called a differential equation.

  • A differential equation is an equation that tells you

  • something about an unknown function in terms of its

  • derivatives. You can have a differential

  • equation involving the second derivative or the first

  • derivative or the fourteenth derivative, whatever it is.

  • You are supposed to find out what x (t) is,

  • given this information. So, one thing is,

  • you can go to the Math Department and say,

  • "Hey look, I got this equation, what's the solution?"

  • and they will tell you. Now, sometimes we have to do

  • our own work and we can solve this equation by guessing.

  • In fact, the only way to solve a differential equation is by

  • guessing the answer; there is no other way.

  • You can make a lot of guesses and every time it works you keep

  • a little table; then you publish it,

  • called Table of Integrals.

  • So, I have in my office a huge table, Mark Caprio has got his

  • own integral, we don't leave home without our

  • Table of Integrals. I got one at home,

  • I got one at work, I may want to keep one in the

  • car because you just don't know when you will need an integral.

  • Okay? So people have tabulated them

  • over hundreds of years. But how do they find them?

  • They're going to find them in the way I'm going to describe to

  • you now. You look at the equation and

  • you guess the answer. Let's make our life simple by

  • taking a case where the forced constant is just 1,

  • okay? It takes 1 Newton per meter to

  • stretch that spring. I let the mass of the object be

  • 1 kilogram. This is just to keep the

  • algebra simple. Later on, you can put any

  • m and k, and we'll do all that.

  • Then what am I saying? I'm saying, find me a function

  • whose second derivative is minus that function.

  • Now, as a word problem maybe it rings a bell,

  • right? Do you know such a function?

  • Student: Exponential. Professor Ramamurti

  • Shankar: Exponential is good.

  • Another one? Yes?

  • Student: Any trig functions.

  • Professor Ramamurti Shankar: Trigonometric

  • functions are good. So I'm going to temporarily

  • dismiss the exponential. I will tell you why.

  • An exponential function is actually a very good guess.

  • You can say x = e^(t), for example.

  • The trouble is, this is growing exponentially,

  • and we know this spring mass system, the x is not

  • growing exponentially. Or you might say,

  • well, let me fix that by making it minus t;

  • then it quickly comes to rest after a time and doesn't move.

  • We sort of feel we want this thing to be jiggling around for

  • a long time. So, the exponential function is

  • even better than this function because it says,

  • take one derivative and I reproduce myself.

  • But that's too much; we only want a function that

  • reproduces itself when you take two derivatives,

  • and that's where the trigonometric functions come in.

  • So, I make a guess, x (t) is cos t,

  • and you can check if you take two derivatives,

  • it obeys the equation, the second derivative of

  • x is –x itself. But do I like this solution or

  • is there something not fully satisfactory?

  • Yes? Student: But you can't

  • help the negative when you [inaudible]

  • Professor Ramamurti Shankar: Pardon me?

  • Student: [inaudible] Professor Ramamurti

  • Shankar: Oh it will be negative x.

  • So, let's do that. So let's take dx/dt = -sin

  • (t). Then, take one more derivative,

  • d²x over dt², derivative of sine is cosine.

  • So, if this iscos (t), but x was cos

  • (t); so it's really minus of the

  • x I started with. Yes?

  • So you have a question or--? Student: I was thinking,

  • you wrote it up there [inaudible]

  • Professor Ramamurti Shankar: Oh,

  • this one? Student: Yes.

  • Professor Ramamurti Shankar: Caught on film.

  • Yes, that's a mistake. Okay?

  • That should be minus. Very good.

  • Look, it makes me immensely happy when you guys catch me.

  • Of course, it becomes an addiction, you try to catch me;

  • sometimes it doesn't work, but keep trying.

  • This time you're absolutely right.

  • That's why you were looking confused.

  • Anything else? Everyone follow the way this

  • very elementary problem is solved?

  • It's solved by making a guess. So, you're right.

  • So, the minus sign was wrong but now you agree everything's

  • working. This is the answer.

  • But no, I meant something else is not quite okay with this.

  • Yes? Student: Your function

  • also works if you added a factor of the sine, also?

  • Professor Ramamurti Shankar: Yes.

  • But first of all, I can add some sin (t)

  • and it'll still work. But more importantly,

  • if I put t = 0, I get x = 1.

  • Why should it be true that I pulled it by exactly one meter?

  • I could have pulled it by 2 or 3 or 9 meters.

  • I want to be able to tell how much I pulled it by at t

  • = 0. Suppose it was pulled to 5

  • meters and released. I want x (0) to be 5.

  • The point is, you can easily fix that,

  • by putting a 5 here. Does it screw up everything?

  • It doesn't because the 5 just comes along for the ride,

  • everywhere. And don't think it's 5x,

  • because this 5 is part of x.

  • So you can multiply your answer by any number you like,

  • and you pick that number, that reproduces your initial

  • displacement. So if you pulled it by 5,

  • you put a 5. So, now we're very happy to

  • describe this problem. It is true, you can put sin

  • (t) but I will, in this problem we don't need

  • it. Here's an answer that does

  • everything you want it to do. At the initial time,

  • it gives you 5 times cos 0, which is 5;

  • and that's what you told me was the initial displacement.

  • Initial velocity, if you take a dx/dt,

  • it's proportional to sin (t) and a t = 0,

  • that vanishes, that's correct;

  • you pulled it and you let it go. So, the instant you released

  • it, it had no velocity. And it satisfies Newton's laws,

  • and that's your answer. I want to do this simple

  • example in totality because this is the paradigm.

  • This is the example after which everything else is modeled.

  • Yes? Student: How do you know

  • that it's not sin (t) at the end?

  • Professor Ramamurti Shankar: Ah.

  • You need a t = 0, if the answer was sin

  • (t); well sin (t) vanishes a

  • t = 0. And cos (t), it doesn't

  • vanish, has a robust value of 1. But I want a value of 5,

  • I put a 5 in front. Look, you guys may have a few

  • more things to say about this problem because this is not the

  • final and most general solution to the problem.

  • It is a solution that certainly works for the one example I had,

  • which is a mass being pulled to 5 meters and released.

  • So, carry this in your mind. This is really to me the most

  • important thing I can tell you right now.

  • The way F = ma is applied in real life is half the

  • world is working on the left-hand side,

  • finding the forces that act on bodies under various conditions.

  • In this example, the force is due to a spring

  • and by playing with a spring you can study the force it exerts

  • the various displacements of various amounts by which you

  • pull it, and in this example if you pull

  • the spring by an amount x, it exerts the force

  • -kx, and you can measure k

  • and you put the label on your product and you sell the spring

  • with a given k. Other people are measuring

  • masses of various objects and they've decided the mass of this

  • is whatever, maybe 1 kilogram. You put the two sides together

  • and you find the response of the system for all future times.

  • And what this tells you then, you can now plot this.

  • This is a graph that looks like this;

  • it goes from 5 to -5 and repeats itself and it's very

  • nice. That's what we think the spring

  • would do. Now, you think this is a real

  • spring or am I missing something?

  • This is the whole story? Yes?

  • Student: There's friction.

  • Professor Ramamurti Shankar: There's friction

  • because we know if you pull a real spring it's not going to

  • oscillate forever. It's going to slowly stop

  • shaking and pretty much come to rest.

  • That just means--Who do you want to blame for this equation?

  • Do we conclude now there's something wrong with Newton's

  • laws? Or what's the problem?

  • Student: The k coefficients should not be

  • [inaudible] Professor Ramamurti

  • Shankar: Pardon me? Student: The k

  • coefficient is wrong. Professor Ramamurti

  • Shankar: The fault is not with the k.

  • Yes? Student: We didn't take

  • into account all the force that's friction and the amount

  • of force. Professor Ramamurti

  • Shankar: That is correct. So, we will say we have missed

  • something. There's another force acting on

  • this mass, besides the spring; that's the force of friction

  • that'll oppose the motion of the mass.

  • So, you can say one of two things.

  • Either you can say something is wrong with Newton's laws,

  • or you can say we've not applied Newton's laws properly

  • because we haven't identified all the forces.

  • So, if you never knew about friction, you do this and you

  • find, hey, it doesn't work. The mass comes to rest,

  • whereas the theory says it should oscillate forever.

  • So, you are at the fork. Either you can say Newton's

  • laws are right; I'm missing some forces,

  • I'm going to look around and find out what they are.

  • Then, you will derive friction from that.

  • Or maybe it's time to give up Newton's laws.

  • It turns out, even if there is no friction

  • this is not the correct law of physics.

  • The correct law is given by Einstein's relativistic

  • dynamics. This does not satisfy the

  • relativistic dynamics; by that I mean,

  • if you really pull a real mass by 5 meters or whatever and

  • release it, its motion will not be exactly what I said;

  • or it'll be off by a very, very, very tiny amount that you

  • probably will not discover in most laboratories.

  • But if the mass begins to move at a velocity that is comparable

  • to that of light, then this equation will be

  • wrong. So, people say,

  • oh, what kind of business are you in?

  • Every once in a while somebody is wrong.

  • I'll tell you right now. Everything we know is wrong,

  • you know? And it's no secret;

  • you can publish it, you can tell the whole world,

  • this practitioner said everything is wrong.

  • But it's wrong in this limited sense.

  • Is Newton right or wrong? Well, Newton didn't try to

  • describe things moving at speeds comparable to light.

  • He dealt with what problem he could deal with at that time.

  • So, it's a law that has a limited domain of validity.

  • You can always push the frontiers of observation until

  • you come to a situation where the law doesn't work.

  • But the specialty of relativity doesn't also work all the time.

  • If the mass becomes very tiny, it becomes of atomic

  • dimensions, then you need the laws of quantum mechanics.

  • That's wrong too. So, things work in a certain

  • domain and sometimes you abandon the formalism;

  • but don't rush to do that. In this case,

  • the problem is not with formalism but due to friction.

  • So, you have to make sure you got all the forces.

  • Once you got all the forces, it still doesn't work,

  • then of course the experimentalists are very happy

  • because there's something wrong with the theory;

  • then you got to find the new theory, till that works.

  • Okay, so that's the cycle. So it's an ongoing process;

  • physics is still a work in progress.

  • This is work as of 300 years ago.

  • And no one's found anything wrong with Newton's laws,

  • provided you don't violate two conditions.

  • You don't deal with objects moving at speeds comparable to

  • that of light, and you don't deal with objects

  • which are very, very tiny.

  • The notion of what's very tiny will be very clear when you

  • learn quantum mechanics next term.

  • Right now, there is something rough, called "tiny," there's

  • things like atomic dimensions. All right.

  • So now we are going to do more concrete problems,

  • but I wanted to write that up for you as a simple,

  • solvable, worked out -- an example that displays everything

  • that happens in physics. All right.

  • So, by the way, people are always looking for

  • new forces, the electrical forces.

  • You find a charge on a table, it's not connected to a spring;

  • I bring another charge next to it, in this case it starts

  • moving. What do I do?

  • A body accelerating without any force… I can say Newton is

  • wrong or I can say maybe there's a new force between that body

  • and this body, and that will turn out to be

  • the electrical force. Electrical force occurs

  • whenever you take something, you take a cat and you take a

  • piece of amber or you rub it, you do various things,

  • then you put them next to other things and they're attracted or

  • repelled, and you discover you're

  • charging things, and that's how electricity was

  • discovered. But you don't have to have a

  • cat; if you have a buffalo,

  • you can take it with a buffalo, it still works.

  • Then, you got to find out what you really need to find electric

  • forces. But you find there is a new

  • force and you know you've got a whole new force to study;

  • then you got to find out what's the cause of the force.

  • It turns out, it's not the mass of the

  • object, it's something else called electric charge.

  • Then, how does the force vary with distance?

  • Well, it varies like 1/r².

  • You know that. So that's how you slowly study

  • and identify new forces. So, the forces were found in

  • the order of gravitation first, and then electricity,

  • magnetism and now of course there are new forces called

  • "strong forces." There are weak forces.

  • They all act on the subnuclear level;

  • so that's why that business is not done.

  • We are not finished with the left-hand side.

  • We're still looking for new forces.

  • Okay, so now I'm going to start doing problems in two

  • dimensions. Everything I did last time was

  • 1D. You're pulling something,

  • you're pushing something, you got one block connected to

  • a second block to a third block and you pull it.

  • We're done all that. So we're going to elevate this

  • to high dimensions. And as you know,

  • in high dimensions, acceleration is a vector,

  • and so force better be a vector because when you take a vector

  • and multiply it by a number you get something else.

  • And there, by that process, you've gone to higher

  • dimensions. So let's take again simple

  • problems and try to make them more and more complex.

  • And you cannot tell me eventually you agree the problem

  • is complex enough, or I will crank it up as we go

  • along. If at some point you plead for

  • mercy, then we'll stop. But there is no limit to how

  • difficult mechanics problems can be.

  • If you go back and read Cambridge University exams,

  • in 1600 and 1700, there were really difficult

  • problems. That's why they invented

  • quantum mechanics, which is a lot easier than

  • those problems from Newtonian mechanics.

  • You can make it as hard as you like.

  • So, I'm just going to do a few, then once you got the idea,

  • you stop. But again, we're going to start

  • with easy one. Easy one is a table on which

  • this mass is sitting, doing absolutely nothing.

  • And we want to study that guy and understand it in terms of

  • Newton's laws. Because we're in two

  • dimensions, you got to have two axes.

  • So here is my x and here is my y.

  • So, the first rule is if F = ma, we're really saying

  • i times fx + j times fy,

  • is miax + mjay. It's a vector equation.

  • The force has to be a combination of i and

  • j because nobody--By the way, I'm working in the

  • xy plane. I told you I don't need to go

  • to 3D, for a long time. In 3D, in every vector there's

  • some amount of i and some amount of j,

  • that's, what's true for f, that's true for

  • a. You write it out.

  • And I told you, if two vectors are equal,

  • then the x components have to match and the y

  • parts also have to match. You can see that.

  • If I draw two arrows on the plane and I tell you this arrow

  • is the same as that arrow, it's the same.

  • It's got the same horizontal part and the same vertical part.

  • So, this is really two equations, and I'm going to

  • apply them to this block. So, when I look at this block,

  • in the x direction, it's not moving at all.

  • There are no known forces acting in the x

  • direction, and therefore it's a case of 0 = 0.

  • Did I miss something? You guys look a little worried.

  • Yes? Okay, now I look in the

  • y direction. If you look in the y

  • direction, I take this block and I draw its free body diagram,

  • and I write all the forces on it.

  • So, there is the force of gravity, mg;

  • however, this M is a big M or a small m?

  • Well, let's change the figure to small m,

  • so that the rest of the equations are right.

  • The little mg is acting down.

  • If that's all you had, the block would fall through

  • the table; it's not doing that,

  • so the table we know is exerting a force.

  • The standard name for that is N, and N stands

  • for normal. And normal is a mathematical

  • term for perpendicular. You say this vector is normal

  • with that vector, you mean they are

  • perpendicular, and here we mean this force is

  • perpendicular to the table. There are the forces in the

  • y direction. In the y direction,

  • N is a positive force, it's with a plus sign;

  • mg is a negative force, and that's equal to m

  • times ay. It's interesting that in this

  • application of the Newtonian equation, I'm going to say that

  • I know the right-hand side. I know the right hand side is 0

  • because I know this block is neither sinking into the table

  • nor flying off the table. It's sitting on the table;

  • it has no velocity, it has no acceleration,

  • vertically. So that has to be 0 because I

  • know the right-hand side. So, in Newton's laws,

  • quite often either I will know the left or I will know the

  • right; either I will know the force or

  • I will know the acceleration. So, if this is 0,

  • I come to the conclusion N = mg.

  • This is a very simple example, but this is how you figure out

  • how the Newton's laws work in this particular problem.

  • Okay, now I'm ready to introduce another force,

  • and that's the force of friction.

  • As you all seem to be aware, there is generally a force of

  • friction between this mass m, and this table.

  • So, how do we learn there is a force of friction?

  • Suppose someone said, how do you know there is

  • friction? Here is a mass.

  • What experiment will you do that tells you,

  • hey, there is another force called friction?

  • Okay, anybody tell me something you would do?

  • Yes? Student: Maybe put the

  • mass on an incline. Professor Ramamurti

  • Shankar: Okay, but suppose it's on this table.

  • But you are around, you can do what you want.

  • Yes? Go ahead.

  • Student: [inaudible] constant v;

  • in order to move with a constant v,

  • you'd have to apply force. Professor Ramamurti

  • Shankar: That's possible. Student: But moving

  • forward, over the constant v, there's no force.

  • Professor Ramamurti Shankar: That's a very good

  • answer. One answer is,

  • you find that to keep it moving at constant velocity,

  • you have to apply a force. That means the force you're

  • applying is cancelled by somebody else,

  • because there is zero acceleration,

  • there's got to be zero force. Yes?

  • Student: Just apply force to it and push it on the

  • table and it'll stop eventually, and that's how you think there

  • must be a force, that's out there.

  • Professor Ramamurti Shankar: Very good.

  • But even before that, even before it starts moving,

  • you find that if you push it, it doesn't move.

  • Right? I try to push the podium,

  • well, I don't know my own strength, but I'm going to

  • imagine, I push it, it's not moving.

  • But if I push hard enough it will move.

  • What's happening before it moves?

  • I'm applying a force and I'm getting nothing in return for

  • it. So I know there is another

  • force opposing what I do, and that's called a force of

  • static friction. So, static friction is when the

  • body is not moving. So, I'm applying a force

  • F. Let me give it some other name,

  • little f. No, that's also bad.

  • Little f was the name for friction.

  • So, I'm going to apply a force called F_me.

  • You can label these things any way you like.

  • Okay? This particular notation has

  • not caught on because people are not as interested in what I

  • think as I am. But right now I'm going to call

  • it F sub-me. I invite you to invent any

  • notation you want because there's nothing sacred about a

  • notation, at least I want to convey to you that notion.

  • So, I call it F_me,

  • I'm applying to the right, then I know that it's not

  • moving; therefore, there has to be

  • another force that cancels it. But how much is the other force?

  • Yes? Student: Equal to

  • F_me. Professor Ramamurti

  • Shankar: That's correct. So, it is not a fixed force.

  • Static friction is not a fixed force;

  • it's whatever it takes to keep it from moving.

  • It will not be less than what I apply, because then it'll move;

  • it cannot be more than what I apply because then it'll start

  • moving backwards. So, it will do what it takes to

  • keep it from moving. So, static friction is a force

  • which has a range from 0 to some maximum.

  • So, let's call it the force of friction is less than or equal

  • to some maximum. And that maximum turns out to

  • be a number called the coefficient of static friction

  • times the normal force. The force of friction seems to

  • depend on how much normal force the table is exerting on the

  • object, which in our example,

  • if you want in our example, not always, this is always

  • true, but in our example you may

  • replace it by μ_s times

  • mg, because N = mg.

  • So the force of friction seems to depend on how heavy the

  • object is that's sitting on the table.

  • But what does it not depend on that it could depend on?

  • Can you think of something else it could depend on but doesn't?

  • Yes? Student: The amount of

  • surface area. Professor Ramamurti

  • Shankar: It doesn't depend on the area of contact.

  • In other words, if this was a rectangular

  • block, it doesn't seem to matter whether it's doing that or

  • whether it's doing that. You might think,

  • hey, there's going to be more friction because there's more

  • contact. But it doesn't.

  • So, it's interesting not only what it is but what it's not.

  • Okay, this is called a coefficient of static friction.

  • So you guys should remember that the force that friction

  • applies is equal to the force that I apply,

  • up to a maximum of this. And once I beat the maximum,

  • it'll give up and it'll start moving.

  • Once it starts moving, you're going to have a

  • situation like this. I apply F_me

  • this way, and friction will apply a force which now is

  • called new μ sub-kinetic times N.

  • So, the force of friction is different when it's moving as

  • compared to when it's not moving, it's slightly less,

  • once a body is in motion. So, it's always true for all

  • situations; μ sub-kinetic is less

  • than μ sub-static.

  • Okay? So, this is how you

  • characterize friction. You study various objects and

  • you find out friction certainly depends on the nature of the two

  • surfaces. Given that constant,

  • it seems to depend only on the normal force and not on the

  • area. Yes?

  • Student: So is μ sub-kinetic less than the

  • maximum [inaudible] Professor Ramamurti

  • Shankar: Remember, μ sub-kinetic is a

  • fixed number. Student: Right.

  • Professor Ramamurti Shankar: Maybe .2.

  • μ sub-static is a fixed number, .25.

  • If it's .25 for static, it means if the normal force is

  • some mg, up to a quarter of the weight

  • is the frictional force it can apply to prevent its motion.

  • Suppose I apply a force on a body, which is originally

  • 1/10^(th) of its weight; it won't move;

  • 2/10^(th) of its weight it won't move;

  • a quarter of its weight it's a tie;

  • .26 of its weight it'll start moving.

  • Once it starts moving, frictional force will be .2

  • times its weight, not .25, because the kinetic

  • friction is less. This is just the summary of

  • friction. Okay, so far even though I used

  • vectors, most of the action is just in one or the other

  • direction. Yes?

  • Student: [inaudible] Professor Ramamurti

  • Shankar: No, it again seems to--It can

  • actually depend on many things. Friction is something we don't

  • understand fully. But it depends only on the

  • normal force, just the way I described it.

  • Okay. Now, we are going to do the one

  • problem that has sent more people away from physics than

  • anything else. Do you know what it is?

  • You haven't heard about the problem that sends people away?

  • Okay, this is called the inclined plane.

  • A lot of people may not remember where they were during

  • the Kennedy assassination, but they say,

  • "I remember the day I saw the inclined plane;

  • that's the day I said I'm not going into physics."

  • Because this is very bad publicity for our field.

  • You come into a subject hearing about relativity and quantum

  • mechanics and then--or maybe gravitation and astrophysics,

  • and we hit you with this. So why am I doing this?

  • Well, if you can do this and you find it boring,

  • you can move on. But if you couldn't do this,

  • you have no right to complain because this is the entry ticket

  • into the business. You've got to be able to do

  • this. Only then I can tell you other

  • things. Okay, so we have to.

  • We'll do this quickly. I'm watching the clock.

  • I won't linger over this, but we have to do this problem.

  • Okay? So let's quickly finish this

  • problem. So here is the problem.

  • There is a mass, m, sitting on its inclined

  • plane, it'll make some angle here.

  • You want to know what it's going to do.

  • All right, we know it's going to slide down the hill,

  • but we want to be more precise, and the whole purpose of

  • Newton's Laws is to quantify things for which you already

  • have an intuition. So the only novel thing about

  • the inclined plane is that for the first time we are going to

  • pick our x and y axes not along the usual

  • directions, but along and perpendicular to

  • the incline. That's going to be my x

  • and that's going to be my y.

  • And I ask, what are the forces on this mass?

  • Well, I've told you, first deal with contact forces.

  • But the only thing in contact with the mass is the plane.

  • The inclined plane can exert a force, in general,

  • along its own surface and perpendicular to its own

  • surface. But I'm going to take a case

  • where there is no friction. If there is no friction,

  • by definition it cannot exert a force along its own length.

  • So it can only exert a force in this direction and we're going

  • to call that N. Then, there's only one other

  • force; that's the force of gravity

  • which we agreed we have to remember.

  • Even though the Earth is not touching this block,

  • the Earth is somewhere down here, it's able to reach out and

  • pull this block down. So, let me show you the forces.

  • So here is N, and here is mg,

  • these are your forces. And with no friction, that's it.

  • That's the key to doing any problem I give you,

  • in any exam. No matter how stressful,

  • I guarantee you'll get it from Newton's laws,

  • provided you write every force that is there and don't write

  • any force that is not there. So, this is it.

  • These are the two forces, and the mass will do what these

  • forces tell it to do. So, what we want to do then,

  • since my axes are in this direction, I have to take this

  • mg and write it as the sum of two vectors,

  • one which is pointing in this direction and one which is

  • pointing in that direction. So, if I add these two arrows,

  • I get mg. That's called resolving the

  • force into some other direction. Now, the key to all of this is

  • to know that this angle, here, is the same as this angle

  • here. In other words,

  • if mg is acting down, that angle θ is the

  • same as this angle θ here.

  • This is going to be used all the time, so I'll tell you a

  • little trick that tells you how I figured that out.

  • θ is the angle between the horizontal and the incline.

  • Now, you've got to agree that if I draw perpendicular to the

  • horizontal and I draw perpendicular to the incline,

  • the angle between those two perpendiculars will be the same

  • as the angle between the two lines I began with,

  • because perpendicular means rotate by 90.

  • If you rotate both lines by 90, the angle between the rotated

  • lines is the same. The vertical is perpendicular

  • to this horizontal, and this is the perpendicular

  • to the plane. Once you got that,

  • the rest is very easy. So, in the x direction,

  • I'm going to write the following equation:

  • mg times sin θ. So, you've got to understand

  • that this part here will be mg times sin θ,

  • and this part here will be mg times cos θ.

  • You've got--This is something you better get really used to.

  • If a vector is like this and you want the force in that

  • direction, the angle between is θ, then you put the

  • cosine. The component that's adjacent

  • to the vector, is the cosine,

  • this one here is a sin θ.

  • So, I got the horizontal part in the x direction,

  • I got mg sin θ, and that's it.

  • There's nothing else. That's got to be equal to

  • m times ax. Remember ax is not

  • horizontal, ax is along the incline, downhill.

  • In the y direction I write N - mg cos θ =

  • m times ay. That's simply writing down

  • Newton's laws as two laws, one along x and one

  • along y, and putting in whatever we

  • know. What happens next is up to what

  • else we know. Well, we know that this block

  • is sliding down the hill, it's not going into the block,

  • nor is it flying out of the block;

  • it's moving along the block. That's the reason we chose a

  • coordinate to be perpendicular to the block,

  • because that coordinate, the y coordinate,

  • it's not changing. Even though the block is

  • sliding down the hill, when it's sliding down the hill

  • the usual x and y coordinates,

  • if you define them this way, they're all changing.

  • But by this clever choice, it's always at y = 0.

  • So, it has no y coordinate, there's no y

  • velocity, there's no y acceleration.

  • So, we know this ay = 0. So, this is an example where if

  • we know the acceleration, by this kind of argument,

  • from that you deduce that N has to be mg cos

  • θ.

  • Okay? Then, you come to this fellow

  • here. There we find out that

  • ax, you cancel the m, and you find it's

  • equal to g sin θ. So, that's the big result.

  • It'll slide down the hill with an acceleration g sin θ.

  • This is a good way to measure g, by the way,

  • because if you drop something, it falls too fast for you to

  • time it. But if you let it go down an

  • incline, by making θ very small so sin θ is

  • very small and the grading is very low,

  • the body can accelerate with a much slower acceleration

  • downhill, because it's reduced by this factor,

  • sin θ. Okay, here's another thing I

  • should tell you right now. Most of us, when we become

  • professionals, don't put in numbers till the

  • very end. So, if you are told the incline

  • is an angle of 37 degrees, g is 9.8,

  • don't start putting numbers into the first equation.

  • We don't do that and I know for some of you it's pretty

  • traumatic to work with symbols, and sometimes on exams they

  • say, oh, you had all these problems with symbols.

  • But I want to sell you another idea.

  • It's much better to work with symbols than with numbers till

  • the end. Why is that?

  • First of all, if you put numbers in,

  • and I suddenly tell you, "Hey, I was wrong about the

  • slope, it's really 39 degrees and not

  • 37," you're going to do the whole calculation again.

  • But this way, you come to the formula and you

  • say, "What's your θ?" You change your mind,

  • okay, that's the new θ. You change the value of

  • g, you made a better measurement, that's the answer.

  • Then, you can see if your problem had some mistakes in it.

  • Suppose you gotof sin θ.

  • You will know it's wrong because this is an acceleration

  • and that's an acceleration. The units have to match and

  • sin θ, of course, has no units.

  • But maybe you got the trigonometry wrong.

  • Maybe it's really g cos θ that should be the right

  • answer. How do I know it's sin

  • θ? You can do a test.

  • We know, for example, as the incline becomes less and

  • less inclined, you got to get less and less

  • acceleration, because it's going downhill

  • less and less, and when θ goes to

  • small values, ax also diminishes.

  • Or, if you take an incline and you really make it almost

  • vertical, and in the end you make it completely vertical,

  • the block is just falling under gravity.

  • So, when θ is 90, a becomes g.

  • So, you can test your results. So, be prepared for that part,

  • if you are in this course. I will give you some homework

  • problems in which numbers are not put in till the very end,

  • or I may not even ask you to put the numbers in,

  • because the thing is to get this formula because anyone can

  • take the numbers and put them in the calculator.

  • So, that's going to be part of our agreement that we will work

  • with symbols. And if you look at any

  • calculation we do, if you caught me in my office,

  • I'm not putting numbers in very early.

  • I mean, I cannot wait to find out what the answer is but I

  • keep the symbols till the end, then I put the numbers.

  • Then, you can vary different numbers.

  • You can say, "Well, what if gravity was

  • weaker? I go up to Denver and do the

  • experiment. Will I get a different answer?"

  • You will. So, how things depend on the

  • input parameters is interesting. Here's another interesting

  • result. What is missing in this formula

  • that could have been there? Student: Mass.

  • Professor Ramamurti Shankar: Pardon me?

  • The mass. It doesn't seem to matter what

  • the object is. That's an interesting property

  • of the result. Right?

  • And that's because m cancelled out of the two sides

  • of that equation. So, these are all interesting

  • features you would not know if you kept numbers everywhere and

  • you got a = .62 meters per second squared.

  • To me, that number doesn't tell me as much.

  • At the end, I want to know the number but the formula is very

  • interesting. All right.

  • So, now we are going to make it a little more complicated.

  • We are going to now add friction to this.

  • So, here is the block but now there's a coefficient,

  • the friction, and let's say the block

  • is--Let's ask the following simple question.

  • I think it was related to what one of you guys said.

  • Let the block be at rest and θ is an angle I can

  • vary. Maybe there is a little thingy

  • here, you can turn it back and forth.

  • Let me crank up θ and see where it'll start slipping.

  • In other words, there is a coefficient of

  • static friction between the block and the plane,

  • and I want to know, like parking a car in San

  • Francisco, what angle can it take before it starts sliding

  • down. So, let's write the forces and

  • let's write F = ma. So, the forces are again here.

  • Let me cut to the chase and write mg cos θ here,

  • N here, mg sin θ here.

  • What shall I write for friction? Student: The first one.

  • Professor Ramamurti Shankar: Do not write

  • μ_s times N,

  • because the frictional force is anything you need,

  • up to μ_s times N.

  • In other words, if θ was very small and

  • mg sin θ is very small, the frictional force will be in

  • fact mg sin θ. So, it'll equal whatever it

  • takes to keep it from slipping. But if I've cranked up the

  • angle to that maximum angle, beyond which it cannot stay

  • there, let's give a name to that angle that is sin θ*.

  • Well, at the maximum angle friction is doing the maximum

  • thing it can do; then, we can certainly say

  • mg sin θ is then equal to the static friction times

  • N, which is now

  • μ_s times mg cos θ.

  • Again, the mass cancels. So, it doesn't matter what kind

  • of car you parked on the slope, it only depends on the

  • following. But g cancels too.

  • So, whether you park your car on the Earth or park it on

  • another planet, this is going to be the same

  • restriction, which is that tan θ has

  • to be less than or equal to μ_s;

  • equal to μ_s is the critical value,

  • less than that is acceptable. Okay, that's a simple problem.

  • If you tilt it to more than that it'll topple over.

  • So, now let's take another problem where the block has

  • started moving; it's going down the hill.

  • Now, what formula do I write? You notice that I'm not writing

  • every step, every time. Each time I deal with it I'm

  • going to do some things faster. For example,

  • I already resolved gravity to two parts, without going through

  • the whole song and dance, because you know how to do

  • that. Now, I put in the friction.

  • So now, I've got μ sub kinetic not equal to 0,

  • mass is moving. You can ask,

  • "What's the acceleration now?" So, the horizontal equation

  • along x is going to be mg sin θ minus μ

  • kinetic times a normal force, but allow me to write for the

  • normal force mg cos θ, and that's got to be equal to

  • m times ax. Now, can you do this?

  • That's the test. It is not enough to watch me do

  • this and say, okay, he seems to know what

  • he's doing. The question is,

  • can you do this? Would you have thought about

  • this? Do you agree that I'm not doing

  • anything beyond Newton's laws when I do this?

  • Yes, each one of you should ask. If you cannot do that,

  • you have to deal with that now because it's only going to get

  • compounded. On the other hand,

  • if you can do this, you know everything I know

  • about Newton's laws. This is all there is to

  • Newton's laws and how to apply them.

  • So, you cancel the m. So, notice here,

  • this is the normal force, but I've taken the y

  • equation that says normal force is mg cos θ.

  • Put that here. Then, I find acceleration is

  • g times sin θ – μ sub-kinetic cos

  • θ.

  • Okay, here is the result: ax is what I'm calling

  • a now. I want to study this in various

  • limits. The first limit I take is no

  • coefficient of kinetic friction, I get back to my sin θ.

  • Well, let me try the following interesting limits.

  • I crank up the coefficient of kinetic friction,

  • more and more and more. Look what happens here:

  • sin θ and cos θ are fixed numbers.

  • If this becomes larger than some amount, than say 10,

  • what happens to the expression?

  • a is negative. And what does that mean?

  • Student: It's going uphill.

  • Professor Ramamurti Shankar: The block is moving

  • uphill. And do you buy that?

  • So, what's wrong? I mean, I took the equation,

  • right? I told you all the time how you

  • can study various limits. Something is wrong.

  • It says, if kinetic friction is very large, the block will move

  • uphill. Student: [inaudible]

  • Professor Ramamurti Shankar: That's not the

  • reason. Yes?

  • Student: It can't be greater than the static one.

  • Professor Ramamurti Shankar: No,

  • that's not the reason. I mean, you can also vary the

  • angles, maybe keeping k at some value,

  • vary the angles so that this number times that can beat this.

  • Then what happens? Yes?

  • Student: Where do you find the direction of sin

  • θ downhill? Professor Ramamurti

  • Shankar: That is the point. Okay?

  • Let me repeat his answer; that's the correct answer.

  • You're right about kinetic being less than static and so

  • on. But you can always imagine

  • keeping that number constant but changing the angle to the cosine

  • gets bigger and sine gets smaller.

  • At some point, you agree, cosine will beat the

  • sine, because sine is going towards 0 and cosine's going

  • towards 1. But the correct answer is,

  • in writing this forced law, I'm assuming the force of

  • friction points to the left, uphill.

  • That is correct only if the body is moving downhill.

  • But if you got an answer where a was negative,

  • therefore it's going uphill, the starting premise is wrong.

  • So, one very useful lesson to learn here is when you apply

  • formula, don't forget the conditions under which you

  • derived it, and don't apply the result to a

  • case where the answer does not belong to the domain that you

  • assumed you're applying it to. So, the trouble with friction

  • is the following. Friction is not a definite

  • force with a definite magnitude or direction.

  • The magnitude is definite, but direction is not definite.

  • It is uphill if you're going downhill;

  • it's downhill if you're going uphill.

  • It's like the constant opposition to motion.

  • There is force of gravity, the normal force,

  • everything else is a fixed direction;

  • you can draw it once and for all.

  • So in a formula, where the forces are what they

  • are in any context, you can apply the formula in

  • any limit. But if the frictional force

  • assumed a downward motion, don't apply it when the block

  • is going up. Okay, so I'm going to do one

  • more of these inclined planes and then pretty much done with

  • the inclined plane.

  • Okay, this is a problem of two masses.

  • There's a rope that goes over the pulley, let's call this

  • m, let's call that M, and this angle again,

  • it's θ, and no friction.

  • You can mix and match various things but let's keep friction

  • out because I'm trying to do something different here.

  • So, you want to know what these guys will do now.

  • So first guess is, well, M looks so

  • impressive compared to small m, so maybe it'll go

  • downhill. But I realize that's of course

  • fallacy because M is a symbol;

  • the value associated with it maybe 1/10^(th) that of

  • m. But if M really is

  • bigger than m, then am I assured it'll go

  • downhill? Student: The mass has to

  • be greater. [inaudible]

  • Professor Ramamurti Shankar: No,

  • no, I'm saying, if the big mass M is

  • bigger than the small mass m,

  • is that enough to say it'll go downhill?

  • Yes? Student: No,

  • it depends on the angle. Professor Ramamurti

  • Shankar: It depends on the angle because part of the mass

  • is not helpful in going downhill.

  • See, if I had a pulley like this, then you're absolutely

  • right, the big guy wins. Now, it's not clear because

  • this has got all of gravity pulling it down;

  • this has got only part of gravity pulling it down.

  • So, let's skip this direction, it's not very interesting.

  • But let's look at this direction, for which the

  • equation is the following. If I just took that block,

  • it's got mg sin θ acting down, and this rope will

  • exert a certain tension T, here,

  • which I don't know. Under the combined effect of

  • these two, I can say mg sin - T = ma, where a is

  • assumed to be positive going downhill.

  • That's one equation. Then, I come to this fellow

  • here, this block; well, it's got mg acting

  • down and tension acting up. I'm sorry guys,

  • I made one mistake here. I think you know what it is.

  • These should be M, that should be M,

  • sorry for that, and M (Mg sin θ – T =

  • Ma). For m,

  • the equation is T – mg.

  • Now, let's be careful with what I do next.

  • I'm going to say m times a, which is the same

  • a. You should all understand why

  • the acceleration in magnitude is the same.

  • We're not saying the direction is the same.

  • This is accelerating downhill, at some angle;

  • that's accelerating uphill. These are not equations which

  • are vectors; these stand for equations along

  • certain directions. This is the up/down direction

  • here, and that is up the slope, down the slope direction,

  • for that one. But I think you realize if the

  • rope is inelastic, if this mass moves down one

  • inch, that's got to go up one inch.

  • Therefore, if it goes one inch in one second,

  • so will the other guy, they'll have the same velocity

  • and they'll have the same acceleration.

  • So, that's why there's only one unknown acceleration,

  • a. Once you realize that you're

  • done because now we know what to do.

  • Whenever you see a –T and a + T,

  • you add them and you get g times M sin θ -

  • m. On this side,

  • it's going to be M + m times a.

  • Can you see that? So, let me short circuit one

  • step, and divide by M + m.

  • So, that's the formula here. And again, you got to ask,

  • does it make sense? Now you notice that for it to

  • be positive, and to go downhill, it's not enough if M is

  • bigger than m. M sin θ should be

  • bigger than m; that's because M sin θ

  • is the part that's really pulling downhill.

  • M cos θ is trying to ram it into the inclined plane,

  • and that's being countered by the normal force.

  • The second thing to notice is: can I use this formula when

  • m is bigger than M sin θ?

  • In other words, can I use this formula when

  • a becomes a negative? After all, I did all my

  • thinking saying it's going down, but can I use it when it's

  • going up? Students: Yes.

  • Professor Ramamurti Shankar: Yes,

  • and the reason? Anybody want to say why that's

  • okay here? Yes?

  • Student: No friction. Professor Ramamurti

  • Shankar: Right. Student: And friction is

  • a force, depending upon which [inaudible]

  • Professor Ramamurti Shankar: That's correct.

  • All the forces I drew here, they're not going to change

  • their mind when you change the body's motion.

  • Gravity is always going to pull down.

  • So, all these forces are fixed in magnitude and direction,

  • and therefore once you got a formula for positive a,

  • you can apply it to negative a, but you cannot do it

  • when you have friction because in friction,

  • you assume the direction of motion.

  • All right, so now I'm going to leave the plane.

  • This is it. We are finished with the plane.

  • You are not finished because, oh don't clap here,

  • I got a homework problem, by popular demand.

  • Two masses, one pulley and friction.

  • That's to really say we're finished with the plane.

  • Once you can do that, you've graduated from this boot

  • camp; you can handle anything else.

  • But that's not something I want to do in class.

  • That's something you do in the privacy of your room,

  • put all these things in and crank it up.

  • And you also have enough time. But that's a very--that's got

  • some interesting angles you should think about.

  • Okay, so now I'm going to do physics that basically involves

  • rotational motion. But let me do one Mickey-Mouse

  • problem, and let it go, because I assigned it to you.

  • The typical problem is: there's a tree here,

  • there's a tree here, and that is--there are two

  • ropes and there's a backpack hanging here with some weight.

  • You are told what that angle is; you're told what that angle is.

  • You are supposed to find the tension on these two ropes.

  • I think it's fairly simple so I won't do the details but you

  • know what you're supposed to do. You take these two tensions,

  • break them into the horizontal parts, with sines and cosines,

  • and into vertical parts; horizontal parts that cancel

  • because of this mass. This backpack is not going

  • anywhere horizontally; the vertical parts of these

  • two, which will be in fact additive, should cancel the

  • weight of the object. That gives you two simultaneous

  • equations for T_1 and T_2,

  • and you can fiddle with them, and solve them.

  • Well, by the way I should tell you, if you got some equation

  • like T_1 sin θ _1 =

  • T_2, sin θ _2,

  • or that – T_2 sin θ is 0,

  • and another equation involving T_1 with some

  • numbers and T_2.

  • Some of you seemed to have trouble the other day when there

  • were sin θs here; whereas, if these had been

  • numbers like 3 and 4, you knew how to solve these

  • simultaneous equations. But I want you to realize that

  • the solving of the simultaneous equations for

  • T_1 and T_2 involve

  • forming combinations of these coefficients to eliminate either

  • T_1 or T_2.

  • So, when you add them, one of them drops out.

  • And that doesn't depend on whether these are trigonometric

  • functions or any other numbers, they're eliminated in the same

  • way. Likewise on the right hand

  • side, instead of some numbers like 3 and 4,

  • you have the weight, W, of this mass;

  • well, if you multiply this equation by 3,

  • multiply everything by 3 and juggle them.

  • So, you should get used to eliminating coefficients which

  • are not simply numbers, which are symbols,

  • sines, cosines, but you know how to--If you had

  • a sine here and a cosine there, for example,

  • how will you combine them? Maybe you'll multiply this by a

  • cosine and that by a sine, and if you subtract it'll drop

  • out. You've got to do stuff like

  • that. Okay?

  • But it's important that in a Physics 200 Level course you are

  • able to solve those things. Okay, now I'm coming to the

  • interesting problems in which motion involves going in a

  • circle. So here's one problem.

  • This is a string on which there is a mass and the mass is going

  • around in a circle. This is like an amusement park,

  • so that they have these dangling little,

  • tiny, baby rockets and you sit there

  • and start spinning, and instead of being vertical,

  • it starts lifting up, and you want to know why it's

  • lifting up. So, if this angle here is

  • θ, and this is some mass m, and it's going around

  • in a circle with velocity v,

  • and the radius of the circle is say r;

  • we want to find some relation between this angle θ and

  • these other parameters in the problem.

  • So what do I do? I apply F = ma,

  • that's it; and I'm going to get everything

  • by applying F = ma, to this guy.

  • So, here is that mass. The tension on the rope can

  • only be along the rope. The rope is not a rigid object.

  • It cannot exert a force perpendicular to itself.

  • It can only pull it along each direction.

  • That's the tension on the rope. There is mg,

  • and that's it. Under these things it's going

  • to obey F = ma. So, what I do with this tension

  • is I trade this guy for two forces.

  • This is T. If that angle is θ;

  • let me see, that angle is θ;

  • so this is T sin θ, and this is T cos θ.

  • So, you trade that oblique T for two Ts that

  • are in the actual simple directions.

  • The fact that it's not going up or down vertically tells me

  • mg = T sin θ. Students: [inaudible]

  • Professor Ramamurti Shankar: Oh,

  • I'm sorry, yes.

  • Okay, then in the other direction T sin θ is the

  • horizontal force, that's got to be mass times

  • acceleration. At the instant--but I've shown

  • the object, because it's going in a circle, it has an

  • acceleration, in this direction,

  • with size v²/R. Therefore, if you divide this

  • equation by that equation you will get v²/Rg = tan θ,

  • and that's the angle that the string will assume.

  • So, if you know the velocity you can do this problem.

  • Okay, then here's another interesting problem.

  • When you go on these roads, if you're going in a circle--so

  • for this is your car, I'm looking at you from the

  • top, you're going in a circle. Anything wrong?

  • Going in a circle, you're accelerating towards the

  • center; someone's got to provide the

  • force, that's what Newton says. That force is the friction

  • between your tire and the road. In fact, it's a case of static

  • friction. You might think it's kinetic

  • because the car is moving, but it's not moving in this

  • direction. So, it's a static friction that

  • keeps you from slipping. So, you need a certain static

  • friction; so that N times

  • μ static -- that N is equal to mg

  • -- must be equal to mv²/R.

  • If you don't have the static friction, your car will not be

  • able to make the curve, it'll fly off.

  • So, what people have done is to find a clever way in which you

  • don't have to have any friction and you can still make the turn;

  • that is, to bank your road. So imagine, now you're going

  • into the blackboard and you're turning left,

  • and the road looks like this, going away from me.

  • Now, I don't have to draw this thing.

  • This road is a track, a racetrack going around and

  • round like this. And I gave it some angle

  • θ. This road has no friction so it

  • can only exert a force that way. But let me resolve that force

  • into two parts, one like this and one like

  • this. Again, if these two have an

  • angle θ, that is the same angle

  • θ there, and there is the mass of the

  • car this way. So, what do I find?

  • I find N cos θ = mg and N sin θ = mv²/R.

  • Therefore, if you again take the ratios you will find tan

  • θ = v²/Rg. I think this may be a little

  • fast now but you guys can do the algebra.

  • What that means is the following.

  • If you want the car to go around the bend,

  • at a certain speed, 40 miles per hour or so many

  • kilometers per hour, and the road itself is part of

  • a circle of radius r; it doesn't have to be a full

  • circle. At that instant it has to be a

  • part of a circle. Then, if you bank your road at

  • that angle, you don't need any friction to make the turn.

  • You got to understand how we beat the system.

  • If your road instead of being flat is tilted,

  • then even though the road can only exert a frictionless normal

  • force, now part of the normal force is

  • pointing to the center. In fact, this picture is not

  • drawn to scale. If I drew it that way,

  • this part of the vector is horizontal and directed towards

  • the center of the circle, on which you orbit.

  • So, that's how you do the banking.

  • In any good system of roads you will find there is banking.

  • Okay, I'm going to do the last problem because it's in the

  • homework, and maybe I'm a little late today, I apologize for

  • that. The last problem is a very

  • famous, very important, and that's the loop-the-loop

  • problem. As you know,

  • the loop-the-loop is when you come down on a track,

  • you go on a circle, and for awhile you're upside

  • down. And the famous question is,

  • "Why don't you fall down?" What's the trick?

  • Does it violate Newton's laws, and so on?

  • We'll find we can understand it fully with Newton's laws.

  • But we'll find out that in order for this thing to pull it

  • off, its speed at that instant has to have a minimum value.

  • And the application of Newton's law to this object is very

  • simple. Here is gravity.

  • What about the track on which it's going?

  • What's the direction of the force the track exerts?

  • Student: Down. Professor Ramamurti

  • Shankar: Pardon me? Down.

  • Anybody said up? A track cannot exert a force

  • upwards, down. So everybody,

  • track, mg and they're all acting down.

  • So you see, that's when you get a little worried.

  • Okay, this guy's up there, all the forces are down.

  • So, we got to ask, so what does that mean?

  • Well, I have to agree they're all down.

  • I say T + mg is the downward force,

  • that's got to be mv²/r.

  • Therefore, T = m times (v²/R – g).

  • So, what does that mean? If this T comes out

  • negative, you're dead, because if it's negative,

  • you want the track to exert an upward force and it cannot.

  • In a real amusement park they have other T-brackets and

  • so on to support you. But if you really believe in

  • the laws of physics, you don't need any of that,

  • you just got to make sure this number is positive.

  • For it to be positive, you needbigger than

  • or equal to Rg. So, you got to make sure when

  • you're on the top, you have this minimum velocity.

  • If you go faster than that, that is just fine;

  • if you go faster than that, T will be some positive

  • number. So, here's the interesting

  • thing. Have we escaped the pull of

  • gravity? How come we are not falling

  • down? And the track is not helping

  • us, it's also pushing us down. Yes?

  • Student: The velocity vector is moving towards the

  • left and the force; well, the full force is

  • changing the direction of the velocity towards the center,

  • but it's still going along a circular path.

  • Professor Ramamurti Shankar: Okay.

  • Student: You can make that for forced radials today.

  • Professor Ramamurti Shankar: Yes.

  • So, the point is, if you have an apple,

  • with two forces, both pushing down,

  • that apple is accelerating down,

  • and accelerating down means really falling towards the

  • ground, towards the center of the circle here.

  • In this example, it is definitely accelerating,

  • but the acceleration is not added to zero velocity,

  • in which case it'll pick up more and more speed;

  • it's added to a huge horizontal velocity.

  • So, in a tiny time, you give it a little velocity

  • like this, your new velocity points at a new angle.

  • That just means your body has come here and has stopped going

  • like that and started going like this.

  • Therefore, downward acceleration does not mean

  • coming closer to the ground, it only means your velocity

  • vector is changing its direction.

  • On the other hand, if you had no velocity vector,

  • downward acceleration means what you think it is.

  • That is why, if you drop an apple from here,

  • it has an acceleration due to mg;

  • if the track was pushing it down even more,

  • it'll fall towards the ground and hit the ground.

  • Here it actually has an acceleration but it doesn't mean

  • that you are any closer to the center.

  • Going around in a circle is an example of constantly

  • accelerating towards the center but not getting any closer.

  • That's the way vectors work and you should think about that.

I would ideally like to have a class with maybe just ten

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B1 中級

4. 牛頓定律(續)和斜面(4. Newton's Laws (cont.) and Inclined Planes)

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    yukang920108 發佈於 2022 年 06 月 26 日
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