physics30

(4:12) I have four objects here and they all have different moments of inertia. They’re all going to rotate about an axis perpendicular to the blackboard, so to speak, and we are going to massage(?) each one of them to predict their periods. Let’s first go to the rod. So we first go to the rod. We have the rod here. This is point P, and here is the center of the rod. The rod has mass M, and it has length L. So we have here, Mg. I don’t have to worry about this anymore, I simply go to this equation, and I want to know what the period is of this rod, of this oscillating rod. All I have to know now, what is the moment of inertia about point P, and I know already that b in that equation equals L/2. So what is the moment of inertia of oscillation about point P? I have to apply now the parallel axis theorem, which also have to do in the exam, which says the moment of the inertia of rotation about the center of mass, which in our case is the C. the axis has to be parallel -- so this axis is perpendicular to the blackboard, and this axis perpendicular to the blackboard, plus the mass of the rod, times the distance between P and C, squared, plus M times this distance squared, so that is b squared, and b squared is one quarter L squared. What is the moment of inertia for rotation of the rod about this axis? I looked it up in the table. I happen to remember it now because I’m lecturing 8.01. Two months from now I will have forgotten. So I remember now that it is 1/12 ML², + 1/4 ML², that becomes to 1/3 ML². And so the period T becomes 2 Pi times the square root, of this moment of inertia which is 1/3 ML² divided by b Mg and b is (1/2) L for this geometry, so 1/2 L Mg. And notice indeed as I anticipated it, you always lose your M, you also lose one L here, and so you get 2 Pi times the square root of (2/3) L / g. So that is the period we predict for the rod. So let's write that under here, 'cause we are going to compare them shortly, so this is 2 Pi times the square root of (2/3) L divided by g. The rod that we have here is designed in such a way that the period is very close to 1 second. That was our goal. So T is as close as we can get it to 1.00 second. So if you substitute it into this equation, T equals one, you will find that the length of this rod, if it is really pivoting at the very end, should be about 37.2 centimeters.

(8:02) So we did as best as we can when we made this rod. There’s always uncertainty, of course -- how you drill the holes and where you drill the holes. So I will say the value that we actually achieved is 37.2, probably there’s an uncertainty of about 3 mm, so 0.3 cm. That’s what we have. That is an error of one part in 370, that’s make that, round that off, that’s 1% error in the length. Since the length is under the square root, the one percent error becomes half a percent error. So the period that I then would predict, is about 1.000 plus half of the percent, so that is 0.005 seconds. So that then has a 1/2 % error. This is my predicted period. And so we are going to make ten oscillations. Of the observed oscillations, we’re going to get a number, my reaction time is not much better than tenth to the second. So we are going to get a number there, we divide this number by ten, so we can always calculate the period then, and then we get a much-improved error, of a hundredth of second because we will divide this number by ten. And this is also going to be divided by ten. And let’s see how close we are able to get this to the 1.0 second.

(9:39) So here is the rod. Turn this on -- the timer. We offset the rod, and I will start it when it stops somewhere. Now! 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Not bad. 9.92. Well, it is in the prediction, 9.92. And so this becomes to 0.992 plus or minus the 0.01. And you see that is well within the prediction that I made. Now all these four objects were designed in situated that they have exactly the same period of 1 second. And now comes the question: How do the dimensions relate to each other in order to get them a period of 1 second? So let’s now calculate the period for the other three objects, for the ring, for the disk, and for the pendulum.

(10:54) Let’s start with the pendulum. The pendulum, center of mass is here. Here is point P, I give the pendulum length little l. You see from the backboard over there. So b equals l, that’s the separation between P and C. The moment of inertia about point P is very easy now. This has no mass, the mass is all here. Mass M, so that must be M l². So I go to this equation, so I ask what is the period of pendulum, and you're not surprised that you find 2 pi times the square root of l/g. We have seen that before, but of course, it also comes out if we do it in this more complicate way. So here we get 2 pi times square root of l/g.

(11:56) Let’s now do the ring. This is the ring pivoted about the point P here. Here is the center of mass, right in the middle, in the middle of nowhere, right here is the center of mass. And so the distance b, is the radius of the ring, so we have to calculate the moment of inertia about point P, again we use, we have to use now, the parallel axis theorem, it is the moment of inertia of rotation about this axis through point C, through the center of mass. That is MR². R is the radius of the ring. All the mass is at this circumference, all at the distance R from the center of mass. And then we have to add to that, according to the parallel axis theorem, the mass times this distance squared, and this distance is R, so you get MR². So it is 2MR². So what is the period of an oscillation? I_P -- that is 2MR² -- divided by b Mg. b is R. R Mg. I lose one R, I lose an M, and we’ve seen this before, I derived this in lecture number 21. I remember everything by lecture number, believe me or not. So the period now is the same as the pendulum with length 2R. This is the ring. So here we have 2 Pi times the square root of 2R divided by g. I’ll make a comparison very shortly, I just want to finish them all, and I now would like to do the disk. Last, but not the least.

(14:09) For the disk, all we have to do now is to calculate the moment of inertia. This was very close to you, the problem you had on the exam. Here we have the disk, solid disk, this is point P, this is the center of mass but now it’s solid. So again b is R, the separation between the C and P, and the moment of inertia for rotation about point P is now the moment of inertia for rotation about the center of mass, which you look up in a table, in the case of the exam, it was given on the font cover, 1/2 MR², that’s the moment of inertia for rotation about the center of mass. Now we have to add M b², and b equals R, so we have to add MR². So we find 3/2 MR². So what is period of oscillation? 2 Pi times 3/2 MR² divided by b Mg, b is R. R Mg. And that equals 2 Pi times the square root -- I lose my M as always, and I lose an R -- I get here 3/2 R divided by g. So let’s write that down here. So we have 2 Pi times the square root of 3/2 R divided by g. So we have them all four there, and so we can now make a meaningful comparison. We want the periods to be the same. So we can hang on to those numbers. So we don’t need this anymore. We want the periods to be the same. We already have established that the period of the rod is close to the 1 second. So we are not going to measure them anymore, we just want to compare them, whether indeed they have the same period of oscillation by making them oscillating in unison. But we want to know what the relative ratios are of these dimensions the way we designed it.

(16:35) So let’s first go to the rod then, make the comparison between the rod and the pendulum. So if you look at the rod, and we use the pendulum as our standard with length l, then you see, the period will be the same if 1 1/2 R is l. Oh sure in the rod, if 2/3 L is l. So for the rod, 2/3L =l, so L = 1 1/2 l. It has to be exactly 1 1/2 times the length of the pendulum. So this length has to be exactly 1 1/2 times this length to the center of the mass, which is the center of that billiard ball. And that’s what we tried to the best as we could.

(17:24) Let’s now go to the ring. For the ring to have the same period of 1 second, 2R has to be the same as the length of the pendulum. So 2R=l. So we can put it in here. For L, this has to be 1 1/2 l. And here, this now, which is 2R, has to be l. Very nonintuitive, that this length here is the same as the diameter of the ring. Not at all obvious. So now we go for the disk, so now we require that 3/2 R, 1 1/2 R, we want that to be l. So we want R to be 2l/3. So we want the diameter 2R to be 4l/3. And when you look at the disc, it is hard to see that is it exactly 4/3, but you can see that it is longer than the pendulum, should be 1/3 longer. But it is not as long as the rod, because the rod is 1 1/2 times the length of the pendulum. So we can now complete that picture. And we have now that the diameter here. 2R is now 4/3 l.

(19:04) And so what we can do now, we can play with them, we can oscillate them simultaneously and let’s see how well they track each other. Uh, there’s no sense in giving you the dimensions. The rod was 37.2 cm. Let me write that down, because we calculated that, so this was 37.2 cm. And I think that translates into, for the pendulum, 24.8 cm. For the others, I leave it up to you to calculate the dimensions. Yes, 25, 24.8 is correct. So, timing is not useful anymore. Let’s just see how these two go together. Let’s offset them and let them go. They go pretty much in unison. If you wait long enough, of course, you will see that there’s a difference. You can never make them exactly the same. But they track each other nicely.

(20:05) We can now also use the rod and the disk. They track each other beautifully, they both very close to 1.00 second. And we can have the disk versus the pendulum. And you see they track each other very nicely. But wait long enough, and you will see that of course the periods will be different. So this is my last word on physical pendulums. But you may see it again on the final, not maybe, you can almost count on that. I'm telling you.

(20:48) OK, I want to discuss now some other interesting oscillation, again simple harmonic. And that is liquid in a U-tube as you see there. If I have here, the tube, which has everywhere -- it's open on both sides, and everywhere the same cross section. And I put the liquid in here, in equilibrium, just like that. And the liquid has mass M, and has density ρ, the area of the tube is A, and the length of the liquid is l. So this is l. I’m going to offset it, the liquid, and I want it to oscillate. And I want to see if I can calculate the period of the oscillation. The total mass of the liquid that I have.... Total mass, is the volume, which is the area times length times ρ. I’m going to offset it, so that this is higher over a distance y, so this then is lower over a distance y. So this distance is also y. Same as that. The liquid now is here, and then I release it and it will start to oscillate. Well, when it starts to oscillate, there comes a time that the liquid, the whole liquid is going to slosh back and forth, and so everywhere in the tube, the velocity at any moment in time will be the same because the cross section is not changing. It is the same everywhere. It has a certain velocity here, v, and it's the same as the velocity here, as the velocity there, as the velocity there. And that of course is y dot. That’s the first derivative of that position here.

(23:19) I’m going to write down the conservation of total energy, mechanical energy. I assume that there is no energy loss, although there probably is some fiction inside the liquid it will probably generate some heat that will cause some damping. You will see that when we are doing the experiment. For now I will assume that’s not the case. So what now is then the total energy of the system that is sum of the kinetic energy plus the potential energy? And if we assume that that’s constant, we will be able to find the period of oscillation very shortly as you will see. The kinetic energy of the liquid is easy. That is 1/2 M times the velocity squared, and the velocity we agree is y dot, squared. Now the potential energy. I call the potential energy here, I call that U=0. When the liquid is standing here and liquid is standing here I call that the potential energy 0. The mass that is now above this level here, I call that ΔM, and ΔM is the area times y times ρ. This is how much mass there is here, that was taken the way from here, and it's put here. How much work do you have to do to take this liquid and put it there? But that’s the same when you take this liquid and put it here. And when you bring this liquid which was there, here, then you have moved it up over a distance y, and so the gravitational potential energy increases by ΔM, this is the amount of mass here, times g, times h, and h is y. Mg h, remember that? And that’s increased potential energy. So I move an amount of mass, which is delta M. I move it over a distance y. I bring it here but that makes no difference of course. And so this is the total energy, and this is now a constant. So I’m going to substitute it there, the A l ρ, so I to get 1/2 A l ρ, velocity squared, plus A ρ g, and I get a y squared, equals a constant. Because I have a y here, and I have a y there.

(26:00) We’ve done this before. This is the conservation of energy. And in order to find a period of the simple harmonic oscillation, we take the time derivative. By the way, before we do that, uh, this was ΔM, this is an A, right? Yeah, A ρ, yeah, that’s it. So we lose A, we lose ρ, and we continue with what we have. So we’re going to take the derivative versus time of this equation. This gives me 1/2, l, the 2 pops out, and the 2 becomes a 1, so I get 2 y dot. Then I apply the chain rule, so I get y double dot. I get plus g, the 2 pops out, becomes 2y, and then I get the chain rule, y dot. And that equals 0. I lose my y dot, because I have y dot in both terms. This 2 eats up this 2. And so I find that the y double dot plus 2g divided by l times y equals 0. And that was my goal. So this is clearly a simple harmonic oscillation, because this is a constant. And it will oscillate in the following way: y equals y max times cos (ωt + φ), and this is the angular frequency, which is directly related to the period. ω, angular frequency, equals the square root of 2g/l. And so the period will be 2 Pi times the square root of l/2g. So this is the period for an oscillating liquid. Notice that it is the period that you would have obtained from the pendulum, if the length of pendulum is l/2. Not at all obvious. Not at all intuitive.

(28:08) You see our setup here, I have to know what l is, and that is not so easy because of this radius here. If I measure l on the outside, it’s substantially larger than on the inside. You may not think that it’s a big difference. But it's huge, it's a 9 cm difference between the outside and the inside. If I take the average value between the two, if I take the average, I find 72 cm. And I could be off by 1. If I use this number for l, and I substitute it in this equation, then I find my predicted period, which is 1.204, and because this error that I have of 1, that will give me an uncertainty of about 0.01 second. However, before we start measuring it, and I will do ten oscillations, to get a reasonable accurate result, I want to warn you. I make a prediction that the period we will measure will probably be larger than this. And I can think of two reasons. First is, that the damping of this liquid will be huge. You will see how quickly it damps. In the past, we have never taken the damping into account, and we won’t do that in 8.01. But the damping has the effect of making the period longer, we've always ignored that, and in most of demonstrations that we did like just now, that was acceptable. May not be acceptable for the liquid. And now there is a second point that I want you to think about it.

(29:56) Is it correct that I take the averaged length? Namely the averaged value between the outer length and the inner length. I don’t think it is. I want you to think about why that is not correct. Look carefully where that l comes into my differential equation and you will probably come up with the right answer. And I claim that the actual l that we should've taken is a little bit larger, I don’t know how much larger, but it's a little bit larger. So that will also make the observed period become larger than the predicted one. So I’m not too optimistic that we will go and hit this the way we want to hit it. But that’s good, because that’s where the physics lies, that you see that there are other factors that have to be taken into account. I’ll turn this one on now, is it on now? Zeros? I’ll make it completely dark in the classroom, cause you are going to see; otherwise you can’t see the liquid. So you see the liquid now? Oh you see these equations too. Uh, OK. It’s zero. So let me try to give this a swing, a large swing it damps so enormously that I really want to get a very large swing. That’s nice. Now! 1, 2, 3, 4, 5, 6, 7, 8, 9, 10! Uh, not bad. 12.18. Not bad. We get a little bit of light, 12.18. So observed 10T, observed is 12.18. OK, my reaction time is 0.1, so T observed is 1.2… Let’s make it 2, plus minus 0.01 seconds. Oh, that’s not bad. It actually, there is a little overlap. If you add this here you can get 1 to 1, and if you subtract it here you can also get 1 to 1. So it’s not bad. I expected it to be a little bit higher. But it's close enough to be happy. Think about my l if you'd been taking it a little larger.

(32:27) Now one more very interesting oscillation: the torsional pendulum. There is a wire there, 2 1/2 meter steel wire. And it is hanging something on the bottom which we are going to offset it and then it is going to oscillate back and forth. That's called a torsional pendulum. And we are going to calculate the period of oscillation of torsional pendulum. They have wonderful properties. They're are in the way like the spring, like the one-dimensional spring. Remember the one-dimensional spring that we have a period, which is independent of the amplitude. Well, within reason, of course. If you make the amplitude too large, then you get permanent deformation of the spring. We'd never have to make any small angle approximations with the spring, as we had to do with the pendulum. Here is the pendulum, the torsional pendulum. There is a bar, and there is a weight here and there is a weight here. I will tell you more about that later. It's hanging here from the ceiling, has a certain length l. This is point P. We're going to twist it, and then we are going to let it oscillate this bar in the horizontal plane. When you look from above, you will see the bar here, see point P here, and then you can offset it over angle θ. And then it will oscillate back and forth.

(34:00) The torque relative to point P is now very similar to what we have with a spring. We have a minus sign, and again it illustrates that it is restoring. Instead of k now we have κ, which is what we call the torsional spring constant, and now we have an angle, which we call θ. So we generate the torque, which is proportional to the angle. Very similar to the linear spring whereby we generate a force which proportional to the linear displacement. Now you generate a torque, which is linearly proportional to the angle. And this is the moment of inertia about point P times α, and α is θ double dot, so we are going to get that θ double dot plus κ times θ divided by I of P (I_P) equals zero. κ by the way is the torsional constant. So we have a differential equation. It's clear that you're going to see a simple harmonic oscillation. This is a constant. So we are going to get θ equal θ_maximum, times cos(ωt + φ). It is getting boring. This is the angular frequency, angular frequency, and angular frequency is the square root of κ divided by the moment of inertia about point P. And therefore the period, which is 2 Pi, divided by ω, equals 2 Pi times the square root of moment of inertia about point P divided by κ.

(36:01) Well, how about κ? κ is a function of the cross-sectional area here A, and the length l. And it is also function of what kind of material you have. Whether you have steel or nylon makes a big difference, that's very intuitive of course. Remember that in an earlier lecture when we stressed a wire to a point that it was breaking, we dealt with the Young’s modulus. We had a wire, and we had a mass hanging at the end of the wire. And we discussed the vertical oscillation. We could stress it and let it go. And then we would get oscillation like a spring. And that spring constant that we found then was Young’s modulus, times the cross-sectional area here divided by the length. And that was kind of pleasing. The thicker the bar, the stiffer it is, the longer the wire, the less stiff it is. Well there is something very similar here, but I don’t want to go into the details of exactly how you derive the κ. It’s a little more complicated. But indeed the same is true, if you make the wire thicker, then κ will go up, and if you make the wire longer, κ will go down. That's immediately obvious. If you have a short rod. And you try to twist that rod it’s clamped at the top, and you twist it, and it’s very short, you will need a tremendous torque for ten degrees. If you make the wire hundred of meters long, and you want to twist it ten degrees it takes nothing. You can immediately see that of course the value for κ the torsional constant is a function of the length, it will go down when the length goes up.

(37:51) We have a wire here which is 2 1/2 meters long, and the thickness of the wire, the diameter, according to the manufacturer, it's a piano string, uh, the thickness is 1/25000 of an inch. And if I calculate κ to the best of my ability, and I find that κ to be very close to 4 * 10^-4 Newton meters per radian. And so all we have to do now is to calculate the moment of inertia of the system and then we can predict what the period of this pendulum is going to be, which is not my goal. You’ll see my goal is going to be a different one. Look at the bar and look at the wire. The wire is so thin that the moment of inertia relative to point P of the wire is very close to zero. Remember it is proportional to R². You can forget about that. Almost all moment of inertia is in this system. I’ll blow up that system for you, here it is, you will see it there. It has on both sides, it has 200 g. It is 0.2 kg. And it has here, 0.2 kg. And this mass is almost negligible. And this distance here is 30 cm. And this distance is 30 cm. So to a very good approximation, the moment of inertia, for rotation about that point P, this rotation, will be this mass times radius squared plus this mass times radius squared. So that it will be twice because we have 2 masses times the mass times the radius squared and that is about 0.036 kg m². And so when I use that into our equation, so I know now what κ is, at least I have a reasonable idea of what κ is, and I know what I of P (I_P) is, it’s really, almost exclusively determined by the cross bar. I will find then, using that equation that the period is very close to 60 seconds. My goal is not to prove to you that it is close to 60 seconds. My goal is to show you that for this dimension, which is very thin, and very long, that we can make the angle θ_maximum, this angle, amazingly large. We are not talking about ten degrees, or 30 degrees. We can go much further. And what I want to test with you is how far we can go. This is one of the great things in life, for you and for me it’s challenge: how far can you go and get the way with it. There comes a time that if we make the angle too large, that we permanently deform the wire. It will not come back to its original position. The same happens with a spring. If you take a spring, it is true that the period of oscillation is independent of the amplitude. But only up to a point. If you go too far, its hoops all no longer hold that you deform it permanently. Then of course, the period will become a function of the extent. And the same is true here, so if we twist it too much up, then of course we will permanently deform it and then the period will not be independent of θ_maximum. Having said that, I will like to start asking you for advice what kind of angle this direction shall we start with. What do you think is reasonable without total torture for the wire? And then we'll write down the times. So we’re now not really interested in testing the 60 seconds, but we would rather like to compare the various angles that we give it. So what is the first one we will try? Any idea? 30 degrees? What? The first try? You are cool! Man! The first one you want 6 Pi? You're out of your mind. I'm willing to go one rotation, OK? You think it’s nothing, it's peanuts for you right? OK I would like to go θ_maximum of 360 degrees, so it’s 2 Pi, and measure the period. In fact, the measure of period takes a minute, and it is not necessary, we can measure half a period. Namely, we wait until the pendulum stops, and we measure the time until it stops again. That’s half a period. Like with a spring, if it stops here and it stops there, that’s half a period. So we'll measure half a period. I don’t know what my reaction time is going to be and maybe not a tenth of the second. Because the moment that it stops is not so well defined. You will see. I’m just guessing, probably a little larger than 1/10 of a second. Let’s give it a shot. Let’s try first 360 degrees. You see, this is black and this is a little red, so we rotate it, one rotation. This is back to here where it was. Now have you seen it? 360 degrees. OK, now I'll first let it go and wait it until it stops. I always do that. Then I start the time. Then it stops again, I’ll stop the time then we have half a period. Let’s first do its own thing. Very slowly, very gently, it should take roughly 30 seconds for half a rotation, so you see now that it’s, now it equilibrium again, because we want that one rotation and so back to equilibrium. And now it is going to stop very shortly. And when it stops, I want to start the timer. Now! OK, so now it goes back, and we wait until it stops again. That gives us half a period. OK. Now! 28.75. 28.8.

(45:08) What are we going to do now? 3 rotations? 5 rotations? Three. Are we in favor of three? Who is responsible for permanent damage to the wire? You accept the responsibilities in life? 3 is a lot. 3 is 6 Pi, man. 6, I can start it because it will take a while before it stops. 3 rotations. First we have to be sure that it is more or less back at the equilibrium, it is always a difficult thing because it is so slow. Yeah, close enough. OK, 3. Shall we go clockwise or counterclockwise? Should make no difference. I went this way first and shall we go back? Yeah, you can sleep with that? OK? OK. 1, 2, 3. 6 Pi. Piano wire. OK, let’s go, I’ll start the timer when it stops. We have some time, 6 Pi. We rotate it 6, three times, and it will rotate it six times back before it stops, I hope you realize it. Was I too late? Thank you for pointing that out. So if you rotate it three times and then let it go, it first go three times back, then it’s equilibrium then it winds 3 times up again before it stops. Look, notice it’s going much faster now but the time that’s the whole thing, should be very close to that number again. 28.5. Whoop, not bad. Shall we now go all the way? What do you want to do now? Break the wire or 10 rotations? You'd love to see that, right? It'd go like mad, ten rotations. Isn't it amazing how mush faster would it go... Sh… Still 30 seconds. Must make sure that I have my equilibrium. This was not equilibrium. I know it’s somewhere here. No, it wasn’t equilibrium. I think this is it. OK, ten, right? Ready? 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Poor wire. Let it go, we’ll see what happens. When it stops, I’ll start the time. Excuse me? Thank you. Thank you for pointing that out. Look at how fast it is going. It is really going wacko. Has to do all that in 30 seconds. Now! So now it has to go back to its stopping it has to make 20 rotations now, 20 rotations in 30 seconds. Ten back to the equilibrium, and then ten to come to a hold. This is going to be your thanksgiving farewell demonstration. Now! 29.2. Fantastic. OK, have a good thanksgiving.