 Hi, I'm Zor. Welcome to a new Zor education. Today I would like to continue talking about oscillations, but in this particular case it's not the same kind of oscillation which we were talking about in the previous lecture. The previous lecture was about reciprocal movement, like on the spring for instance, you have an object at the end of the spring and it's going back and forth along straight line. Today we will talk about oscillations which are related to rotation. So the lecture is called Irrotational Oscillations. Sometimes it's called torsional. It comes with the word torque, which is very important for rotation. Now this lecture is part of the course called Physics for Teens. It's presented on unizor.com, this is the website. There is a prerequisite course called Mass for Teens and whatever information is there is definitely a prerequisite for physics. You have to know mass before you attempt physics. Like for instance in this particular case we will talk about differential equations, which is part of the calculus presented in the Mass for Teens course. So now the website is completely free, unizor.com has no strings attached. There are no ads. So there is absolutely no financial transactions which are basically resulting in going into this website. Completely free knowledge in its purest form. So I encourage you to use the website to watch this lecture because the website contains also very detailed notes for every lecture. It also contains certain problems, exams and there are some other functionalities which are related to basically registering, but you don't have to register if you just want to listen to the lectures. Registration might be needed if you would like to get engaged into supervised education. Then there is some kind of a supervisor or a parent and then there are students which he is responsible for etc. That's a separate issue. In any case, all is free. So rotational or torsional oscillations. Well, first of all, a couple of examples. In a good old times when the watch were mechanical, right now there are many of them are electronics. So it's a completely different story. But in the mechanical world the handwatch had something which was called a balance wheel. It goes this and this like this and there is a spiral spring actually which allows this rotational oscillation. That's rotational oscillation when some kind of a disc or or weight or whatever is going around the circle. That's what makes it rotational and it goes back and forth. It goes this way along the circle and then another way along the circle. That's what makes it oscillating. So that's what oscillations comes from. Rotational oscillations. It's rotating and it's rotating in both directions alternatively. Okay, I will talk about a little bit more, I would say, simple case. As an example, my simple case is some kind of a wire. Let's say it's a steel wire and there is a rod. So we'll consider the rod to be weightless and there are two weights here of mass m. Now there is a really good connection here, rigid connection. Now it doesn't really matter whether we are talking about this particular wire to be like vertical in the gravitation field. It doesn't really matter. Consider it's in a spaceship. So my only point in this case is this is a straight wire. Let's say it's a steel wire and then what you can do, you can actually wind this particular rod around which twists the wire. Now as it twists the wire, there is certain tension in the wire. It's a steel wire so it's trying to basically untwist itself. So if I will wind this thing a little bit and let it go, the wire starts untwisting so it will go the opposite direction. It will pass the point of neutrality and goes even further by inertia. Same thing as with the spring. Spring goes back and forth. It's reciprocal movement. This goes around this way or around that way. So it will basically oscillate back and forth, back and forth. So this is an example which I would like to talk about. So we will do certain calculations and you will see that these calculations are very much like the ones with the spring. So I'll basically come up with some kind of an equation which describes the movement of these masses around the circle and you will see that that's basically the same kind of laws which are in a straight line direction. They are applicable here with some modifications because it's a rotational. So let me just remind something about rotational movement which was presented actually in this course in the mechanics part. So whatever is a force in movement along the straight line is basically equivalent to torque when we are talking about rotational movement. Why? Well torque first of all by definition is radius times force. Now in the general definition these are all vectors but in cases which we are considering we are considering the force always being perpendicular to the radius in which case the vector product which is supposed to be you know this is usually the sign for vector product. So the vector product actually becomes just a product of these two values the radius and the absolute value of force in case these are perpendicular to each other. And that's the case so that's why I will not use the vectors here and I will use just the plane multiplication here. Okay so they are playing in some way equivalent role. Force makes certain mass to move with certain acceleration right. Remember the second law of Newton. Now in terms of rotational movement we would like to use angular position and angular speed and angular acceleration. So if the angle is this it's function of time angle by which we have turned around the circle then the linear speed and linear acceleration depend not only on position of the angle itself but also on the radius right. So what A is? A is second derivative of position and speed is the first derivative of position. In this case instead of position along the straight line of certain point we have an angle which starts at certain fixed location the angle by which we have turned. And now we have if you would like we can obviously check this length. Length is equal to radius times angle and the linear speed is the first derivative of this and acceleration linear acceleration is corresponding with this right. We're just differentiating this. So the part is basically our angular speed and this is angular acceleration. So we will use A here and we will not use the first derivative at all. So I would like to have something like this in terms of torque and mass obviously and angular acceleration rather than linear acceleration. Now how can I do it? Well I will start from this since f is equal to a well mass first mass times a right. Now instead of f I would like to have torque so it's r times now instead of a I would like to have this I will use this alpha. So what do we have? This is torque so torque is equal to m r square right m times a a is r right one second. So it's m a times r okay so it's this r and this r so it's r square times acceleration. So in terms this is i called moment of inertia. So in terms of torque and moment of inertia and acceleration this is formula which is basically equivalent to the second Newton's law for rotational movement. So instead of mass we should use moment of inertia instead of linear acceleration we are using angular acceleration and instead of force we're using torque. So this is equivalent in a rotational I just derived it basically the same thing as I did during some lecture about rotational movement in mechanics. So I will use this equivalent to the Newton's law. So torque is equal to moment of inertia times alpha. Okay this is a little deviation from our test here. I just reminded you this rotational thing. So this is the first thing which I would like to talk about. This is equivalent of the Newton's law. Now if you remember in the previous lecture when we were talking about spring we used Hooke's law. Remember the Hooke's law? Hooke's law is this for linear displacement of the neutral spring. So this is a spring it's a neutral position now we're stretching it by x and the force that's minus here because the force goes to opposite stretching is proportional to displacement where k is some kind of characteristic of a spring. Obviously this is not exact law for any kind of a spring and any kind of displacement. Now it's usually about relatively small displacements but it works anyway. It's enough to make some reasonable calculations. Now what is this in this particular case? We're not stretching any springs right but we are twisting the steel wire but it's really fixed here but now we're twisting this end. It's a steel so it has certain tension whenever we are twisting it and obviously as a result of this tension the steel wire is trying to untwist itself so it actually develops some kind of a torque. It's trying to turn back the rod on which we have the masses. Now it obviously depends on what kind of a steel wire it is how much we have turned it etc etc but again there is a similar kind of law. You can say it's a rotational equivalent of the Hooke's law and it basically says that the torque is proportional to angle of twisting. So if we have turned by certain thing it will be certain torque which is directed obviously to the opposite side because it's trying to untwist itself. So this is the characteristic of the steel wire and this is an angle by which we have turned this rod and twisted the wire. It's really equivalent again it's not universal but for relatively small deviations. We are not talking about winding it and winding and basically bending the whole steel wire yet we're talking about just a little bit like half a circle for instance because that's how by the way the how's it called balance wheel in the hand watch. It's turning but not like the whole thing a few times right. It's turning by 180 degree maybe 270 degree maximum so it doesn't really bend the the spring the spiral spring inside. So here we're not bending really the steel wire we're just twisting it a little bit and then this particular law is working. Now since we are doing these rotational oscillations this is function of time of course right. So this thing is also function of time. This is acceleration. Acceleration obviously is always changing. This is the second derivative of angle displacement angular displacement. Now as we basically did exactly the same thing in the previous lecture I was equating the force expressed using the Hooke's law to the force presented the force presented through the second Newton's law. I basically equated them and that's how I got the differential equation. Here I have exactly the same thing. This is a torque and its expression as in terms of how's it called momentum of inertia and this is the angular acceleration. The same torque is expressed in terms of qualities of this wire and again angular displacement. So we can just equate them and have the differential equation which describes movement of angular movement of this rod. Now in this particular case let's do it this way. Now this is exactly the same differential equation as we had in the previous lecture for a spring but for the spring instead of moment of inertia we had mass and instead of angular displacement we had linear displacement. Basically solution is from the mathematical standpoint is exactly the same and I will use the same approach basically. I have to have some kind of initial condition. Now initial condition is that my initial turn of this rod by angle phi at time zero is something. Is it gamma? Is it angle? And let's say that I just turn it by this particular angle gamma and left alone which means I did not push it in any direction so the first derivative is equal to zero. Now the second derivative obviously is depending on the torque etc. So these are initial conditions and using these initial conditions I will just do exactly the same as in the previous lecture. I derived basically the function which satisfies this differential equation and these initial conditions. So I'll use basically the same function. I'll just write it in different letters involved here but it's exactly the same differential equation. So my function phi of t is equal to gamma cosine square root of k divided by i times t. So again in the previous lecture we had initial linear displacement a and that coefficient was a and instead of moment of inertia we had mass so that was k divided by m. So the function is exactly the same just different letters here. So this is a solution to this differential equation which means it basically describes how our angle is changing with time. If t is equal to zero we have cosine of zero it's therefore cosine of zero is one so times gamma is gamma so in the beginning. Now the first derivative from cosine is a sine with a minus sign with some coefficients and that's why if you put t is equal to zero sine of zero is zero so you will have zero so it satisfies the initial conditions and basically what we're saying is this angle is changing from basically from now cosine is changing from minus one to one so this angle is changing from minus gamma to gamma so it goes to gamma this way and then gamma that way to the opposite direction plus minus plus minus and that's how the whole thing is rotating on this particular wire and two very important characteristics are the period. Now the period is equal to 2 pi the period of cosine is 2 pi but if you are multiplying by some multiplier the period will be divided by this multiplier so I'll put an opposite i divided by k and the frequency how many oscillations per second this is 1 over t again exactly the same as in the previous lecture I explained a little bit more detailed that would be 1 over 2 pi square root of k over i so k is a characteristic of the spring not a spring on a steel wire that's basically the same kind of characteristic as in a spring in a previous lecture it's elasticity how springy or how steely this particular wire is and i is moment of inertia so basically that's it and it will start oscillating back and forth all the time obviously we are talking about ideal case when there is no friction etc etc now one very interesting observation I would like you to make the period the period and the frequency obviously there very similar to each other there is no gamma here what does it mean it means that no matter what my initial turn of the rod around this wire is the period will be the same if I will turn it a little bit the torque will not be very strong right and therefore it will be slower it will be a less distance back and forth so the gamma would be smaller which means my angular deviation will be smaller from the neutral position but linear speed also will be smaller and the time to cover from plus gamma to minus gamma would be exactly the same as if I turn it initially to a bigger angle if gamma is bigger so I turn it bigger my torque is stronger in this particular case because we are twisting the wire a little bit more and that's why the force is stronger and the speed linear speed of this mass as it's rotating around the circle would be will be greater but the time will be exactly the same so no matter how strongly I wind it in the beginning the time the period of rotating would be the same and frequency will be the same number of oscillations per second it's a very important observation and that's why actually the this type of movement is used in mechanical watches because no matter how strong our initial turn is my my wheel my balance wheel inside the watch will have exactly the same frequency of oscillations and as the spring unwinding and the force actually becomes weaker maybe we are not actually turning all the way to the maximum so but as the spring unwinds itself it loses energy right so it doesn't really turn it all the way to the right and all the way to the left let's say it's halfway but still the period of oscillations will be the same and it's very very important for watch because we have to have the same period to have the same amount of time which we are kind of measuring because with every movement of the wheel in one direction it's some kind of a fraction of a second or a second or whatever and it should be exactly the same if we would like our watch to be precise so that's a very important okay I do recommend you to go to the website unisor.com to get to the physics for teens you just click on the course there are some other courses like mass for teens and inside the physics you will have another menu this belongs to the part called waves and it's mechanical waves on the next menu and then you will find the rotational oscillations so thank you very much and good luck