WEBVTT chemistry/physical-chemistry/hovasapian
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Hello and welcome back to www.educator.com and welcome back to Physical Chemistry.
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Today, we are going to talk about the rigid rotator.
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Let us just go ahead and get started.
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Now, before I actually start discussing the rigid rotator, I want to go back and
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actually clear up a possible fuzzy point from our previous discussions.
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Let us go ahead and write that down.
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Before I discuss the rigid rotator, I would like to go back and clear up the possible confusion,
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there may not be any confusion but just in case, I like to make sure that it is pretty clear.
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A possible confusion from the previous discussion being the harmonic oscillator.
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Let me go ahead and draw out this parabolic thing for the harmonic oscillator, the energy levels.
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Just put a couple of them here.
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We have got R = 1, R = 2, R = 3, R = 4, R being the quantum number for the harmonic oscillator and
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these being the particular energy levels.
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I’m going to have a 0 point energy so let us go ahead and this is going to be R=0, 1, 2, and 3.
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Some students are a little fuzzy about what this parabola is and
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why are we using this parabola to represent these different energy levels, about what this parabola is.
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They sometimes think the motion of the particle itself, of the mass of the particle or the mass is somehow parabolic.
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And that is not true.
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It is really easy to lose your way in what it is that is going on in the big picture.
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There is so much math going on, there are so much derivation of equations, manipulation of equations,
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that oftentimes it is very simple to lose your way in terms of remembering what are we trying to do and
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what is it that we are actually trying to elucidate, what are we looking for.
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Let us talk about what is going on.
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They think that the motion of the particle is parabolic.
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The answer is no, the motion of the particle is not parabolic.
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Let me go ahead draw and a line here.
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This is a 0 point, the equilibrium position of the harmonic oscillator.
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You remember the harmonic oscillator is just a particle that moves back and forth like this.
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Let us go ahead and put A and A, this is A1, A1, or – A1.
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For the first energy, the amplitude is that far.
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We put a little bit more energy into the system, the amplitude is going to increase.
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We know that, it will just go a little bit farther .
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A3 and – A3, they correspond to these energy levels.
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The more energy you put in, the bigger the amplitude.
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That is what is actually happening here.
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The motion of the particle is harmonic, it just goes back and forth like this.
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As it deviates from the equilibrium position, what this parabola describes
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is the amount of potential energy that is being stored in the system as making its way out to its farthest point.
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As it goes here, the potential energy goes to here.
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As it goes here, the potential energy goes to here.
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As it comes back and goes to 0, the potential energy falls and all of it becomes kinetic.
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If it goes the opposite way, either here or here, the potential energy of the system rises to here or to here.
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You remember that the potential energy function of this was ½ KX².
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If I want the potential energy of the harmonic oscillator, it is ½ KX².
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This is the parabola ½ KX² is the parabola.
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It is describing how much potential energy this system is experiencing as it deviates at distance X from that.
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That is why we are using this parabola to describe it and these of course are the energy levels of higher energy levels.
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You have bigger amplitudes and the potential energy goes off into infinity.
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Do you remember when we are talking about the particle in a box?
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The particle in a box, it was just something like this.
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This particle is moving back and forth, there is no potential energy in there.
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But the minute it hits the wall, the potential energy shoots up to infinity.
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In other words, it cannot escape that box.
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Here, the potential energy rises parabolically, it does not just boom rise up to infinity that way.
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It rises parabolically.
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As you get farther from the center, there is a potential energy that is building up in the system.
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That is what the parabola represents.
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The motion itself is harmonic.
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It is just back and forth represented by a sin or cos curve.
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This just represents the potential energy function.
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I hope that clears up a little bit.
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Let me actually write that down so we have it.
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The motion of a particle is harmonic which is just a fancy word for back and forth, repeating, periodic, whatever you want to call it.
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The further the mass goes from equilibrium position, which is this position right there, the parabola describes its potential energy.
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It is a potential energy that is important in all of this.
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It is the potential energy function that actually changes the Schrӧdinger equation.
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Recall that V of X = ½ KX².
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As the particle or the mass deviates from equilibrium position at distance X its potential energy increases parabolically.
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Increases as ½ KX².
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I hope that clears a little bit up, if there was confusion.
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If not, no worries.
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Let us start talking about the rigid rotator.
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Now I diatomic molecule oscillates along this bond.
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We already discussed the harmonic oscillator but that is not the only thing that it does.
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A diatomic molecule also rotates in space, rotates this way.
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It is a rotating molecule also, so we have rotational motion.
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We can model that rotational motion by something called the rigid rotator.
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A molecule not only oscillates along its bond, it also rotates about its center of mass.
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Let us say you have a big ball here and a little ball here, the center of the mass is not going to be in the center.
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It is going to be a little closer to over here.
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It can actually end up rotating about that thing like that .
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That is all, nothing strange.
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Let us start by looking at the rotation of a single mass about a fixed center and then we will talk about 2 masses, the actual diatomic molecule.
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Let us start by looking at the rotation of a single mass around a fixed center.
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We have a center like this and we have maybe a mass like that and it is moving a circle.
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That is it, we have a nice, fixed circle that way and the radius of the rotation we will just go ahead and call it R.
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I’m going to start with the basic equation for circular motion and that is going to be S= R θ.
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S is the distance that the particle actually makes.
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It is the distance along the circle.
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R is the radius and θ is the angle and radians.
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If it travels 90° or π/2, let us say something like this.
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From here to here, S would be this distance and θ would be this fact that there.
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Basic fundamental definition of radian measure.
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S= R θ, let us go ahead and differentiate this with respect to θ.
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We have DS D θ = DD θ of R θ.
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R is a constant so it comes out.
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I’m getting a little confuse here.
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We are going to differentiate this with respect to time, my apologies.
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S is a distance and θ is an angle.
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We are going to do DS DT = DDT of R θ.
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The R is constant so it comes out and what we are left with is R D θ DT.
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DS DT is the time derivative of a distance which is the velocity.
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What we get is velocity = R and the time derivative with respect to angle rotation gives me the angular velocity RO.
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Velocity = R × the angular velocity.
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The linear velocity of the particle is equal to R × the angular velocity or the angular velocity is equal to the linear velocity ÷ the radius.
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Here, ω is the angular velocity and we have seen it before.
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Its unit is radians per second.
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Let us go ahead and go.
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We know that the equation for the kinetic energy of a particle is ½ its mass × velocity².
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Let us go ahead and actually put this for V into here.
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We are going to get ½ mass × R ω² = ½ M R² ω².
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This thing right here, the ½ M R², we refer to that as I.
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It is called the moment inertia or more commonly known as it is called the rotational inertia of the system.
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This is called the rotational inertia of the system and it plays the same role that mass does for a particle moving in a straight line.
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We just call that inertia, linear inertia, that is the mass.
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This I is the same thing for bodies that rotate.
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The rotational inertia and we also refer to as the moment of inertia of the system.
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So far so good.
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I, and then we have O².
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The kinetic energy = the rotational inertia × the angular velocity².
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That is an equation for the kinetic energy of a rotating body.
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I want to freshly do it on this page.
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That is fine, I can go ahead and do it on this page.
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We said that this is in radians per second.
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I'm going to go ahead and multiply it by one cycle.
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It means one complete cycle, one complete circle, that is 2 π radians.
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It is 2 π radians, radians cancels with radians and we are left with ω / 2 π cycles per second which is the same as the Hz.
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ν which is equal to the angular velocity ÷ two π is the rotational frequency.
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Ω is the angular velocity.
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ν which is the ω ÷ two π gives me the rotational frequency in cycles per second.
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We are just setting up some variables here, that is all.
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This right here is the rotational inertia, I’m going to keep the ½ separate.
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We will do it that way so the kinetic energy is equal to ½ I ω².
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We are ready to move on.
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For a linear system, we have the kinetic energy is equal to ½ the mass × the velocity² = M² V² / 2M.
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Just do the mathematical manipulation.
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= MV² / 2M.
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What we have is the kinetic energy is equal to the linear momentum ÷ twice the mass.
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This is just kinetic energy in terms of momentum.
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That is all, nothing strange about it.
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Momentum is mass × velocity so the kinetic energy is the square of the linear momentum ÷ twice the mass.
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Now for a rotational system, we have the kinetic energy = ½ the rotational inertial × the angular velocity².
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Again, notice the correspondents ½, ½ mass rotational inertia, linear velocity, angular velocity.
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That is going to equal I² O² / 2 I which is equal to IO² / 2 I.
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Therefore, for the rotation we have something analogous, the kinetic energy is equal to L²
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which is the angular momentum of the rotational system ÷ twice the rotational inertia.
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Again, the rotational inertia was the M R², the mass × the R.
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I is equal to M R².
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Here where L is the angular momentum.
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Notice the correspondence between linear system, and rotational systems, the angular momentum of the system.
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Nice and basic discussion of simple rotational motion, a single mass rotating about a fixed center.
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I can talk about its linear velocity and momentum.
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I can talk about its angular velocity and its angular momentum.
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I can talk about its rotational inertia.
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These are all the equations that represent it.
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I can talk about its rotational frequency and things like that.
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Let us go ahead and talk about a rotating molecule.
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We have a rotating diatomic molecule.
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Let us go ahead and set up the system here.
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We have one of the atoms here, let us make it slightly bigger.
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Let us go ahead and make this one over here a little bit smaller.
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Let us go ahead and call this mass one over here.
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Let us call this mass two and the rotation is going to be in this direction.
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Let us go ahead and pick a counterclockwise rotation.
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And its center of mass is going to be let us say somewhere around right there.
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Let us go ahead and call the distance from the center of mass to mass one, we will call that R1.
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And from the center of mass to mass 2, we will go ahead and we will call that R2.
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We will call a total distance between the masses R.
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This total distance is going to be R.
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R is going to equal R1 + R2.
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That is our basic setup of a rotating diatomic molecule.
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Atom, atom, it is going to rotate about its center of mass.
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If they happen to be the same mass, the same atom, then it is going to rotate right down in the middle.
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That is all, basic physics.
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What is interesting here is the angular rotation is actually going to be the same for both.
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I will go to the next page.
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Let me draw my molecule here, the center of mass, R1, mass 1, R2, mass 2 and this is going to be R.
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Now the velocity 1 is equal to R1 ω 1.
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We also have the velocity of mass 2,the linear velocity is going to equal R2 × ω 2.
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But the angular velocities are the same, they are rotating at the same rate.
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Our O1 is equal to ω 2.
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Velocity 1 = R1 ω and velocity 2 = R2 ω.
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We just call it ω, it is an ω 1 and ω 2.
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I have got the kinetic energy = ½.
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The kinetic energy of the systems is just the kinetic energy of the first mass + the kinetic energy
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of the second mass 1/2 of M1 V1² + ½ M2 V2², that is going to equal ½ M1 R1² ω².
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I hope can keep all of my variables straight here.
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I hope that you will be vigilant and make sure I do not make too many mistakes.
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½ M2, we have R2² and ω² so we can rearrange this.
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It becomes ½ M1 R1² + M2 R2².
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I’m going to pull out the ω² right there and this is going to equal ½.
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I'm going to call this thing ,the M1 R1² + M2 R2² I ω².
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Where I is equal to the M1 R1² + M2 R2².
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This is the rotational inertia of the two body system.
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I just manipulated some equations from the same thing.
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I started with the kinetic energy and I work my way and I wrote it, I expressed it in terms of something × the angular velocity².
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That something is what we call the rotational inertia.
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The rotational inertia of this 2 body system is this right here.
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It is the rotational inertia of the 2 body system.
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Let us see what are we going to do now.
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We count what we have.
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We have the kinetic energy = ½ the rotational inertia × the angular velocity.
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And we have this equation right here.
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Sorry, I tend to rewrite things several times.
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M1 R1² + M2 R2².
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When we look up here, we also have an equation for the center of mass.
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We know where the center of mass is.
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The center of mass is right here and there is a relationship that exists for the center of mass from classical physics.
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It is going to be this mass × its distance from the center of mass and that is going to equal this mass × its distance from the center of mass.
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We have another relationship here.
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We have the center of mass relationship which is going to be M2 R2 = M1 R1.
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We also have the relation that R is equal to R1 + R2.
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The total R is equal to the sum of the radii.
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These are the relationships that we have and I want to play with this a little bit.
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I’m going to start with this one.
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I’m going to solve for R1 and R2.
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R1 is equal to R - R2 and R2 is equal to R - R1.
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I’m going to start fiddling with these equations and see what it is that I can come up with.
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I have the center of mass relation M1 R1 = M2 R2.
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I'm going to substitute one of those other equations that I just created from R.
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It is going to be M1 and we said that R1 is equal to R - R2 is equal to M2 R2.
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I get M1 R- M1 R2 = M2 R2.
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I'm going to move this / here and then factor out.
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I get M1 R1 is equal to M1 + M2 × R2.
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I’m just going to solve for R2.
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I get R2 is equal to M1 R1 / M1 + R1.
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Interesting, let us do the other one.
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For all practical purposes, I do not really need to go through all these derivations for you.
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Basically, what I can do is I can just present the Schrӧdinger equation for the rigid rotator
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and I can present the wave functions and the energy functions for you.
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It is important to actually go through this.
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The stuff that we are going through, please know you do not necessarily have to actually know this or
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even recreate it unless your teacher asks you to.
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That is just not going to happen.
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This is just part and parcel of your basic scientific literacy.
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It is important to see these derivations to at least have seen what a derivation is,
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what kind of mathematical manipulation is involved in this.
00:28:35.200 --> 00:28:39.400
Again, this is a part of your scientific training.
00:28:39.400 --> 00:28:44.900
You need to see these techniques over and over again, that is why we are going through these.
00:28:44.900 --> 00:28:47.300
You do not have to know this, as long as you can follow it,
00:28:47.300 --> 00:28:52.000
that is all that matters and it is all that you going to be expected to do, I promise.
00:28:52.000 --> 00:28:53.700
Let us do the other one.
00:28:53.700 --> 00:29:05.200
We started again with the M1 R1= M2 R2.
00:29:05.200 --> 00:29:13.200
This time we are going to go M1 R1 = M × R – R1.
00:29:13.200 --> 00:29:16.200
We are going to solve this one for R1.
00:29:16.200 --> 00:29:22.300
We get M1 R1 = M2 R – M2 R1.
00:29:22.300 --> 00:29:24.500
I’m going to bring this over here and factor.
00:29:24.500 --> 00:29:28.900
Again, I'm going to get M1 + M2.
00:29:28.900 --> 00:29:33.800
This time it is going to be × R1 = M2 R.
00:29:33.800 --> 00:29:40.200
When I solve for R1, I end up with.
00:29:40.200 --> 00:29:47.300
It is going to be M2 R ÷ M1 + M2.
00:29:47.300 --> 00:29:49.200
I made a little mistake here.
00:29:49.200 --> 00:29:52.700
We always have M, R, and things floating around.
00:29:52.700 --> 00:29:55.800
Let me take a second and make sure we got everything straight.
00:29:55.800 --> 00:30:04.800
M1 R1/ M1 + M2, M2 R ÷ M1 + M2.
00:30:04.800 --> 00:30:07.900
So far so good.
00:30:07.900 --> 00:30:20.500
We said that I, the rotational inertia, is equal to M1 R1² + M2 R2².
00:30:20.500 --> 00:30:23.000
I’m going to take these values that I calculated.
00:30:23.000 --> 00:30:38.700
This is R, I have to make sure that all of these things are correct here.
00:30:38.700 --> 00:30:40.000
I do not want to make any crazy mistakes.
00:30:40.000 --> 00:30:41.100
So far so good.
00:30:41.100 --> 00:30:47.400
We are going to plug in R2 and R1 into the equation for the rotational inertia.
00:30:47.400 --> 00:30:55.600
We are going to end up with M1 R1² R1.
00:30:55.600 --> 00:30:57.800
I’m going to square this.
00:30:57.800 --> 00:31:15.400
It is going to be M2² R² / M1 + M2², that × that, + M2.
00:31:15.400 --> 00:31:18.400
For R2, I’m going to plug this value in.
00:31:18.400 --> 00:31:41.500
It is going to become M1² R² ÷ M1 + M2².
00:31:41.500 --> 00:31:43.700
I’m going to go ahead and pull out the R².
00:31:43.700 --> 00:31:55.000
I’m going to express this as M1 M2.
00:31:55.000 --> 00:32:02.200
M2, I’m going to separate that out, / M1 + M2.
00:32:02.200 --> 00:32:06.300
Again, this is just some mathematical manipulation here.
00:32:06.300 --> 00:32:11.000
+ M2 × M1 × M1.
00:32:11.000 --> 00:32:23.400
I’m expressing the M1² / M1 + M2².
00:32:23.400 --> 00:32:29.400
And all of that I'm going to multiply by R².
00:32:29.400 --> 00:32:35.500
I'm going to express this in a slightly different way.
00:32:35.500 --> 00:33:01.800
This is going to be M1 M2 ÷ M1 + M2 × M2/ M1 + M2 + M1 M2/ M1
00:33:01.800 --> 00:33:18.900
+ M2 × M1/ M1 + M2, all of this × R².
00:33:18.900 --> 00:33:21.600
This is I, right?
00:33:21.600 --> 00:33:29.600
Well guess what, this right here M1 M2/ M1 + M2.
00:33:29.600 --> 00:33:44.700
M1 M2/ M1 + M2 that is equal to μ, the reduced mass.
00:33:44.700 --> 00:34:02.600
I put this and this together and I get M2 + M1/ M1 + M2 × R², this is just 1.
00:34:02.600 --> 00:34:10.600
I have the rotational inertia = μ × R².
00:34:10.600 --> 00:34:18.500
Where again, μ is the reduced mass.
00:34:18.500 --> 00:34:23.000
We saw this already when we discussed the harmonic oscillator.
00:34:23.000 --> 00:34:27.600
We ended up taking this 2 body problem and reducing it to a 1 body problem.
00:34:27.600 --> 00:34:31.500
Where now, we are no longer talking about the mass of 1 or mass of 2.
00:34:31.500 --> 00:34:35.400
We actually combined it in this thing called the reduced mass.
00:34:35.400 --> 00:34:47.300
Μ = M1 × M2/ M1 + M2.
00:34:47.300 --> 00:34:58.600
We combine the two masses and now we can treat this as a single particle of mass μ rotating about a fixed center.
00:34:58.600 --> 00:35:02.000
This is what we have done.
00:35:02.000 --> 00:35:05.300
Let me go back to black here.
00:35:05.300 --> 00:35:19.900
Where μ is the reduced mass and R = R1 + R2.
00:35:19.900 --> 00:35:23.100
It is just the distance between those two masses.
00:35:23.100 --> 00:35:50.600
What we have done, we have again turned a 2 body problem into a 1 body problem.
00:35:50.600 --> 00:36:19.700
In other words, we can treat 2 masses rotating about their center of mass as a single mass,
00:36:19.700 --> 00:36:36.500
which we call μ rotating in a circle of radius R.
00:36:36.500 --> 00:36:46.700
Where R is given by this and μ is given by that.
00:36:46.700 --> 00:36:48.900
Let us go ahead and move over to the next page.
00:36:48.900 --> 00:37:07.900
Earlier, we found that the angular momentum is equal to the rotational inertia × the angular velocity.
00:37:07.900 --> 00:37:15.200
Which is just like the linear momentum is analogous to linear momentum equaling mass × linear velocity.
00:37:15.200 --> 00:37:20.700
Angular momentum = rotational inertia × angular velocity
00:37:20.700 --> 00:37:29.000
We also found that the rotational inertia is the μ × R².
00:37:29.000 --> 00:37:43.000
We also found that kinetic energy is equal to angular momentum² / twice the rotational inertia.
00:37:43.000 --> 00:37:45.700
This is very important relationship for rotating systems.
00:37:45.700 --> 00:37:50.200
The angular momentum is given by rotational inertia × angular velocity.
00:37:50.200 --> 00:37:57.000
The rotational inertia is actually given by the reduced mass × the R², the distance between the two masses.
00:37:57.000 --> 00:38:00.900
Again, we treat this as if it is a single mass.
00:38:00.900 --> 00:38:03.400
A reduced mass rotating about a fixed center.
00:38:03.400 --> 00:38:10.800
The kinetic energy of this rotational system is equal to the square of the angular momentum ÷ twice the rotational inertia.
00:38:10.800 --> 00:38:15.800
Again, the rotational inertia is given by this μ R².
00:38:15.800 --> 00:38:19.900
There is no potential term here.
00:38:19.900 --> 00:38:30.100
There is no potential energy term, notice.
00:38:30.100 --> 00:38:36.700
For some diatomic molecule that is rotating in free space, in any orientation actually does not matter,
00:38:36.700 --> 00:38:38.900
there is no potential energy term for this.
00:38:38.900 --> 00:38:57.600
There is no potential energy term because there are no external forces on the masses.
00:38:57.600 --> 00:39:02.600
If we want to, we can certainly put these molecules in a magnetic field and create some external forces.
00:39:02.600 --> 00:39:14.900
But as it is, it is just a rotating molecule, there no external forces.
00:39:14.900 --> 00:39:25.700
The orientation of the rigid rotator.
00:39:25.700 --> 00:39:29.500
When we say rigid rotator we are just talking about the thing that is actually turning,
00:39:29.500 --> 00:39:33.100
the diatomic molecule or whatever it is that we happen to be discussing.
00:39:33.100 --> 00:39:39.200
The rigid rotator which is just a molecule.
00:39:39.200 --> 00:39:52.400
The orientation of the rigid rotator does not affect the energy because there is no external force.
00:39:52.400 --> 00:39:59.800
It does not matter how it is rotating, it does not change the energy of the rotating system.
00:39:59.800 --> 00:40:28.100
Since there is no potential energy term, the hamiltonian of the rigid rotator is just the kinetic energy operator.
00:40:28.100 --> 00:40:34.800
It is just K, the kinetic energy operator.
00:40:34.800 --> 00:41:00.900
The kinetic energy operator which is going to equal –H ̅/ 2M but now it is twice the reduced mass and δ².
00:41:00.900 --> 00:41:02.800
This is just the initial discussion.
00:41:02.800 --> 00:41:07.800
In the next lesson we will continue on and talk about the Schrӧdinger equation.
00:41:07.800 --> 00:41:09.600
Thank you so much for joining us here at www.educator.com.
00:41:09.600 --> 00:41:10.000
We will see you next time, bye.