WEBVTT physics/ap-physics-c-mechanics/fullerton
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Hello, everyone, and welcome back to www.educator.com.
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I’m Dan Fullerton and in this lesson we are going to talk about moment of inertia.
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Our objectives include determining by inspection which set of symmetrical objects has the greatest moment of inertia.
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Determining by what factor an object’s moment of inertia changes if its dimensions are increased by a consistent factor.
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Calculating the moment of inertia for various objects
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and stating applying the parallel axis theorem which we use to find a moment of inertia about a different point for an object.
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Let us talk about types of inertia for a moment.
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An inertial mass or translational inertia is an object's ability to resist a linear acceleration.
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When we talk about things rotating, you also know that if you try and rotate something it has some resistance to being rotated as well.
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It has a resistance to the angular acceleration, we will call that the moment of inertia or rotational inertia.
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It is an object's resistance to rotational acceleration.
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Objects that have most of their mass near their axis of rotation have a smaller rotational inertia,
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an object that have most of their mass farther from their axis of rotation.
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Now inertial mass, we are going to give the symbol M, we have been using that.
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For moment of inertia, we are going to call that capital I or rotational inertia.
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The rotational analogue is linear inertia.
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Let us start our exploration by looking at the kinetic energy of a rotating disk.
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If we look at this disk and if it starts to spin, it is pretty obvious that it is in motion.
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Therefore, it must have some kinetic energy but we do not know how to deal with that yet.
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All we have talked about is the kinetic energy of objects moving translationally at different points in space.
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Now parts of this are rotating, parts are maintaining their position.
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The way we are going to do this is, let us start by defining the direction for angular velocity ω.
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Let us assume that is spinning about its center point with some angular velocity and we will draw a point for a center here.
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Let us take and let us start by finding just the kinetic energy of a little tiny piece of our uniform disk here.
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As we do this, let us figure out that the entire disk has a radius capital R, we have a vector RI,
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to our little piece of mass of our disk MI, that is moving with some velocity at this point in time VI.
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With that, I think we can start to look at the kinetic energy of just that little tiny piece of disk.
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The kinetic energy of that little piece I is going to be ½ × its mass MI × its speed Vi².
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We also know for objects rotating that linear velocity is its angular velocity × its radius.
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We can write this as the kinetic energy of that little piece I is ½ × its mass × the square root of its angular velocity × RI².
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But then if we wanted the total kinetic energy of the entire disk, we have to add up this kinetic energy for little tiny pieces of the disk to get the total.
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Our total kinetic energy for it rotating is going to be the infinite sum for all those little pieces I of KI,
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which is just going to be the sum over I of ½ MI ω² RI².
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As I look at this, I have a couple constants here that I can pull out of the summation.
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The ½ is a constant, it looks like the ω, the angular velocity is going to be the same for any points.
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We can pull those out and say that this is going to be ω² / 2 to say our total K=ω² / 2 × the sum overall I of MI RI².
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For my next step, what I am going to do is I'm going to define some constant capital I, constant for this problem, which we are going to call the moment of inertia.
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The sum of all MR² we will call I.
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I could rewrite this then as our total kinetic energy is going to be ½ I, some of them are squared, ω².
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When I do that summation, our total kinetic energy is ½ the moment of inertia × the squared of the angular velocity.
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If I know the moment of inertia, I do not have to worry about all these little pieces.
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Notice how similar this is to our formula for translational kinetic energy ½ MV².
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Instead of linear velocity, we now have angular velocity.
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Instead of inertial mass, we are using rotational inertia or moment of inertia and we still have the ½ vector.
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We have added one more variable that switches when we go to the rotational world instead of talking about mass and we talk about rotational inertia or moment of inertia.
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How do we calculate this more generally?
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Moment of inertia is the sum of all MR² for an object or if you got an infinite sum, we can take the integral of R² × the differential of mass.
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The mass of one tiny piece.
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As you go through this unit, it is helpful to memorize some moments of inertia of common objects.
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Some of those include, well here is the general one, for anything you can use the sum overall MI R².
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For a disk ½ MR², for a solid sphere 2/5 MR², for a rod about its center point 1/12 ML², L being its length.
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For a hoop that is MR², for a hollow sphere its 2/3 mass × squared radius and for rod about its end it is 1/3 ML².
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We will derive a couple of these but as you go through the course, it is usually helpful to memorize a couple of these to save you some time on some problems.
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If you are asked to derive it, you already know the answers so you know you are doing things correctly.
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These ones you will definitely want to know, the solid sphere, hollow sphere, hoops, and disk,
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they are all good ones to have in the back your mind.
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Let us take a look at calculating MR² or moment of inertia with a couple of these.
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Find the moment of inertia I of two 5 kg bowling balls joined by meter long rod of negligible mass when rotated about the center of the rods, rotated from right there.
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Almost looks like a barbell.
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Compare that to the moment of inertia of the object when rotated about one of the masses.
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Before we even start, let us think about which one of these we think is going to be tougher to rotate.
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It is going to be tougher to rotate this about its center point or this about its end?
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Just by common sense and experience things I have seen in my life, I'm going to guess that this one is a little bit hard to rotate.
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We are going to find that out.
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Let us start here on the left, we will call this M1, this is M2, we will call this R1 and we will call R2.
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Our moment of inertia is the sum of our MR² which is going to be M1 R1² + M2 R2².
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The mass one is 5 so 5 kg in that distance if the whole thing is in meter, R1 must be half of meter so 0.5 m² + M2 5 × its length 0.5².
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Moment of inertia is going to be 10 × 0.5² or 2.5 kg m².
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Same setup, it rotates in about a different point on the right.
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Moment of inertia is still the sum of all our MR² so we are going to have M1 R1² + M2 R2² where this is going to be R1.
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It looks like R2 is going to be 0, moment of inertia is going to be 5 × 1m² + 5 × 0² or just 5 kg m².
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Same object but rotated about a different point, we have twice the moment of inertia here on the right which makes sense.
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As we said before we started here that we thought that it would be hard to give a rotational acceleration 2.
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How about a slightly more complicated object?
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Find the moment of inertia of the uniform rod about its end and about its center.
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As we do this one, what we are going to do as far as a strategy is we are going to break the rod up into little tiny pieces
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with some mass VM at some distance R from our rotation point.
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We will define the linear mass density as the total mass of our rod divided by its length.
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If we do that then, the differential of mass, the amount of mass that in that little tiny bit,
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DM is going to be the linear mass density × dx as we integrate from 0 to L.
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If we do this about its end, our moment of inertia is going to be using our formula R² dm which is going to be the integral from 0 to L.
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Our R is just our x coordinate that is x² and dm we said was λ dx.
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This implies then that the moment of inertia is going to be equal to, λ is a constant in this case so we can pull λ out of the integral sign,
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integral from 0 to L of x² dx, integral of x² is x³/3 evaluated from 0 to L which is going to be λ L³.
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But we also said that λ was M/L so our moment of inertia I is going to be, we will replace λ with M/L, we still have an L³/3.
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I can make a ratio of 1 that becomes L² so we end up with ML² / 3.
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There is the moment of inertia of that rod rotated about its end.
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Let us do it about its center.
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We have a different starting point, instead of rotating about this end, we are going to rotate it about the middle.
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Same basic calculation but set up just a little bit differently.
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Moment of inertia is R² dm which is now going to be the integral from -L /2 to L /2 calling our center point here 0 of x² λ dx,
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which implies then that our moment of inertia is going to be, we can pull our λ out again and
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we are going to have x³/3 evaluated from-L /2 to L/2, which is going to be λ /3 × we will have L /2³/3 - -L /2³/3.
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Our moment of inertia is going to be equal to λ/3 all of this is going to be equal to λ³/8 - - λ³/8, 2 λ³/8, or λ³/4.
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Once again, we can take a look at our λ which we defined as M/L.
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If λ is M/L that will be M/L, for λ we have still got a L³ and we got a 12 down here L that becomes L².
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And I end up with 1/12 ML², much smaller moment of inertia to rotate about that center point.
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You can verify those as when I said you probably have to memorize from our previous formula screen.
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All right looking at another object, let us take a look at a solid cylinder.
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Find the moment of inertia of a uniform solid cylinder about an axis through its center.
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This is kind of our soda can that is spinning through its point on the center.
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We are going to assume it has a uniform density because it is a uniform solid cylinder.
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Its volume mass density is going to be its total mass divided by its volume, which would be its mass divided by, its volume will be the area × its length which is π R² L.
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Our strategy is going to be as we integrate it, we are going to take little tiny pieces of the can and think of them as very thin slices.
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As we integrate from the tiny once all the way out, we are going to get our total cylinder.
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We need to figure out the differential of mass that is in one tiny piece of the can of that size.
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To take that, imagine we take this hollow can, what you are going to do is cut it and spread it out to find its mass.
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To do that then the differential of mass, the mass included there is going to be the area of the rectangle of material we make × its thickness.
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Its area is going to be the circumference 2π R × its length L × the mass density and then the thickness
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is going to be our little dr is we integrate from 0 all the way out to R.
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We are going to make these infinitesimally thin.
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Our differential of mass inside our little hollow, it is in that piece of little hollow cylinder is 2π RL × our volume mass density × dr the thickness.
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As if we have cut that and spread it out to make a tiny thin rectangle of material.
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Once we got that set up, the actual integration piece is pretty straightforward.
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Our moments of inertia is the integral of R² dm which is going to be the integral from R =0 to R,
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the radius of our entire cylinder of R² and our dm we just defined as 2π RL ρ dr.
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Our moment of inertia then, we can pull out our constants.
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2π is a constant, our volume mass density ρ is a constant because it is uniform, L is a constant.
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That will leave us with the integral from 0 to R of, we have here R³ dr which is 2π ρl.
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The integral of R³ is R⁴/4 evaluated from 0 to R which implies then that our moment of inertia is going to be 2π ρ L.
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We will have R⁴ / 4 -0⁴/4 which is 0.
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But we also know that our ρ is M/π R² L.
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We will substitute that in so this is going to be equal to, we will not imply, we will say that is equal to, we have 2π, our ρ is M/ir² L.
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We also still have our L R⁴/4 so this implies then that our moment of inertia is going to be, let us see what we can cancel out of here.
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We have got an L, we have got an L, we got a π and π, R² and R⁴ that becomes R in the second, 2 becomes a 2.
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I end up with MR² /2 moment of inertia for our solid cylinder.
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You get the idea of the procedure you go to define moments of inertia of these continuous or more complicated objects.
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Let us take a look at the parallel axis theorem, this is a really cool,
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helpful formula that will help you with moment of inertia when you are not talking about the center point.
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If you know the moment of inertia, I of any objects through an axis that intersects the center of mass of the object, we will call that axis L.
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You can find a moment of inertia around any axis that is parallel to that current axis of rotation we will call L prime.
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If this is our object of some sort, we know its moment of inertia through L.
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We want to know its moment of inertia through L prime at some distance D, away in parallel to that initial one.
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Assuming the initial one goes to the center of mass, the way to find the moment of inertia about L prime is just going to be the moment of inertia
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about that center of mass + mass × the distance between those two axis².
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Let us take a look at how we can use that to solve a problem.
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Find the moment of inertia of a rod of mass M and length L about one end of the rod using the parallel axis theorem.
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We have already done this sort of problem before but let us find about the end since we already know what it is, about its center.
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About its center, we know that the moment of inertia about the center of mass is 1/12 ML² around about its center.
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And this distance must be L/2 because we are going to move from here to here.
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Our distance D is L/2 so the moment of inertia of the rod about its end is the moment of inertia about its center of mass + mass × the square of its distance,
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which is, we have moment of inertia about the center of mass 1/12 ML² + mass and our D is L /2² which will be ML² / 12 + ML² / 4,
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which implies that the moment of inertia about the end is going to ML² / 12 + this will be 3 ML² /12 to give ourselves a common denominator
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which is 4 ML² / 12 or ML² /3.
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Another way you can find a moment of inertia about an object once you know its moment of inertia about the center of mass,
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as long as that new axis is parallel to the one where you know already, it is nice, straightforward, easy calculation.
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Alright, to calculate the moment of inertia of a hollow sphere with a mass of 10 kg and a radius of 0.2 m.
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Here we are going to assume that you have memorized your moment of inertia for common objects.
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So this would be 2/3 M R² for a hollow sphere which is 2/3 × mass 10 kg and our radius 0.2 m² or 0.27 kg m².
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How about for a long, thin rod?
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Find the moment of inertia for a long, thin rod with the mass of 2 kg and a length of 1 m rotating about the center of its length.
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Let us take a look and assume it is uniform so that is 1/12 ML² about its center which is 1/12 × 2 kg × 1 m² or about 0.17 kg m².
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What is its moment of inertia when rotating about its end?
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That is just going to be 1/3 ML² or 1/2 × at 2 kg × L² 1 which is just going to be 0.67 kg m².
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An object with uniform mass density is rotated about an axle which may be in position A, B, C, or D.
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Rank the objects moment of inertia from smallest to largest based on axle position.
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From smallest to largest, we are going where it is easiest to accelerate it rotationally towards its toughest to accelerate rotationally.
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As we know it is going to be easiest when you got it at the center point here, C we have the smallest moment of inertia.
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As you move away from that center point, it gets tougher and tougher C, B, D and finally if you are rotating it about A.
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I would go C, B, D, A, for the ranking of the moment of inertia from smallest to largest.
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Alright let us do one more, a uniform rod of length L has a moment of inertia I 0 when rotated about its midpoint.
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A sphere of mass M is added to each of the rod, what is the new moment of inertia of the rod ball system?
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Over here, moment of inertia is I 0, here we need to figure out its new moment of inertia.
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The moment of inertia on the right is going to be the moment of inertia of the rod + we have to add up our M².
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The moment of inertia will be I 0 + we have M × its distance from our center point that is going to be L /2.
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We got L /2² + same thing on the right hand side ML /2².
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I is going to be I initial + ML² /4 + ML² /4.
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Our moment of inertia is going to be I initial + ML² /2.
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Alright, hopefully that gets you a good start on moment of inertia or rotational inertia.
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We will be using it quite extensively in the next few lessons.
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Thanks for joining us at www.educator.com and make it a great day everyone.