9. Rotations, Part I: Dynamics of Rigid Bodies
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Computeranimation
Transkript: Englisch(automatisch erzeugt)
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So, this is a new topic called the physics or dynamics of rigid bodies. So, a rigid body is something that will not bend, will not change its shape when you apply forces to it. And one example could be a dime or a coin or a meter stick.
00:21
Of course, no body is absolutely rigid. You can always bend anything. But we'll take the approximation that we have in our hands, a completely rigid body. A technical definition of a rigid body is that if you pick a couple of points, the distance between them does not change during the motion of
00:40
the body. So, the dynamics of rigid bodies in three dimensions is fairly complicated because what I want you to imagine now is not a point mass but some object with some shape like this guy. You take this, you throw it up in the air, you can see it does something fairly complicated. You can see it. In fact, we can characterize
01:05
what it does using two expressions. I think you can guess what they are. One is, if this had been a point mass, then the point mass will just go back and forth or up and down. It will do what's called translations. It will go from a place to another place,
01:22
maybe on a straight line, maybe on a curve. But as it goes, the point mass has no further information. You've got to tell me where it is and how fast it's moving. That's the whole story. But if you have a body like this, it's not enough to tell me where it is because if I throw this up in the air,
01:40
you can imagine that as it travels along some parabolic path, one instant the eraser may look like this, the next instant the eraser may look like this. So, we say that it is translating and rotating. It's a non-obvious theorem that anything the body does from
02:03
one instant to the next instant can be achieved by first translating it, then doing a slight rotation to bring it to the configuration you want. In other words, you can go from here to here by a translation followed by a rotation. Now, translation is what we've been studying all the time
02:24
because for point particles there is no notion of rotation. I take a point mass and rotate it, you don't even know it's rotating, it has no internal variables. So, that was the problem we've been focusing on. But now we're going to enlarge our study to objects which have a size and which have a shape,
02:41
therefore which have an orientation. So, it's not enough to say the football is here. The football could be here, but it's long, pointy, and could be up to down or side to side. So, what we want to do is to break the problem into two parts. We want to focus on a body which is only rotating but not translating.
03:03
Once we've sharpened our skills, we will then put back the ability of the body to translate. So, we're going to focus initially on a body that cannot translate. So, what that means is you take the body and you grab one point in it. Then you can ask, what can it do if I grab one
03:23
point in it? Well, any motion it does will be such that that point is not moving. Now, in three dimensions, if you take a football and a certain point is held fixed, it can gyrate in many ways, but at any instant you can show what it'll be doing is to be doing a rotation around an axis,
03:42
passing through the point where you're grabbing it. Because if it rotates through an axis, points on the axis don't move, and the point where you're grabbing it better lie on that axis. Now, that's also pretty complicated. You can see why I gave you a little precautionary thing, because it's hard to visualize these things.
04:01
So again, I want to take the simplest possible problem, then make it more and more complex. Now, you might say the simplest thing you've been doing all semester long is to take things in one dimension. But that turns out to be too low. If you're in one dimension and you've got a rigid body, By the way, a rigid body means not only
04:22
extended but actually rigid. For example, if you take a cat or a snake, a snake you can grab at one point, but the snake is not simply doing rotations around the one point, right, trying to bite you, trying to move around, because snake is not a rigid body. Dead snake could be a rigid body. Living snake,
04:40
definitely not a rigid body. Now, you've got to remember that I've given you a break by studying rigid bodies, but real bodies can actually not only wobble and twist and turn, they can also vibrate and they can move their arms around. That's much more difficult. So, we won't go there and we'll try to take the simplest rigid body, but I said if I take a
05:01
rigid body in one dimension, it cannot rotate at all because there's not room in one dimension to rotate. It can only go back and forth. So, to show you there is some new possibility, I have to go at least to two dimensions. Here's one example where one dimension's not enough. So, I'm going to imagine rigid
05:20
bodies which are living in the plane of the blackboard. Now, one way to do that is to take a piece of metal, just cut out a shape, and that's a rigid body. Now, of course, this has a slight thickness, because anything you make of matter has some thickness, at least one atom thick.
05:41
But imagine that thickness is negligible, so it's like a piece of metal, a piece of foil or something, but it's rigid, and this is the rigid body we're going to study. This rigid body can move in the plane of the blackboard and rotate, but we're going to grab it at one point. So, let's pick some point here. If I grab it at that point and I ask you,
06:01
what can this fellow do? So, imagine skewering it through the blackboard. You say, okay, do whatever you want. You realize that all it can do is rotate around an axis, penetrating that point. So, let's say at t equal to 0, the body looked like this.
06:23
A little later, it's going to look rotated. I have to tell you what it is doing at a given instant. I have to tell you how it's oriented. So, what we do for that is we draw through the point of rotation some line. You can pick any line you like, pick a line like that, and we say what angle
06:42
that line makes with a standard chosen direction, usually the x-axis. So, if I give you θ, I think you have to think about it and you have to agree, I have told you where the body is. You can reconstruct everything, right? Because that point is nailed, it cannot move.
07:02
You take the body, you turn it by an angle θ. In other words, here's the question. Suppose you're not in this room, you're in another room, and I want to tell you what the rigid body is doing. What information do I have to communicate to you without seeing the picture? I claim if I tell you how the line was drawn in the body,
07:21
which is something you do once and for all, a ruled line, then I say the ruled line makes an angle θ with the x-axis. I believe you can imagine in another room what this body is doing. And that's the meaning of saying I've given complete information. Of course, that's not a complete dynamical information because it also depends on how fast it's rotating. So, it's not complete, but to simply say the analog
07:41
of where the body is in one dimension is this angle θ. So, what we are going to set up, you will find, is an analogy between one-dimensional translation and two-dimensional rotations. You will find the analogy is very helpful. So, I'm going to use a piece of blackboard where I'll
08:03
keep drawing that analogy. I do this because the number of things you have to remember will be reduced if you map this problem mathematically to the problem of translation in one dimension. In other words, in one dimension, when a body is moving, x tells you where it is, right?
08:21
I give you a number, 3, a π or minus 14, you know where it is. For this rigid body, living in the plane of the blackboard, this angle θ tells you what the body is doing. So, even though the body is in two dimensions, you need just one angle θ to tell you what its orientation is. You don't need two numbers.
08:41
A single number tells you what the rigid body is doing. So, a little later, it could be rotated by an angle δθ. That's the analog of saying it moved from x to x plus δx. First thing you've got to do is how do you want to measure θ? For x, we know,
09:01
we have agreed we're going to measure in meters. θ, the standard preference for people at elementary stages, is to measure it in degrees. You measure it in degrees, and when the body completes one full revolution, you like to say it has turned by an angle of 360 degrees. Now, that is not the preferred way to measure angles in more advanced
09:23
dynamics. We're going to use something else. You can imagine what it is. You know what I'm driving at. Anybody know where I'm going? Radians. So, I'm going towards radians, and you can ask yourself, why would anybody think of a radian? What's wrong with 360 degrees? I mean, all the textbooks, I mean, all the novels say, you know, I was going down the
09:42
wrong path, then I did a 180, right? So, everyone knows. In fact, I saw a book by a violinist who said, I was going down all the wrong things, then I did a 360. So, you realize that 360 is not really a way to reform your life.
10:02
But among the physicists, it struck a chord. In physics, a 360 is really not a trivial thing to do. It turns out the particles in the world are divided into bosons and fermions, and half of the particles are bosons and roughly half are fermions. Electron is a fermion. The light quantum boson is a photon.
10:22
For all the guys who are photons or bosons, if you turn them by 360, they come back to what they are doing. Electron, it turns out, when you turn it by 360, all the numbers are going to minus those numbers, and only when you turn by 720, you come back to technically what you're doing.
10:41
So, it's quite possible this person had in mind the behavior of maybe she was a fermion, but that's what it said. I did a 360. I've caught the 360 many times, so if you use it, you better back it up with the kind of information I gave, otherwise you're just off by a factor of two. Anyway, 360 is very nice. I think the Babylonians like 360 for a lot of reasons.
11:02
It's divisible by a whole bunch of numbers. It's easy to partition into smaller and smaller pieces. I believe in some engineering circles, a circle is 400 something else, some other engineering degree. So, you can decide what a circle is, but we like a radian. So, let's ask, why do we like a radian? For that purpose,
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I want you to take a point here at a distance r from the center and let it move to a new location with an angle Δθ. You agree that if the rigid body is rotating, this guy will be traversing a circle of radius r, because it's a rigid body.
11:43
It cannot change the distance from this point. So, it'll be going around in a circle, and I'm asking you, how big is that segment, that arc length, Δs, that it has traveled when it's rotated by an angle Δθ? So, one way to calculate that
12:01
is to say, look, if I did a full circle, I would have done 2πr. r is the distance of this point in the rigid body from the center. I would have gone at a distance 2πr, but I've gone Δθ, so the fraction of the circle I've covered is Δθ over
12:20
360, if θ is measured in degrees. So, let me indicate that in this fashion. So, we can write this as Δθ times r, π over 180 Δθ times r. If you measure θ in
12:46
degrees, this is what you have to do to convert an angular traverse to an actual distance along the tangent to the circle. The linear distance you travel is connected to the angle by which you rotate by the
13:00
distance r from the center and this number here. So, people say, look, why don't we make life easy by calling θ times π over 180 as a new angle? Let's call it Δθ times r. So, Δθ,
13:20
without any little zero on top of it, stands for radians from now on. It's related to θ, so without the Δ's, θ itself is θ in degrees times π over 180. Then, the advantage is the
13:41
distance you travel, linear distance you travel, bears a simple relation to the angle you travel by the arm length r from the center. So, you don't have to carry this 180 and π and everything, you can write the simple result. But now, when you do this, you are measuring it in radians. So, you've got to have an idea
14:01
how big is a radian. Well, a full circle, if you go a full circle, the distance you travel should be 2πr. So, a full circle will be Δθ equal to 2π. So, a circle is with 2π radians. 2π is like 6, so 1 radian is roughly 1 6 at 360, which is 60 degrees.
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The real answer is close to 58 something degrees. It looks like an odd thing to pick. It's odd only if you start with 360 degrees, but if you met people from an alien culture, you just got to go to the supermarket to run with these people. Those people may not be using 360 at all. 360 is a human artifact.
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Nothing special about 360 that's independent of our own cultural biases. It may be connected to how many fingers we have, it may be connected to how long a year is, and so on. But there's nothing natural about the number 260. The 2π, on the other hand, I believe, would be discovered by any advanced civilization. If you haven't figured out π, you're not advanced civilization. If you haven't figured out
15:01
two, you're way behind some people. So, you know two and you know π and that's what other cultures and other planets and galaxies will be using. That's the unit we like to use. You'll find over and over people using units which are natural to us, but later on, if you want to communicate with people in another planet,
15:20
you will have to use different lengths. For example, the height of a human or the length of Napoleon's arm, they're not very good units. If you don't have Napoleon in your background, you don't know what you're talking about. But if you measure distances and how big it is in terms of an hydrogen atom, hydrogen atoms all over the universe, that's a natural length scale to use.
15:41
This is a very popular issue in the old days when they sent off a rocket or they buried something underground for aliens to find out. You want to show how tall we are, no use giving the answer in meters or feet or anything. But if you can somehow relate your size to the size of a hydrogen atom, then that number will make sense to people in all civilizations because hydrogen atoms conveniently are
16:02
manufactured and scattered all over the universe and they have the same size. Okay, anyway, that's the long thing about radians. That's why we like to use radians. You have to know a few popular angles in radian measure because I'll be using only radians most of the time. You've got to know a circle is 2π,
16:21
π is half a circle, and π over 2 is a quarter a circle. So, when you turn by 90 degrees, it's π over 2. So, that's something you should know. Once you write it this way, let's get another result that's very useful. Take the distance it travels along the tangent by the time over which it does that.
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This is the actual tangential speed, the magnitude of the velocity in the tangential direction. I think you all agree with that. That's what you and I would mean by speed. If at an instant, this piece of the sample just broke and flew off, it'll keep flying with that
17:02
speed. This thing, we have a symbol for it, ω, and ω is called the angular velocity, and it's measured in radians per second. So, you can calculate angular velocity for various things.
17:24
For example, what's the angular velocity of the Moon as it goes around the Earth? Well, angular velocity will be one revolution every 28 days, right? One revolution is 2π radians. At 28 days, you do 28 times 24 times 3600. It's that many radians per
17:44
second. Oh, there's a chainsaw blade and the chainsaw is spinning at some number of RPM, maybe 3000 RPM. You can take revolutions per minute to revolutions per second. But you've got to remember, every revolution is 2π radians. So, the angular velocity of a
18:02
chain saw blade that's spinning with frequency F is 2πF. Do you understand that? F is the number of complete revolutions it does. Therefore, you pick a point on it, it does a traverse of 2πF radians. So, the dictionary has got a
18:21
second member here, which is the velocity, which is dx dt, is now replaced by angular velocity, which is dθ dt. This is an analogy, by the way. This is to help you say ω is like what velocity used to
18:42
be in the old days when things were moving back and forth. But now they're going round and round, but even here there's a notion of velocity. But it's not in meters per second, it's measured in radians per second. Now, here is something you guys got to keep very clear in your head. I can only mention it to you, but I cannot do it for you. This ω is an analogous thing to
19:00
the original ordinary velocity, but there's also an actual velocity here, the good old velocity. That's the tangential velocity of points which are part of a rotating rigid body. That is not part of my dictionary, but it's a very useful result to know the tangential velocity is the angular velocity times the distance of that point from the center of the rotation. So, the rigid body as a whole
19:25
has a single angular velocity. That's what it means to be rigid. The whole thing is rotating. You cannot say the angular velocity is so-and-so here and so-and-so there. It's a property of the entire body. But the linear velocity is not the same everywhere. Points near here in the
19:40
same time go that far, and points near here in the same time go that far. Therefore, the linear velocity increases from the center. The angular velocity is constant, and a rigid body cannot have two angular velocities. It can have only one angular velocity. Okay, then what we do is we say, okay, let's take a problem
20:02
where the angular velocity is itself changing. Fine? That's the analog of saying that it's the rate of change of velocity, and here we have a quantity called α, which is the rate of change of angular velocity, or the second rate of change of the angle. And that's denoted by α. So, if you took a change,
20:26
and you took one of these rotating circular saws, and it was spinning at a certain speed, and due to friction, let us say, it was slowing down, or when you turn on the motor it's speeding up. Suppose you just turned on the saw motor and it's speeding up in its rotation.
20:42
You've got to be very clear that the acceleration of that, of parts, this is the saw blade, okay? I'm trying to draw for you one of these saw blades, and it's spinning. If you pick a point, say, on the circumference, you remember long back we
21:02
learned, even if it's going at a constant speed, constant angular speed, it has a tangential velocity which is ω times r, and because of that it'll have an acceleration towards the center, centripetal acceleration, which is v square over r, and you can write it as ω square r square over r, ω square r.
21:23
That's one acceleration it has even if it's spinning at a constant speed. But I'm saying it can also have a tangential acceleration if on top of everything else the angular speed itself is changing. So, let's take our time digesting that. This is not too hard, but I want you guys to
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understand. If it's spinning at a steady ω, we have learned anything going in a circle has an old-fashioned linear acceleration toward the center, which is v square over r, but we have realized v in the tangential direction is ωr, so if you put that in, if you want you can call it v tangential, that gives me ω
22:00
square r. That's just the v square over r. But if you put the brakes on this by speeding it up or slowing down, it'll have an addition, an acceleration in this direction. So, an acceleration of a point on the edge or anywhere else can have two components. The tangential one will be there only if you speed up or slow down the rotating disk, but the radial one will
22:23
always be there as long as it's rotating. Okay, once you've got this dictionary, you can have a whole bunch of analogous quantities. For example, if you focus on problems where the body has a constant angular acceleration α, what can you say?
22:43
Well, you can already say that θ will be the θo plus the initial angular velocity plus ½ αt square. This is because θ is a constant. If x had a constant acceleration a, we all knew x was equal to
23:02
the starting x plus initial velocity times time plus ½ at square. Similarly, the angle by which it rotates will be the initial angle, initial angle velocity and time, plus something involving t square. This is only if α is a constant. Just like I hope you guys
23:21
know, this is true only if a is a constant. All right, now I'm going to pursue the analogy and ask, what is the kinetic energy of a rigid body when it's rotating?
23:41
Yes, here? Here? Oh, here? This one? Okay, I'm saying if you have a rotating disk,
24:03
take any point on the rotating disk. Even if the disk is going at constant angular velocity, it has a tangential velocity, and the direction of the velocity is constantly changing. This is the oldest thing we learned. So, it has an acceleration towards the center. That's not an angular acceleration. It is just a standard
24:21
acceleration toward the center of anything going in a circle. That's always going to be there, but on top of it, if you speed up or slow down this rotation rate of the disk, you'll also have a tangential acceleration, which is α times the r. Not clear yet? Not clear yet? That's the acceleration lay
24:44
people would know. If you're on a rotating platform and somebody sped up the rotation rate, you would notice you're going faster. That's the acceleration in the tangential direction. The acceleration toward the center is a more sophisticated one, we have learned, and it's due to the fact that velocity is a vector and not a scalar. So, if the direction is changing, there's an acceleration
25:02
we associate with that, and it's as real as any other acceleration. It needs a force to make it happen. Maybe the better way to ask you is this. Suppose this little speck here is you, okay? You happen to be stuck on a rotating chainsaw blade, and you're clinging to it for
25:23
dear life. What force do you have to apply to stay onto that rotating blade? This may not happen to you and me, but if you're an organized crime, this is not a very unusual situation. You can find yourself intermingled with all kinds of machinery, and you want to know
25:40
what do I do to stay on? Well, Newton tells you the answer. Newton says, if you want to be on this rotating chainsaw blade, your acceleration toward the center has to be paid for, and if the blade is speaking of speed, that's got to be paid for too. That's the net acceleration. That times your mass will be the force that has to be applied to you.
26:01
So, you will cling onto it with that force, and that force will be applied in turn by the third law, by the disk on you. For example, suppose it's a top view of a rotating platform. You're just standing there and the platform is rotating. You can ask, what force does it, what frictional force does it need, do I need,
26:22
to stay on the platform? Well, you need the force to provide you an ma, and I'm telling you, your a has two parts. Even if the merry-go-round is spinning at a steady rate, you will need to have a frictional force directed towards the center to bend you into a circle. Now, if on top of it, the merry-go-round starts increasing its speed or decreasing its speed, you'll also need a tangential
26:42
component of force to provide the tangential acceleration. Okay, now this is kinematics of rotation at constant angular acceleration. It's identical to this one mathematically, because mathematically, if you know d two theta dt squared is alpha,
27:02
the answer is just like d two x over dt squared equals a. You change the symbols, you get the same answer. So, you can imagine a variety of problems I don't feel like doing. I think you guys can easily do that. I'd rather focus on more difficult ones. For example, a chainsaw blade spinning at 3600 RPM is decelerated at some rate
27:22
α. How long does it take to come to rest? Well, for that you will write the second formula, ω equals ω zero plus αt. That's the analog of v equals v zero plus a t. And you start with some ω knot, you put on the brakes, α will be some negative number, you see at what time this
27:42
ω goes to zero. That's when it'll stop. Another result which I will use today, in fact, is the following. ω square is ω zero squared plus two α theta minus theta knot. That's the analog of v square equals v zero square plus 2ax minus
28:03
x knot. This says if the body has an acceleration α, then it's going to pick up speed, of course. And the final speed square is related to the initial angular velocity square plus two times the acceleration times the angle rotated.
28:21
So, this is the analogy. So, we don't have to reinvent all of these things. They just come from the fact that things which are mathematically the same have the same mathematical results. They obey the same equation, they have the same results. All right, now let's find the kinetic energy of a rigid body. It's got mass, and if it's spinning, all the little atoms making
28:42
up the body are moving. They've got their own ½ mv squared, and we want to ask, what is the total ½ mv squared summed over all the particles? For that purpose, it's convenient to take the following simpler rigid body. We take a rigid body by taking a mass m1 at a distance
29:02
r1, connect it to some central hub with a massless rod, and you take another mass, and you take a third mass. This is m2, it's at a distance r2, this is m3 at a distance r3, and so on. You can make a rigid body by modeling it in this simple
29:22
fashion. If you are a little more abstract, you can take a continuous rigid body and do the same thing, but let's start with the simple thing. Here's a rigid body, three rigid rods with masses m1, m2, m3, and it's nailed here, and there is a skewer going through the board,
29:41
and the things can only rotate around that. By the way, I should tell you the convention for ω. In one-dimensional motion, velocity is positive if you're moving to the left, and negative if you're moving to the left. With angular velocity, we've got to have a convention on what's positive and what's
30:00
negative. This thing which is spinning can only spin clockwise or counterclockwise. It turns out to be that ω is positive for counterclockwise. That just happens to be the convention people use. In fact, you notice the way I drew the rigid body, when I did Δθ, I had it rotating counterclockwise.
30:23
That's the preferred direction for measuring the angle θ from the x axis. Okay, now let's take this rigid body. It is rotating with a common angular velocity. I hope you all understand. That's the meaning of being rigid. So, everybody's got the same angular velocity ω.
30:41
So, what's the kinetic energy of this object? Well, it's the kinetic energy of each mass summed over all the masses I, mi, vi squared. Now, what's the velocity of each object? If you take this body, m1, its velocity is
31:00
necessarily perpendicular to the line joining it to the point of rotation. This one is moving this way. It cannot move in the line joining it, because if it is, it's not a rigid body. A rigid body, all distances are maintained. So, you don't come near the center and you don't go far from the center, and you don't go far from the center, so all you can do is rotate around the center. So, the velocity of this guy,
31:22
v1, is equal to ω times r1. The velocity of this guy, v2, is the same ω times r2, and so on. So, you can then add it all up, and what do you get? You get one-half sum over i, mi, ri squared ω squared.
31:43
And you have to do the summation over all the bodies. In my example, summation goes from 1 to 3, you may have 1 to infinity or 1 to a million, it doesn't matter. That's the kinetic energy. You guys follow this? This is not a new kinetic
32:00
energy. This is what we know to be kinetic energy, just one-half mv squared for every piece that makes up the rigid body. For convenience, I've taken the rigid body to be made up of a discrete set of masses. But now, you notice, as you go from 1 to 3, m1 and m2 and m3 are three different numbers. We don't know what the masses are.
32:20
r1 and r2 and r3 are three different numbers. We don't know how each one could be at a different distance from the origin. But the ω doesn't have a subscript i. There's nothing called ω for this one and ω for that one. There's only one ω for everything, so you can pull it out of the sum and write it this way. Then, the answer we write as identical to
32:43
r1 and r2 is equal to one-half i ω squared. And we give this i a name. i is called the moment of inertia.
33:02
So, continuing with the analogy, this board is going to be used for analogy, we used to have k equal to one-half mv squared. Yet, we're going to have k equal to one-half i ω squared. And i is called the moment of
33:21
inertia. So, the moment of inertia is determined not only by the masses that make up the body, but how far they are from the center. If all the masses just fell on top of the center, the body would have no moment of inertia. It'll weigh the same, but the moment of inertia would vanish. And likewise,
33:41
if the masses spread out, the moment of inertia is more. For example, if I'm standing around here and you come along and decide to spin me. If I'm standing like this, you try to spin me. If I stick my hands out, my moment of inertia has changed, because some of my mass now is further away from the axis of rotation. So, moment of inertia depends
34:03
on the arrangement of the masses. We're taking a rigid body. It's not like ice dancers who do this. We're talking about a rigid body, therefore moment of inertia is constant. But it requires a calculation to find the moment of inertia.
34:20
And here's another important thing. If I decide to rotate the body in a new location, say around this point, moment of inertia will change, because all the distances r will change. The moment of inertia, if someone says, here are the masses, here's where they are, please find me the moment of inertia, you will say, I cannot do it. You will say,
34:42
I cannot do it till you tell me the point around which you plan to rotate the body. Unless that is given, I cannot tell you what the moment of inertia is. So, the moment of inertia is with respect to a point. There's nothing called the mass with respect to the point.
35:01
The mass is just the mass. The moment of inertia depends on the point around which you're computing the moment of inertia. It's a variable. It's also worth knowing that if the body is being rotated about mass m three, then m three doesn't contribute to moment of inertia. It's out. Anything sitting on the axis doesn't contribute. Things further away contribute more in proportion to their mass
35:23
and proportion to the square of the distance from the axis. So, this is my dictionary here. So, who is playing the role of mass? Mass is played by moment of inertia. It's got units of kilogram meters squared, and no one has given a name for that.
35:41
For example, Newton times meter is called a joule. You can ask, well, this is kilogram times meter squared. Maybe it's named after somebody. So far, it's not been named after anybody. It's just called Newton meter squared. So, this is going to be the analog of mass in our world of rotations.
36:04
You remember, in the world of rotation, we defined something called momentum, which is mass times velocity. Here, we're going to call something called angular momentum, which is going to be I times ω. This I is called the angular momentum. One extremely important point
36:31
that you guys should notice is that all the concepts I'm using, like mass and energy, they are the same old concepts that we learned earlier on. The fact that it's a rotating
36:42
body doesn't change anything. Kinetic energy still means the same. This k, though it looks so different, is in fact the ½ mv² of every part of the body summed over the bodies. All right, so this L is called angular momentum. It'll take a while to get used to this, but right now, it's a product of the moment of
37:01
inertia and the angular velocity. So now, we are looking for the most important equation in mechanics, which is F equals ma. That's going to be equal to some mystery object. Now, you should be able to guess what goes on the right hand side. Okay, you should know by now,
37:22
by analogy, it can be written as dL dt, if you like, the rate of change of momentum, or it can be written as dL dt times I times α. It doesn't matter. I hope you understand that if you take rate of change of momentum, you will get ma, because when you take the derivative, m is not changing. Likewise, here, if you take
37:42
dL dt, I is not changing, ω is changing to give you dω dt is α. I is not changing because as the body rotates, of course particles in the body are moving around, but the distance from the point of rotation is not changing. Even though the bodies are moving to new locations,
38:01
the r squared is the same, because they're just rotating. That's why in the time derivative, you don't have to take the time derivative of the moment of inertia. But we don't know who's going to play the role on the left hand side. Okay, so we're going to find that. We're going to find out the analog of that. So, I'm going to tell you,
38:22
there are many ways to find this out. But it's a lot more subtle than generally appreciated on what the correct law is. You can hear the law stated quite often, but some of the reasonings behind it are incorrect. So, let me tell you one way to get to the law. I hope you know what I'm
38:42
looking at. I'm trying to find out what is the entity that's responsible for the rate of change of angle momentum, or what's the analog of the left hand side for ma. I know the right hand side is Iα by analogy. Who's on the left hand side? First of all, it's pretty clear that if you
39:00
want the angle of momentum of the body to change, ω has to change. You can see if something is spinning around at some fixed ω, you want to change it, you've got to push or pull or do something. So obviously, you're going to apply a force, but the thing that comes to the left hand side is not simply the force, but something else. And we'll find out what the something is. And that's done by the following device.
39:21
I'm going to use the good old work energy theorem that says when force acts on a body, the work that it does, which is the force times the distance it travels, is equal to the change in the kinetic energy.
39:43
That is the standard work energy theorem that we established. So, if you took a rigid body like this, and let's apply it to the rigid body, I'm just going to take a simpler case of a rigid body with these two masses, m2 and m1, it doesn't matter. I apply a force now,
40:02
F, perpendicular to that body at a distance r from the point of rotation. And let it turn the body. During that time, let's say the body moves by
40:22
an amount Δθ. Maybe I should draw you a better picture for this because this is very, very important. So, here is the body. I just took a simplest one with two masses, m1, m2. I'm just going to push it here
40:40
and I decide to push it in this direction. And it's at a distance r from the point of rotation. And let it turn by a small angle and come to this position, and this angle is Δθ.
41:00
What is the work done by this force? I'm going to write it down, but I want you to think about it. The work done by the force is the force multiplied by the distance over which it has moved. And basically, I'm asking you, what is that distance it has moved? And if you remember anything at all from what I did earlier, it's approximately this arc
41:21
length, which we have seen, is r times Δθ. In other words, if you apply a force F and you rotate the body by Δθ, the distance you actually move is rΔθ. So, this is force times distance.
41:45
Now, that's got to be equal to the change in kinetic energy. It's got to be equal to the change in kinetic energy. Kinetic energy, I remind you, is ½ I ω square. Therefore, the change in kinetic energy is going to be ½ I times ω
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square minus the initial ω square. By the way, these are two very neighboring instance in time. If they're neighboring instance in time, during that brief period, we may take the body to have a constant acceleration.
42:22
Therefore, α is constant. So, this formula you can also calculate when α is constant, period. In a real body, when it's rotating, α need not be constant, but for a small enough interval, as small as you wish, we may apply the formula that works when α is a constant. In that case, you will write it as
42:42
½ times 2α times the rotation angle. Right? Look at this formula here. I'm referring to this result. ω final square minus ω initial square is 2 times α times the change in angle.
43:00
So, I want you to then equate this work done to this change in kinetic energy. I want you to equate this guy to this guy, and we get our great result that F times R is equal to ω times α.
43:20
Iα is ma. That's what I'm looking for. So, this fellow here, Fr, is the analog of force, and it's called a torque. It's very important to notice that the torque due to this force is the value of the
43:41
force times the distance from the point of rotation. And it's only the external force that I'm talking about. There are also internal forces in a rod. Every part of the rod is being dragged along the way in another part of the rod, but that analysis in terms of forces can be done, but it's very tricky and
44:00
complicated. It's much easier to look at the energy where the energy change is only due to external forces. If you only look at forces, you've got to be careful. This pivot point is also applying a force on the rod, but it's not moving and not doing any work. So, the only external force is the one I'm applying, and it's responsible for the change in the kinetic energy. There are other forces inside
44:23
the rod. In other words, this mass is going to pick up speed, not because I'm directly pushing it. I'm probably pushing this part of the rod and that's pushing the part next to it, and it propagates and then finally pushes this one. So, I'm not directly applying the force law to this mass. I'm saying whatever is happening internally, I don't care.
44:41
The external force times distance is a change in the kinetic energy of an object. Yes? No. I said this quantity, the change in kinetic energy is the final kinetic energy,
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½ ω square minus the initial kinetic energy, where final initial referred to when the rod was there and when the rod had rotated by a small amount. Okay? Right?
45:23
Here. Now, I use this equation. The change, the ω squared at the end minus the ω squared at the beginning is 2 times the acceleration in angular acceleration times θ minus θ times the change in theta over that time, right? In other words,
45:42
that formula can be used for even finite periods of time, provided α is constant. We are applying it for a tiny period of time, small enough for us to believe that the acceleration during that tiny period is some number α. So, this ω squared minus ωr squared is really if you want a change in ω
46:01
squared, and that's proportional to α times the change in angle. So, α is not a constant. α can vary from instant to instant, but at that one instant, depending on the forces, there will be some α, and this thing called torque is I times α. Now, I think it's very clear that if I had several forces acting on the body,
46:21
then the torque will be F I r I. And there's just one final caveat to this thing, one final qualification, which is that in practice, the forces that are used to rotate a body may not always be perpendicular. You can apply a force in some
46:42
other direction, like that. Now, you've got to realize that torque is the ability to change the rotational state of a body. And what we're realizing is that it depends on the force you apply, and it depends on how far away the force is from
47:00
the point of rotation. So, if you have a door, if you're trying to open a door, here is the door, here are the hinges, and you want to open this door by applying a force. The question is, where do you put the doorknob? Okay? You suddenly invented the door.
47:21
You thought about hinges, but you haven't quite figured out where to put the doorknob. You've got to ask yourself, well, I have complete freedom. Maybe it'll look nice if I put it there, right next to the hinges. Then you realize that there are parts of the universe, like moron land, where if you go, all the hinges, all the doorknobs are here, and they're saying,
47:42
you know, we are applying a lot of force, we're not getting anywhere. What's wrong? Well, what we're realizing is that whereas force is everything in linear motion, force is not everything in rotational motion. If you want to get your money's worth, you've got to take the doorknob as far as you can and put it near the end.
48:01
Okay, then you figure that out. Then you come along and you say, okay, I've put the doorknob at the right place to open the door, then you apply an enormous force this way. And again, frustrated. Again, you're getting no results. You say, I'm going as far as I can from the hinge. And it dawns on you that if you really want to get something going, you should really take the
48:22
line of separation between where it's rotating and where you're applying the force, and apply a force perpendicular to it. Any part parallel to it is not doing anything unless you're trying to rip it off the hinges. You don't want to do that. So here, this force has got a useless part and got a useful part.
48:41
And the useful part, the useless part, which is trying to pull the rod off the hinges, is going to be balanced by a force from the hinges because the rod is a rigid rod. It's not going to move. But the one perpendicular to the rod can in fact turn the rod without affecting the rigidity. So, we need an extra factor here, which is sin θi, where θi is defined as the
49:00
follows. If you are at the point ri and you apply a force F that way, θ is the angle between the line joining the point of application of the force and the direction of the force. You are here and you're separated by this vector, and it's the angle between
49:20
that separation vector and the force, that angle θ, can vary from force to force. For this force, for this force, Fi, the angle could be θi. That's the final formula for the torque, okay? So, we have built up all the pieces and this is the final answer. This is the definition of torque, and with this torque,
49:41
we can now complete the dictionary and say what we want on the left-hand side is the torque, where the torque is defined as Fi ri sin θi. F sin θ is just a component of the force perpendicular to the separation r. That's what F sin θ does.
50:01
I hope you can see that. If you've got a force acting this way, F cos θ is this part, which is no good, and F sin θ is that part, which is good for rotations. Also, if you want, in terms of work done, you want the force times the distance traveled. The distance traveled is only
50:22
in this direction. There is no distance traveled in this direction. So, the distance traveled is F dot product with the direction in which you move, and that applies only to the force perpendicular to the separation and not to the part that's parallel. Anyway, I hope we understand intuitively why this is the
50:42
guy that's responsible for rotating objects. If you want to rotate something around some axis, you've got a certain amount of force. You're best off applying the force as far as you can from the point of rotation. In other words, suppose we want to rearrange this table here, keeping this end fixed, but want to turn it so it's
51:01
this way. Then, I'm saying, I think we all know we want to start pushing here. Also, we don't want to apply a force this way or that way. We want to apply a force perpendicular. If for some reason we tie it to a rope and can only pull in that direction, then the component of the force this way is to be discounted, no use. This is the part that does the
51:20
turning and that's why that's the part that also does the work. Any force perpendicular to the motion is not going to do any work. Any force perpendicular to the displacement is not going to do any work. This rigid body can only be later in that position, so the displacement is in the tangential direction, and only the force in the
51:41
tangential direction contributes the work. And in my work energy argument, that's what comes in. This is the reason why you come up with this definition of torque. Okay, so we've got most of the pieces for this. I will write one more thing by analogy. The work done by a force
52:01
that displaces the body by an amount dx is this. The work done by a torque will be τ times, I think you can guess what I want to put there, times dθ or Δθ. When a torque rotates a body by an angle Δθ, then the work done is τ times Δθ.
52:23
You can see that even here. You can write the work done as F times r times Δθ, but that is the torque. That's the Δθ. So, it's very useful to know. But if you guys are going to reason by analogy, I would ask you to know the
52:42
counterparts. Only then you can do the work done by problems. So, let's take, let's get used to this formula here, ΔW is τΔθ. I'll show you one example. Suppose I have a rod hanging from the ceiling.
53:00
Well, let's say it's massless and there is a mass m at the bottom and it's got some length L. It's a pendulum. I'm going to take this pendulum and I'm going to bring it to this position by an angle θ0, which is the, unfortunately θ0 here stands for the final angle.
53:23
I want to know how much work I do. That's what we're going to calculate. So, we take some intermediate position when we are here. What is the force acting on this guy? This rod is massless. The force on this is mg. If that angle is θ,
53:46
you can see that angle is the same θ. So, mg cos θ acts this way and mg sin θ acts that way. To hold the pendulum from falling when the angle is
54:03
θ, I have to apply a tangential force in this direction of size mg sin θ. I hope you guys can see that. The mg cos θ is going to be the force that is provided by the rod because the rod is a rigid body. It won't let the mass move in this direction. It'll provide whatever force
54:21
it takes to keep the mass moving radially. In the tangential direction, to keep it from sliding back to where it began, I have to apply a force mg sin θ. So, the torque that I apply is mg sin θ times L. So, I did all that by taking the force, finding the component
54:42
of the force perpendicular to the thing, and balancing it with the countering force. You can equally say the torque that I have to apply is really mg, which is the force, R, which happens to be L here, and sin of the angle between them. No matter how you write it, that's the torque. Let's find the work done by
55:02
that torque when I rotate it from the starting angle to some final angle. That'll be mgL sin θ dθ from 0 to some final angle θ0. I'm just saying the work done is τΔθ, summed over all the angles, that means integrated.
55:23
So, you've got to go back to your calculus thing, and you've got to remember that the integral of sin θ is minus cos θ. And you want to start from θ equal to 0 to θ equal to some final angle θ0, and if you crank it out, you will find it's mgL times 1 minus cos θ.
55:44
So, that's the work done to take a pendulum which is horizontal, which is vertical, and turn it to an angle θ0. I'm going to do a cross-check on this to tell you that it works out from another point of view. This mass was down here,
56:02
now it's climbed up there. So, you can ask, what happened to the work I did? Going back to work energy theorem, let's take this mass here. Forces were acting on it. One was the rod, but the force of the rod is always perpendicular to the motion of this bob here, so it couldn't do any work. The only work was done by me.
56:22
What happened to the work I did? The work I did has to be changing the total energy of the body, but the bob was not moving before and it's not moving after, because I gave just the right force to cancel gravity. I didn't give it any speed. So, it must be that the potential energy of this bob when it's here should explain the work I did.
56:42
That's exactly what it is, because let me rewrite this as mg minus mgl, let me see, plus mgl cos θ.
57:03
Or maybe we can see that it is mg, sorry, mg plus mgl plus mgl plus mgl, this one. Well, I think it's best to write it as L minus L cos θ, and you guys can see what's going on. L is this distance here.
57:22
L cos θ is that distance, and L minus L cos θ is the increase in the height of this bob. This height here is in fact L minus L cos θ, because if that is L and that is θ, this is L cos θ till that point.
57:40
So, this is just mg times the height, and that's where your work went in. So, what's the point of this exercise? This exercise is to get you used to the notion of calculating work done, not as force times distance, but as torque times angle. And here's a problem where the torque itself is changing
58:00
with the angle. The torque is itself a function of θ, so you have to do an analysis of the torque. Okay, so now we've got all the machinery we need. We've got all the machinery we
58:21
need to do kinematics or dynamics of rigid bodies in this analogy. So, let me summarize what it is. We've got a rigid body. It's got a moment of inertia. It's got an angular velocity. The product of the two is called the angular momentum, L. And to change it, you've got to apply a torque. To find the torque,
58:41
find all the external forces acting on the body, multiply each force by the distance from the point of rotation, or the component of each force perpendicular to the line of separation, times the distance of separation. That's the Fr sin θ and add it up. Again, when you do the torques, you've got to give it a sign.
59:03
If a torque tends to rotate a body counterclockwise, we will take it as positive. And if the torque tends to rotate a body clockwise, we'll take it as negative. So, it's possible that on a given body, that could be torques trying to move it in different directions. So, for example, in the simplest rigid body,
59:22
made of, say, two masses, you could be pushing this way and I could be pushing that way. You're all trying to rotate them in opposite directions. The torque you apply would be your force times that distance. This is called positive because you're going counterclockwise. The torque that I'm applying will be my force times the smaller distance,
59:41
and that'll be considered negative because it's clockwise. And you'll add them up, you'll get a net torque. The net torque will cause an acceleration of this dumbbell-like object. What acceleration will it cause? It'll be my force times my distance. Sine theta happens to be one.
01:00:00
minus your force times the distance over which you're doing it times the moment of inertia. The moment of inertia for the simple problem is m1r1 squared plus m2r2 squared if this is m1 and this is r1, and this is m2 and that
01:00:20
distance is r2. So, this is the simplest problem you could do in rigid body dynamics. Take a couple of forces, work out the torques, divide it by moment of inertia, and get the angular acceleration. If the torque is a constant, it's hard to maintain the, what do I have to do to keep the torque constant? I've got to start running
01:00:41
around with this as it rotates. And you could imagine doing that. Then the torque will be constant, α will be constant, α constant is like a constant, and you can do the whole bunch of problems you guys did before. Okay, now there is one technical obstacle you've got to overcome if you want to do
01:01:01
rigid body dynamics, and that is to know how to compute the moment of inertia for all kinds of objects. If I give you 37 masses, each with a distance r from the point of rotation, it's a trivial thing. Do mr squared for each one and add it. But here's the kind of objects you may be given. Not a discrete set of masses, but a continuous blob.
01:01:22
Just like in the case of center of mass, when you had a bunch of masses, you took m1x1 plus m2x2 plus m3x3 and so on. But if you had a rod which is continuous, you have to do an integral. So, you will have to do an integral here also. So, let's take certain rigid bodies and try to find their moment of inertia, which are rigid bodies which
01:01:43
are continuous. So, here is a ring of radius r and mass m. It's not a disc in spite of what it looks like, it is a ring. And I want to find its moment of inertia. So, you have to think about what you will do.
01:02:00
So, you think for a second. I'll tell you the way I will do that is to partition it into tiny pieces, each of which is small enough for me to consider a point. This tiny piece has a mass Δm I, and I goes from 1,2,3,4 to whatever, 10 million till I go around the circle.
01:02:22
Moment of inertia would be, I use Δm I to tell you it's a tiny mass for this section. But what is the r2? If I'm planning to rotate it around that point, plan to rotate around that point, it is the same r2 for everybody. So, you can pull it out of
01:02:41
the summation and you just get mr2. So, this is a very easy problem. This is as easy as this problem where you clearly had only one mass and all of it was the distance r from the point of rotation, it's mr2. Here, you've got the mass spread out, but luckily it's spread out in such a way that every part of it is the
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same distance r from the center, so it's very easy to do the summation. Pull out the r2, do the sum over mass and get the total mass. Then, we're going to do, any question about this? Everything is going to be the same. So, I'm planning this. So, for a ring, this is the moment of inertia, just mr2. But now, I want to do the
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following. I want to take a disk now of radius r. I've got to do the sum over mr2 for every mass in the problem. So, here's where you have to organize your thinking.
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I plan to do the moment of inertia through the center. First of all, if I didn't tell you moment of inertia through where, you cannot even start. I tell you, I want it through the center. If I want it through the center, we have the ability then to think of this as made up of a whole bunch of concentric rings, each one at radius r and of
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thickness dr. If I can find the moment of inertia of this tiny shaded region, I sum over all this certain, all the little washers, annualizes, annualite that I have to fill up the whole disk. So, what's the moment of inertia of this annualist?
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Well, I already told you, if the annualist has a mass m, then it's mr2 is the moment of inertia of that annualist. So, the question is, what is the mass of the shaded region? Clearly, the r2 to use is the r at which it is sitting.
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But what's the mass I should use? For the mass I should use, I argue that if the total mass is that, that times πr2 is the mass per unit area, then I need the area of this shaded region. That's the tricky part. Have you got that? Yes?
01:05:01
Student 2 πr dr. Professor Ramamurti Shankar Right. So, let's ask why it's 2πr dr. If you take this annualist and you take a pair of scissors and you cut it out and you open it out, you'll see that this is going to look like a long rectangle. Its thickness will be dr, its length will be 2πr.
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So, 2πr dr is the, that's its contribution. Here is the dr. Now, you've got to add it up over everything. Adding it up is what we do by doing the integral. I don't want to do this
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integral. You guys can, you can at least guess what's going to happen. There's going to be no π in the answer. π is gone. You've got r, r squared is r cubed, r to the 4th over 4, and that'll, if you combine it with that, you get this MR squared over 2. The moment of inertia of a disk is MR squared over 2.
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That's because if I came and told you, moment of inertia, suppose I made a calculational mistake somewhere and I said, it's MR squared. Will you know I'm wrong? And why will you argue that it cannot be the right answer?
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Yes? In other words, his answer is, the correct answer is, if it is MR squared, it means the entire mass is at a distance r squared. But we know some of the mass is a lot closer to the center. In fact, some of it is right at
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the center. So, you cannot get the same capital R squared as a contribution from all the pieces. Some will be at zero distance, some will be at the full distance. So, whenever you do the moment of inertia of a disk, it's got to look like MR squared times the number, which is definitely less than 1. It turns out to be half. If I got a third or a fourth, you wouldn't know it's obviously wrong. But if I got 1 in front, or worse, if I got 2 times
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MR squared, then you know I made a mistake. Okay, now the last of the objects I want to look at today is a rod. I want to take this actually,
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the example that's done quite often, but that's the first example, but I will do it now. So, here is a rod. Okay, it has length L, mass M, and I say,
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what's the moment of inertia? Do you know what you could do? What's the first question you will ask when I say, what's the moment of inertia? Around what point, right? Till I tell you that you cannot begin. So, I will say,
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okay, let's find the moment of inertia around this center. What you should imagine in that case, you stick a nail through here, and the guy rotates like that on a big circle with that as the center. If you stuck the nail through this midpoint, it's a different story. The answer is different. So here, what I do is I take a distance x,
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I take a sliver of thickness dx. That sliver is small enough for me to consider it as a point mass. Therefore, the moment of inertia of that tiny portion will look like the square of the distance times the mass of the tiny portion. The mass of the tiny portion looks like this.
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Let me see. That's the mass per unit length. That's the length. That's the mass of the object. Then, I have to do the
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integral from 0 to L. Well, x squared will give me x³ over 3. That will give me mL squared over 3. So, that's the moment of inertia of the rod. Again, it looks like mL squared. It's got to have a number in
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front of it that's less than 1, because if you're measuring from here, the furthest you can be is L. A lot of the guys are a lot closer, so the weighted average of the square of the distance cannot be all L squared. It's got to be something less than L squared. The fact that it's one-third is, of course, what you do all the work for. Pardon me?
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You forgot the L here? Also, the dimension for the moment of inertia should be, in the end, mass times length squared. Yes? You're not happy with that. You agree?
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Okay. Look, sometimes I do get these things wrong, so you should always try to keep an eye on this. The last thing I want to do is to find the, well, when I say last thing I want to do, I don't mean I'm reluctant to do it. I'm just saying I'm near the end of the day.
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So, the final thing that I will now calculate is the moment of inertia around the center. Let's see what's going to happen. It's the same integral, x squared, dx over L, m over L, but the range of integration for x will now go from
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minus L over 2 to plus L over 2. Then, I simplify my life by writing it as 2m over L, times x squared dx, from 0 to L over 2.
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This is something I encourage you to do all the time. If an integral is an even function of x, namely negative x contributes as much as positive x, then you can take half the region of integration and double the answer. You can do it only if it's an even function. So, x and, yes? Oh, I'm sorry.
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You're absolutely right. The way to do that would be x squared dx over 2m over 2, here. See, you keep trying, you will catch me. That's right, that's correct. Okay, so let's do this now. 2m over L, okay, now I'm really anxious because this L cubed,
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it's the x cubed over 3, but the thing that's being cubed is L over 2, and that's giving me mL squared over 12. So, the moment of inertia is not a fixed number. It got a lot smaller when I
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took it around the center of mass. When I took it around the left edge, I got mL squared over 3. I went to the midpoint, I got mL squared over 12. I think it's pretty clear that in going from here, I reduce the answer. You might say, why not keep going? It will get even smaller. That's where your common sense should tell you that if you got some answer from there, you're going to get the same answer from here and you've
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hit the best possible moment of inertia you can. You should be able to argue intuitively, and if you do the calculation, it'll come out to be the same. Now, the thing to check if you do this, the moment of inertia around one end is the moment of inertia around the center plus the mass times L over 2
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squared. This is something you can actually verify. Just take the numbers I gave you and you'll find that this answer plus m times L over 2 squared will be equal to that answer. And this happens to be a very general theorem I'll talk about
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next time. If you know the moment of inertia through the center of mass, you're done. Because if I want the moment of inertia through any other point, I take the moment of inertia through the center of mass and add to it the total mass of the object and the square of the distance by which you made me move my axis, which here happens to be L over 2. I will of course prove this.
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I don't want to give you anything without proof. We can actually, in our class, actually satisfy ourselves that this is true. Not only is it true for a rod, it's even true for a disk. It's true for any object in two dimensions. If you've got the moment of inertia of this creature through the center of mass and
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you want it through that point, just take the answer through center of mass and add to it m d squared, where d is that distance. That'll be proven next time.