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visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: The purpose of

these recitations, small group recitations, is so

that we can get out the key concepts over

the week and what I call the essential

understandings– what are the really important

points for the week so that when the

first quiz comes, you will know how

to deal with it. So let’s start with that. But you’re going to help

me think through this. So take a minute

or two, write down on a piece of paper

two or three things that you think are the most

important things that you heard, saw, read this

week about this course. Let’s report out. I want one from a number of you. Who wants to volunteer here? AUDIENCE: Using different

reference frames? PROFESSOR: Say it again? AUDIENCE: Using different

reference frames. PROFESSOR: Using different

reference frames. I’m going to write that,

once I get the chalkboard. Using– I’m going to write it

as multiple reference frames. Close enough? What’s your name? AUDIENCE: Christina. PROFESSOR: Christina? What do you have? AUDIENCE: All points on

a rigid, rotating object have the same rate of rotation. PROFESSOR: She said, all

points on rigid object that’s rotating, all points have

the same rotation rate. So this is rotation and

translation of rigid bodies. I’m going to generalize

what you said a little bit, because

somebody else tell me, what can you say

about translation? So rotation, key point is, all

points share the same rotation rate. How about translation? Two different points

on an object– what can you say about it? AUDIENCE: They follow

the same paths. PROFESSOR: Parallel paths. They go through the exactly

same parallel paths. So those are two key things

we remember about that. How about another point? AUDIENCE: [INAUDIBLE] PROFESSOR: OK. This is actually

quite important. I’m going to write it

slightly differently. We need to talk about this. And that is that rotations–

to be absolutely correct, finite rotations

are not vectors. I want to come back

to that in a minute. Lots of possible

confusion around that. One more– or

actually, there’s more. AUDIENCE: The MLM strategy? PROFESSOR: MLM. So I mentioned that last time. That’s problem

solving my way, which is first M is figure out the

motion, describe the motion. That’s kinematics. The second term is

L. What is that? Laws. By the physical laws. And the second M? Do the math. So motion, laws, and math. There’s something else here. Well, you may have

just decided it’s going to encompass in that. But I want to go a

little further than that. What did we talk a lot about

yesterday in the lecture? AUDIENCE: Different

types of acceleration. PROFESSOR: Accelerations and

velocities and translating and rotating frames. Translating and rotating

frames– running out of room here but you get the point. All right, that’s

a pretty good list. If I’d been coming up

with a list on my own, what would have

thought was important, that would’ve captured

most of those things. Certainly this is really

important this week. And we definitely need to

learn how to use translating and rotating frames. And you’re absolutely in

trouble if you don’t know this. This is just sort of

fundamental to the whole thing. And then this is a subtle point. Let’s start right

there for a second. Who has a textbook? It doesn’t actually

really matter. Let me borrow your notes. Rigid body, got the

print on the front. I’m going to rotate it twice. The x-axis and call

this the z-axis. It comes out top

actually pointing at you. So I did that right. So now I’m going

to do the rotation. Now this one first. And then what was

the other rotation? AUDIENCE: Backwards. PROFESSOR: Different

answer, right? Totally different answer. You can’t add angles as actors. Doesn’t work. And it’s just– the

way I think of it, mathematics is

largely done to help describe the physical world. Newton and all those people were

figuring out– needed calculus to describe the

motion of the planets. Vectors were invented

to do analytic geometry. And it doesn’t work for angles. You just can’t use

them for angles. It’s just the vector math

that they figured out just wasn’t quite clever

enough to include angles. However, vectors can be applied

to positions, velocities, accelerations, and

angular velocities and angular accelerations,

but not angles themselves. That’s the basic thing you

need to learn from that. Let’s use some multiple frames. We’re going to do that today. We’re going to now apply

this and this and this today to do some problems. And let me see where I

want to go first with this. So I have a problem

that I wanted to do. And it’s a circus ride. There’s an arm. And that arm is rotating. Attached to the arm

is a cross piece. And a passenger can sit in

each one of these things. And this is

basically horizontal. You’re looking down on it. So you’d be riding around

in these cups at the circus and it’s going

around and around. And I want to know the velocity. What’s the velocity of

point B in the O frame? And so this has to do– one

of the things on this list might be to get

the notation down. So this is the velocity. This is the point. And this is the frame. So what can you write down? Just take 30 seconds. See if you remember. Write down the general

velocity formula that was put up

yesterday– vector velocity formula for a point in

a moving frame that’s moving in a fixed frame. Came up with a general formula,

had two or three terms in it. And we’ll walk our

way through it. I realize I did something

maybe slightly out of order. So hold that thought. You’ve written down

what you’ve got. We have to do something before

you can actually write that. We haven’t actually picked

our reference frames, have we? So think about

that for a second. How would you set

up this problem? What would you make

translating reference frames, your rotating,

translating frame– where would you assign it? Think about it for 30 seconds. Who’s got to take a

shot at it for me? Where would you pick reference

frames for this problem? What your name? AUDIENCE: I’m Ben. PROFESSOR: Ben. AUDIENCE: O and along the cross? PROFESSOR: Here, for sure. This is your inertial

frame– not moving, right? And? AUDIENCE: Two axes on the cross? PROFESSOR: So you

would put one up here? OK. I’m going to line

up with the cross, and I’m going to stick

out here and call it x2. And then there’d be a y2 here. And it rotates with the cross? All right. That’s good. Now go back to that equation. Now give me the velocity,

the general expression. I don’t want you working out the

details, just what set of terms would you plug things into now

to get the velocity of B and O? Then we’ll evaluate the terms

and talk about it, using now what we’ve decided here. OK, somebody help me out. What’s on the right hand

side of this equation? First term, Mary. AUDIENCE: Velocity of– PROFESSOR: What’s your name? Steven? AUDIENCE: Velocity of

A with respect to O. PROFESSOR: Velocity of A

with respect to O. All right. That’s the velocity at this

point in this frame, right? What else do we need? What’s your name? Andre? AUDIENCE: Yeah. [INAUDIBLE] PROFESSOR: I hear a velocity

of V with respect to A. And what is that– is that

influenced by rotation? Can you describe what

you mean by the velocity of v and A physically? AUDIENCE: [INAUDIBLE] PROFESSOR: So it’s as if you

were sitting on that frame, right? Does the rotation have anything

to do with what you see? No. So I sometimes remind

myself right here this is omega equals 0. And you can set the omega

equal to 0, what you would see is what this term is. Do we need anything more? Name? AUDIENCE: Christina. PROFESSOR: Sorry, you

gave it to me once before. It’s going to take me awhile. AUDIENCE: It’s the

rotational motion of B spinning around in there. So it has to do with the

omega as seen in the reference frame, the origin, cross product

with r from the in regards to the x2 xy. PROFESSOR: And we have the name

of that frame to help us out. This is then frame

A, x2, y2, z2. If you really wanted

to write [INAUDIBLE] We just call it frame A. so

this is would be rB as in NA And these are all vectors and I

often forget to underline them. Do we have it right? Anybody want to add

to that, fix it? Correct it? Steven, right? AUDIENCE: [INAUDIBLE] PROFESSOR: I left

it vague on purpose. We need to figure that out. He asks, is it

omega 2 or omega 1? Really important point we want

to make today about what omega this is. We’ll get to that. Yeah? AUDIENCE: Well, if they’re

rotating in the same direction, wouldn’t it be added in

both omega 1 or omega 2? PROFESSOR: Well, OK. Let’s talk about it right now. Are we agreed that this

is the right formula? Then let’s set about

figuring it out. And we can talk about

this term first. So we want to know, this

is the rotation rate of this arm out here

in the base frame. That’s what the notation says. And we know that the rotation

rate of this first arm in the base frame is this. And we know that

the rotation rate of this thing with

to– now this has gotten a little

complicated, because this isn’t quite exact enough. This is omega 2 with

respect to this arm. That’s what’s given

in this problem. So this is omega 2 with

respect to the arm OA. Yeah? AUDIENCE: So does

that mean it’s omega 2xz from coordiante system B? PROFESSOR: No, coordinate

system A x2y2 rotates. And if you’re sitting in

there, you wouldn’t see it. So this is correct. It’s the rotation

rate as seen in O. So we need to figure

out what that is. And I’m telling

you in this case, you were given–

you might have been given the rotation rate in O. You weren’t. You were given the rotation

rate relative to here. So I’ll write it as W2 with

respect to this arm OA. So how do you get– we

need omega in O is what? Help me out here. AUDIENCE: Is it O– should

there be a small b at the bottom [INAUDIBLE]? PROFESSOR: Good. But what is it? Let’s deduce it. If my arm here, this

is the first arm. And this is the at AB link. Now if omega with respect

to this arm, this thing weren’t moving, no rotation

rate relative to this, the whole thing would

be straight, right? And it’s going around like this. What’s the rotation rate

of the link out here? Omega 1. And now this arm’s not

moving, but this is rotating relative to it at omega 2. What’s the rotation rate

of the link out here? Just omega 2. If I put the two together,

what is the rotation rate of this arm, this second link? AUDIENCE: [INAUDIBLE]. PROFESSOR: Omega 1 in,

certainly in O plus omega 2. It’s not with respect

to A. I’m just going to call it with

respect to the R maybe. Even this notation is

failing a little bit. But you get what I mean. It’s omega 1 plus omega 2. And let’s just write it

as omega 1 plus omega 2. And what direction is it in? It’s a vector. So one of the things we

have to pay attention to are unit vectors. Yeah? AUDIENCE: [INAUDIBLE] PROFESSOR: So this

is capital I hat here and capital J hat there and

coming out of the board, K hat. Now, this is certainly

K hat, capital K hat. This one, though, is relative

to– it’s the rotation rate of this thing. Here is a reference frame. What’s sticking out this way? A little k2, right? But is it parallel to capital K? Always parallel to capital K? So they’re the same thing. If unit vectors in

this are parallel, they amount to the same thing. So we can put capital

K, lowercase k, anything we want here and it’s correct. Now we’ve got an

answer for that. So when you’re given– when the

one thing’s attached to another and you’re given

the– if out here you are given the rotation rate

in the base frame, you’re done. But if you’re given

the rotation rate relative to some

other moving part, then you have to add them up

to get the true rotation rate. That’s the bottom line message. All right. So we started– we’re

trying to figure out this expression here. And we started with one

of the harder terms. And we need to figure

out– to finish it, though, let’s do this over here. We have velocity of A in– and

we have the velocity of B in A with no rotation. And we have omega B in

O. And let’s finish that. We know what omega is now. Whoops. It’s not omega. The third term is omega

B and O cross RBA. So we’ve gotten the

first bit of this. Let’s finish the problem. This is omega 1 plus omega 2

times k hat cross with what? We need a length. I’ll call this L. It’s L long. So what is RB respect to A? Yes? AUDIENCE: L X 2 hat? PROFESSOR: L X 2 hat. And I’ll call that Lj2. The coordinate is x2. The unit vector would

be I2, not a j, an i. The unit vector is i2. OK, great. Now what is k cross i2? j2. So we get omega L omega

1 plus omega 2 j2 hat. That’s that term. And we need to figure

out our other two terms. What’s this term? Remind yourself of the meaning. This is the velocity of point

B with respect to the A frame, which is attached to it. It’s on a rigid body. AUDIENCE: [INAUDIBLE] PROFESSOR: He said omega

2 times L. She says 0. Any other? I hear another 0. Why 0? AUDIENCE: Because it’s rigidly

attached into the ride, if you’re moving around. It’s not moving on the

ride versus strapped in. PROFESSOR: Right. So this term is always from

the point of view of a person riding on the frame. Riding on that frame– so

you won’t ever see rotation from inside the frame. You’re just moving with it. So that’s called, in

the Williams book, he calls this term the rel. It’s the relative

velocity between these two points and no rotation. So what is that in this case? I hear 0. Everybody agree it’s 0? It’s a rigid link. Two points don’t move. So now we’re just

left with this one. And now, one of

the points I really wanted to drive home today is

in fact this problem is one that, depending on

how you set it up, you can think of as actually

having multiple rotating frames. And if you do that,

what’s the correct way to add up the parts so you

get to the right answer? Because we’ve left

this one for the last. And I want to make

sure you go away knowing a formula you can always

use, and it’s going to work. And the formula

we can always use is the one that’s of this form. Every one of these problems,

including multiple links and things, you can

build up by doing a sequence of this problem

again and again and again, until you get the whole answer. So we’ve actually

done what I would call the outer problem first. We’ve worked out this thing. We have to do the

inner problem now. We could have done it

in a different order, but I need to know the

velocity of this point. And just to get you in the habit

of using the vector equation, that we have, I want to

know the velocity of A in O. And I’m going to

attach a rotating frame to this arm, x1, y1. It rotates with this

arm at that rate. And I want you to use

that frame to solve for the velocity of this point. And that would be–

velocity of point A in O would be the– this frame now

is an O little x1, y1, z1. It’s a rotating frame, right? Because the O’s are

going to get confusing. Better not call it O.

We’ll call this rotating one– in Williams, he

uses a lowercase o, but it’s hard to

do on the board. Let’s call this c. So this is a frame, C, x1, y1. So this frame will

be my c frame. So I want to know the

velocity of point A. It’s the velocity of what? If you get stuck, use

that top formula up there, put in the right points. So what’s the first term mean? It’s the velocity of the–

this time the rotating frame, does the rotating

frame translate? We have a rotating frame. Does it have any

translational velocity? No, but you still have

the right to turn it down and set it equal to 0. So what’s the right term? How do you write it? Right? It’s the velocity of

my reference frame. It’s the transitional

velocity of that reference frame in the O frame. And that’s what it is. And in this case,

it’s 0 plus velocity of A with respect to c. And I’ll remind you again. It’s as if you were

now rotating with it, and you’re sitting at c,

looking at A. What’s its speed? 0. Plus omega– what omega? Seen where? Measured from where? Measured with respect

to what frame? I hear on O, cross

with– we need length? We’ll make this length

capital R, scalar. So what’s the

cross product here? What’s the unit vector? Correct unit vector? Not x. x is the coordinate. The unit vector is– AUDIENCE: [INAUDIBLE] PROFESSOR: Right? That turns 0. That turns 0. This turns 0. Omega 1, And what’s the

unit vector associated with this omega 1? k cross i 1 j hat. And so we have R omega 1 j1 hat. And now, we should

be able to write out the full answer of

the velocity of B in O is the velocity of A,

which is R omega 1 j1 hat plus velocity of B

with respect to A, which was 0 plus this term,

which we figured out as L omega 1 plus omega 2 times j2 hat. So we have two or

three sub– this kind of a hard problem, actually,

for the first time out. Because it has a number of

subtle concepts built into it. You actually have

two rotating bodies. How do you deal with them? Well, you do sequential

applications of that vector velocity formula. Yeah? AUDIENCE: So I was

wondering why we made another coordinate system

that’s rotating with the arm to solve for the velocity

of A [INAUDIBLE]. PROFESSOR: She asked,

why did we bother to make this another frame. The problems are going to

get nastier and nastier. I could have asked you, when

I walked into class, what is the velocity of point A.

And you would have said, well, obviously R omega. Why are we going to

all this trouble, when everybody knows from high school

physics that it’s just R omega? And the answer is

is because we’re going to get to doing

really nasty problems. And I want to make sure you

understand all the subtleties about how we get these. So we started simple, but I

did it the long, hard way. Because later on, if I’d

walked in at the beginning and just asked you right

off the bat, what’s the velocity of this

point– go for it– you guys would have

failed miserably. It’s not much

harder, but it takes two sequential applications

of what you think is obvious when you

walked into class. So that’s why. We’re just doing it the

hard way, so that you get all the little nuances. Yeah? AUDIENCE: So why [INAUDIBLE]

there for omega 2, we have it with respect to arm

AB [INAUDIBLE] with respect to arm OA? PROFESSOR: Why is it that way? AUDIENCE: [INAUDIBLE]

with respect to arm AB and when you wrote

[INAUDIBLE] with respect to arm OA? PROFESSOR: When I wrote

the equation for– AUDIENCE: For the omega 2. There you’re saying

it’s with respect to OA, and there you say

its respect to AB. PROFESSOR: Oh, I see. Because this is wrong. AUDIENCE: [INAUDIBLE] AUDIENCE: My question deals with

j1 and j2– are they the same? PROFESSOR: So he has a question. And that’s the final, subtle

point I want to get to today. Good question. He’s saying, are these the same? Are they different? How do we deal with it? So a question for you. In general, if

you’re asked or given a problem like we just did,

and you arrive at a solution, is it OK to give an

answer where you had unit vectors in multiple frames? And neither of these unit

vectors are in the base frame. And yet, the answer

we’re claiming is that this is the

velocity of B in O. And here we’ve got unit vectors

that are not in the base frame. Is it a legit equation or not? What do you think? See a lot of no’s out there. I think we better figure it out. So we have unit on the arm. This is my c and O here. On the arm, I have

frame that rotates with it that has unit vectors in

the direction of the arm of i1 and j1. So here’s i1. It’s unit long. Here is the angle theta. Here’s j1 and the angles. And I want to know– this

is i1 and this is j1– can I express i1 and

j1 in terms of capital I, capital J, the unit

vectors in the base frame? I want to express them

in terms of unit vectors that are in this rigid,

non-moving, non-rotating, inertial frame. So down here, this is

the i hat direction and this is the j hat

direction, right– not moving. So this is just a

unit thing, unit long. Can I project it onto its

i component and capital J component? All right. So i1, it looks to me

like cosine theta I hat plus sine theta J hat. Do you agree? Just standard trick, right? And this one, takes me a

minute to figure this out. Which is the theta here? This is theta. That’s 90 minus theta. So this must be theta, right? There’s a theta here. And if this is unit

long, what’s that? That projection there is–

so j1 has two components, minus sine theta I

plus cosine theta J. And I highly recommend

you write that one down. Make sure you can

drive it yourself. You’re going to need it again

and again and again and again. Now, could we do the

same thing for– could we convert J2 to the base frame? And it rotates, so this x2 can

be at any arbitrary position. But in order to do the problem,

you have to pick a position. And then you’d have

to do draw an angle. And then you’d have

to apply this formula. And so you’re going

to end up with an i2 and some cosine phi capital

I plus sine phi capital J. And the same thing, j2 is

minus sine phi i plus cosine theta J. So we’ll do a trivial

example, solve a trivial case. What is the

instantaneous velocity at the moment that the

coordinate system is lined up as we see, and B is

sitting right here? So we’ve got to go

look at our answer. Where was our final answer? Velocity of this

guy here, right? What would be the

contribution of this term? We have to take each

term and convert it to the base system and

capital IJ terms, right? You do it one term at a time

and add up the components. So how do you break this one

down and put it into capital I, capital J components? AUDIENCE: Substitute? PROFESSOR: Yeah,

what’s the answer? So j, if it’s

lined up like this, j2 is importing

in what direction? Up. And what is that in this system? Just capital J. At this instant in

time, that’s just capital J. Trivial

calculation, because this angle is 90 degrees. Plug in 90 degrees, this term

goes to 0, this term goes to 1. J2 is capital J. And what about the

other term, J1? You just got to–

it’s J1, right? So you just gotta

go with the flow. It is is. You’d substitute this

in for J1 right here, and you’d have R1

omega 1 cosine theta sine theta and j and k terms,

plus this thing, capital J. And you have just

converted the answer, which was in terms of

unit vectors in rotating to different rotating frames. You’ve converted it all

down to the base frame. AUDIENCE: [INAUDIBLE] PROFESSOR: Oops, I’m sorry. I just made a mistake. You guys got to get

better catching me. That now make sense? So phi is the angle that

the j2 unit vector makes with the inertial frame, right? And theta is the angle

that the j1 or i1 make with the inertial frame. Yes? AUDIENCE: Phi is

0, though, right? PROFESSOR: In this

case, phi is 0. Does that still

work out over there? Sine of phi is 0 and cosine

of 0 is one and you get j. AUDIENCE: Then why did we

didn’t plug in anything for j? PROFESSOR: We did. There isn’t a simple

answer for it. And so you have to use

the full expression. I just got lazy and didn’t

want to write it out. The answer is this. Stick in 30 degrees if you want,

and then you’ll get numbers. So real important point

that we discovered is that the answers are correct,

expressed in rotating unit vectors, expressed in

different unit vectors, different rotating ones. This is correct, because

you can take this and you can reduce it

down to the base frame. So you will be– usually in

problems that you’re given, you’ll be asked to express

the answer in terms of unit vectors in the base frame. Or you’ll be told

you can leave it in whatever is your comfortable

set of unit vectors. Most of the time, the

first ones you’ll arrive at are the ones in terms of the

rotating coordinates that are easier to use. The more natural answer

falls out in terms of these. Good. All right, and we’ve got

three, four minutes left. What have I confused

you with here? So key concepts–

what have we– what hasn’t been clear or

maybe we didn’t cover it yet– another point. AUDIENCE: So the

reason we chose those as the starting reference frames

instead of I hats and theta hats? PROFESSOR: Only because

at the beginning of class, we talked about it– which

frames do we want to use, and then we chose those. Could we have used a

polar coordinate system to do this problem? Sure. Twice You do it once in each– AUDIENCE: Is there

a way to know up front which one would simplify

down to the inertial i hats and j hats more simply? PROFESSOR: The easiest way? Is there a way to know upfront? No. That’s just experience. Work lots of problems, and

you get good at picking frame. We can probably, with

time as we meet and talk about these things,

we’ll come up with some sort of general

insights about how to do that. Yes? AUDIENCE: Is this

picture up in parentheses supposed to be those

coordinate systems? PROFESSOR: This picture

is the coordinate system of that first arm. AUDIENCE: OK, so is that

supposed to be phi up there? PROFESSOR: Yeah. Wait a minute. No. This is the first arm. That is theta and

these are ones, right? The 2 system would be phis AUDIENCE: OK. I was just wondering if that

was the ride or if that was not. PROFESSOR: No. This is point A, if you will. Well, it could be. It’s lined up with

point A. This is A. AUDIENCE: Because I thought

we decided that the phi was– PROFESSOR: This

is point A, right? That is point A.

And this is arm CA. AUDIENCE: So is this one

here the origin of this one? PROFESSOR: Well, look at

whatever the unit vector is. The unit vector in this system

is lined up with that arm. So this is just a breakdown

of these unit vectors so I could draw the

angles and figure out the sines and cosines. You could draw a similar

picture for i2 j2’s. And then it would be phi’s. Good question. Yes? AUDIENCE: So, since

you can choose between Cartesian and

polar coordinates, could you set one in

Cartesian, one in polar, you can mix and match it

or– is that beneficial in some problems? PROFESSOR: Polar is– I don’t

have time to show you today. But for planar motion

problems, which are things confined to a plane,

they rotate, axis of rotation’s always in the k direction, which

is all the problems that you ever did in 801 Physics. You didn’t do general

things actually. But for planar motion problems,

cylindrical coordinates, actually, you still need the

k to describe the rotation, right? Polar coordinates,

cylindrical coordinates are oftentimes

really convenient. And they’re easy to use

because you’ve learned them a long, long time ago. And you know the relations. But you can make it a rotating

x1, y1, z1 rotating system and it will all work out. We came up with this

little formula here, right? This could just as

easily have been r hat. And this could have just as

easily been no difference whatsoever in a planar

motion problem, when you attach an xy system

that rotates with it, or I call it r and theta. These are the same direction. R is in the direction of i1. Theta is in the direction of j1. So use it when it’s

convenient, and it’s convenient a lot of times, especially that

nasty acceleration formula. In polar coordinates,

it reduces down just to the set of five terms. Memorize it and just tick them

off– Coriolis, centripetal. You see them right away. You know what they are. But there are certain problems,

even in planar motion problems that polar coordinates don’t

work for– doesn’t work for. [INAUDIBLE] And think about that. It’s actually a simple problem. Put a dog on a marry-go-round. The dog’s running in

a random direction on the merry-go-round. And the merry-go-round

is turning at some rate. And you only want one rotating

coordinate system, r and theta. You can’t do the problem

with polar coordinates. Think about it. Go away, think about why not. I’ll tell you the

answer in words. You go figure it out. You can’t describe the

velocity of the dog in polar coordinates. The dog is running around. If the dog’s fixed on the

rotating thing, than polar coordinates work. If the dog’s running, you

can’t do that velocity. So you need a more

sophisticated coordinate system.