R2. Velocity and Acceleration in Translating and Rotating Frames

September 29, 2019

The following content is
provided under a Creative Commons license. Your support will help
MIT OpenCourseWare continue to offer high quality
educational resources for free. To make a donation or to
view additional materials from hundreds of MIT courses,
visit MIT OpenCourseWare at 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
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.

Leave a Reply

Your email address will not be published. Required fields are marked *