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 ocw.mit.edu. PROFESSOR: OK, let's get
on with today's lecture. And I want to look at a
variety of different problems, different classes of problems. We're going to look at four
different classes of problems and talk about the way you'd
go about approaching them. We need very few
fundamental laws. We use the same fundamental
laws again and again. And the issue is really, how
do you go about applying them? One is the sum-- this
fundamental law's for rigid bodies. So the fundamental laws,
they're always true. The sum of the external
forces, vectors. Time-rated change
of the momentum of the system with respect
to an inertial frame. And we recognize this
mass times acceleration. Newton's second law. The other one is the
summation of external torques. And that one is
the time derivative of angular momentum with respect
to a point which you choose-- this is an A-- plus the
velocity of A with respect to the inertial frame crossed
with the momentum with respect to the inertial frame. So that's the torque equation. Now we have two special
cases of this where the second term goes away. One is when you are at
the center of gravity, in which case by definition
the velocity of that point and the momentum are
in the same direction, the cross product disappears. So there's two special
cases we use all the time when it simplifies to that. One is when you're at
the center of gravity. The other is when,
for whatever reason, this velocity is
parallel to that, and the cross
product goes to zero. And sometimes the velocity
is just plain zero. So there's some
certain cases that we use lots of times when
that term goes away. But sometimes it doesn't. You just have to put up with it. So those are our two
fundamental laws. And the question really is
how to go about applying them. So I'm going to look at
four classes of problems. Let's name them right now. Pure-- and we'll do simplest
to hardest-- pure rotation about a fixed axis through G,
through the center of mass. Pretty trivial. We can all do those
kind of problems. A second class, pure rotation
about a fixed axis at A, which is not equal to G
not at the center of mass. Again, pretty simple problems. Third class, no
external-- I'll call it no external constraints. We'll have to give an example
to see what I really mean. Well, I'll give you an example
right now, which we'll do. You have this hockey puck with
a string attached to it force. And this whole thing is
on a frictionless surface. So it's constrained so it
can't go through the surface. So no external constraints
in the direction that the motion can happen. So this is a 2D problem. Can move in the x,
y, and rotation. But it's not touching. There's no things
constraining it in the directions
of movement, which are allowed in the problem. That's just too much
words to write up here. So this kind of problem. And the fourth are problems with
moving points of constraint. You won't see a textbook with
sections broken out like this. I'm using my own terminology. I just made it up
last night as I was finishing off the lecture. But I'm trying to
give you some insight, and this is the way I
think about these things. So let's quickly
go through examples from the first couple
of types because they're especially easy. So what are we saying? They're pure rotation
about a fixed axis through G. Center of mass
is somewhere along this axle. The axle is fixed so
object can spin around it. And these kind of problems
are particularly simple, so I'm not going
to dwell on them. But here's class one,
class one problems. And it's basically-- these
are rotors, typically rotors, of almost all kinds. So for these problems, you use
momentum, angular momentum, with respect to
the center of mass. You can express it as IG times
omega x, omega y, omega z. All these problems
can be expressed with a moment of inertia matrix
times the rotation vector. And it will give you HX, HY, HZ. And you can use that
second formula up there. The torques are dH dt. So it's fixed through G. So
the second term doesn't appear. So to do these problems,
the sum of the-- see, I'll just give an example here. For 2D planar motion, they
get especially simple. Then, for 2D planar motion,
that means omega here is 0, 0, and we'll let z
be the rotation axis. And this H then with respect
to G just becomes Izz omega z. All those 2D planar motion
problems boil down to this. And dHG dt then is Izz with
respect to G omega z do, or more familiar notation. So all those planar
motion problems, axis passing through G like this. For those problems,
does this matrix have to be with respect
to principal axes? Do you have your XYZ
coordinate system? To do this style
of problem, does that actually have to be
expressed in principle coordinates so
that it's diagonal? It doesn't have to be. You're worried about
rotation about z. You'll find that that
statement's still always true. Now, there may be-- the
thing could definitely have imbalances and have
unusual other torques. That will fall out
in the problem. But for the motion around the
axis of spin, this [INAUDIBLE]. Where you only have a 1
omega z component, just the one component, you will
get-- it'll work out just fine. Yeah. AUDIENCE: If you do have some
kind of Ixz and Iyz terms, you would end up with more-- PROFESSOR: You will
end up with-- first of all, if you have the
off-diagonal z terms, xz, yz terms, when you multiply that
out, you will find components of H that are in the z direction
as well as perhaps in the x and y. And those other two
components tell you that H is not pointing in the
same direction as omega, right? And that it tells you instantly
that the thing is dynamically unbalanced and will
have other torques that are trying to bend that axle. So they'll always appear. All right. Let's move on. I want to make sure we
get through this today. Class two problems. These are basically--
these are the pure rotation around some point
that's not through G. And again now, this is a fixed--
key here is fixed axis at A. It's not moving. This is what makes
these problems simpler. For these kinds of
problems, you do the sum of the
torques with respect to A, the external torques. And because that point's fixed,
the second terms don't appear, the v cross p terms. And you can write these as I,
a moment of inertia matrix, times whatever
the rotations are. In order to define the mass
moment of inertia matrix, you must have chosen
a set of coordinates attached to the body. And then with those
coordinates, you've computed the moments of
inertia for the body. And if you chose wisely, you
get principal coordinates and you only get diagonal
entries on the matrix. If you chose unwisely,
you will get other stuff. But it's still a valid mass
moment of inertia matrix. It just gives rise--
you have to deal with a bunch of other terms. So this will still
yield the same answer. How do you get I
with respect to A? These are opportunities when
you can use parallel axis. Yeah. AUDIENCE: Isn't I
omega just H dot dH dt? PROFESSOR: Excuse me. You need to finish that out. This is H. So the
time derivative of H is the time derivative
of this expression. You have to figure out
the moments of inertia and the rotation rates. And you may get multiple
terms, only one of which-- let's say that
this will work out. You will get multiple terms. You will get torques that are
not in the direction of spin. Again, these might
be unbalanced. On the other hand, it may
be, for 2D problems, which is common-- so for the
2D planar motion, which most of the problems we do
are, then what you would do is you're saying omega
is, say, 0, 0, omega z, which simplifies that. We're multiplying this
thing out quite a lot. And if I with respect
to G is diagonal, then that means you your G
you chose principal axes. But then how do you get
to I with respect to A? For 2D problems,
really simple ones, how do you get to I
with respect to A? AUDIENCE: Parallel axis. PROFESSOR: That's
the classic case for using the
parallel axis theorem. So for these 2D planar motion
problems-- and planar motion problems, then you
can use parallel axis. I'll do an example. So this is kind of
the set up for this. An example of this
on the homework. What problem on the homework
is just perfect for this? It's 2D, planar motion,
about a fixed point. AUDIENCE: Circle with
the square cutout. PROFESSOR: Yeah, the cylinder,
this [INAUDIBLE] disk with the square cutout with
the pin at the top turning it into a pendulum. That's the fixed point. You can figure this out. You can use parallel axis. And we're going to do an
example right now that's almost identical to that problem. And we started it last time. So the example I want
to work is very similar. It's basically this
problem is that pendulum. So let's just work
it through quickly. We had done the setup last time. So last time I basically
derived an example of the parallel axis theorem
for my little stick here. And I'll give you some
geometry, some values. So here's G. There's the point
I wanted to rotate about, A. The distance between
these two points is d. That's going to pop up in
my parallel axis theorem. We've got a set of
coordinates here. My G is, of course, in
the center of this thing, the geometric center. So I have a body
fixed set of axes, which I'm going to call z. And my x is in this direction. So that would make my
y going into the board. So the y is kind of like that. And this has some properties. The dimension in
this direction means A. The direction in
this dimension is B. So it's a width of A, a
thickness B, and a length L. So when you compute, these
are-- ah, symmetry now. So the axes that I've
chosen for this problem are they principal axes. There's three planes of
symmetry in this problem, and I've got one principal
axis perpendicular to every one of them. And all three pass through
the center of mass, and they're orthogonal
to one another. So those conditions
are all satisfied for these to be principal
axes for this rectangular body and its uniform density. So I'll give you the--
for bodies like this, in your book or any
book on dynamics, you can look up then
the properties, the Izz. And we're going to
spin this thing. This thing, we're going to
have it rotating about-- which one am I going to use? Yeah, around the z-axis. That's how I'll set it up. That's the way I
drilled my hole, so it's going back and forth. So that the wide part
of it's this way. So Izz with respect to G M L
squared plus a squared over 12. Iyy. OK, now just to make a
point, the dimensions L 32.1 centimeters,
a 4.71, b, 1.25. And eventually my
d, this offset would be, for example, where I've
drilled that hole, is at 10.2. The point I want
to make here, lots of times it would be
nice if I could just make the simplification
to call this a slender rod and be able to ignore these
a and b dimensions in this, just to get quick answers. Do you think that's
slender enough? It's not even 10 times--
the length of the width here isn't even 10. It's probably six or seven. So the key issue then,
if you look at this, is really what's the ratio
of a squared to L squared? That would tell you something
about the relative importance of the a squared term
to the L squared. So let me tell you about that. So a squared over
L squared is 0.022. b squared over L
squared is 0.002. So even with this
kind of fat stick, the approximation of ML
squared over 12 is pretty good. It's only 2% off. And this approximation,
L squared over 12, is less than 2/10
of a percent off. So for roughly slender
things, we oftentimes just say ML squared over 12
for spin about their center. We now need to
apply parallel axis. I want to spin this around,
let this rotate around, not around G, but now around
a point off to the side. So we worked out last time
that Izz with respect to A is IzzG plus Md squared. So in this particular
problem, this Izz about G is approximately
ML squared over 12. So just by way of example,
to see what we might find out here, is let's let d equal L/2. That would be if
I move this-- if I put my hole right
at the very top, how would this thing behave? I don't have a hole right at
the top, but I have one close. So this is just because the
numbers are easy to work. What happens if you put
in L/2 into this formula? Well then IzzA is ML squared
over 12 plus M. L over 2 squared is L squared over 4. And that's ML
squared over 3, which is a number you'll run into
again and again and again in mechanical engineering
because examples like this are used a lot. The mass moment of inertia
about a slender rod pinned at its end,
ML squared over 3. So let's take this problem a
little more towards completion. The sum of the external moments
with respect to A dHA dt. And that's going to be d by dt. In this problem, the
only rotation is omega z. So this is going to be Izz
with respect to A omega z. And that just gives us IzzA
omega z dot, or more familiar, IzzA theta double
bond, if you want. Here's our problem. It simplifies to
this slender rod. And let me do the
more general case. Here's my rod. The pivot point that it's
going around is here. This is d still, and this is G.
And it's swinging with respect to this point. So here's the angle theta. So it swings about this
point that you've fixed, and that point is d above
the center of gravity, center of mass. So what are the external
moments about this point? There's no torque
right at the point, but our free body diagram
of drawing this as a-- here's our point of rotation. Here it is displaced
through an angle theta. The weight of the object
can all be concentrated, thought of, for the
purposes of the free body diagram as acting through G. So here's the mass at G,
gravity acting down on it. And the length of this arm here
about which it's swinging is D. So the torque about this is--
and it's pulling it back-- minus Mgd sine theta. Probably you've seen
that many times before, including the recent homework. And that must be equal to
Izz about A theta double dot. So we have an
equation of motion. Just collecting
the terms together. So this is a oscillator
that, for this problem, has no external excitation. This is its equation of motion. And I need theta double dot. But is it linear? Is it linear? No, it's not a linear
equation of motion. But for sure, it
is an oscillator. And for anything
that vibrates, you can have lots of
nonlinear problems that exhibit vibration. You can think of them--
you can pose problems where you say, OK, what's their
static equilibrium position? And think of a very small motion
about that static equilibrium position. You can always linearize about
the static equilibrium position and be able to come up with a
linearized equation of motion that at least from that you can
calculate the natural frequency of the system for small motions
around its static equilibrium position. So in this case, it's
pretty easy to do. And you've seen it
before for small theta. Sine theta is approximately
equal to theta. So we are going to
linearize the equation. This theta equals 0 is a
static equilibrium position. So we're linearizing around 0. And around 0, that's
the approximation. So you just substitute
that in, IzzA theta double dot plus Mgd theta equals 0. There's your linearized
equation of motion. I want an estimate of
the natural frequency. So find omega n. So this is basically
entering into solving differential equations. But I let mother nature
tell me what the answer is. I do the example, and
I say it oscillates. Looks a lot like
sine omega t to me. Plug in [INAUDIBLE] to make a t. Let's find out. Some theta amplitude
sine omega t. Plug it in. So you plug that in [INAUDIBLE]
and you get minus omega squared IzzA plus Mgd. And all of this, you
can factor out the theta 0 sine omega t equals 0. That's what you get. And in general, theta,
that's not 0, or else you'd have a trivial
problem, not moving at all. But in order for this equation--
and this can be anything. We're doing the 0 minus 1 and
plus 1 depending on the time. So in order to
satisfy this equation, this part inside of the
parentheses has to be 0. And when you just
solve that, you find that omega squared
equals Mgd/IzzA. And the reason I've gone to
the bother of working this out in detail right to the end
is that every one degree of freedom rotational oscillator
that you will ever encounter-- sticks, wheels with
static imbalances. Let me show you this one. It's an oscillator too. Any one degree of
freedom oscillator, rotational oscillator,
pendulum-- basically all pendula-- this
is the formula for the natural frequency. So it's going to be of
that form for any pendulum. So any 1dof pendulum. This is the generic answer. So that cutout problem for
today has to come down to this. Or this is the distance
between the mass center and the point of rotation. And that's your mass
moment of inertia about the point of rotation. So that's worth
knowing that one. OK, keep moving. Now things will begin
to get interesting. These latter two classes
are harder conceptually, but once you have a
solution method for them, they're not all that hard. This one is the problem I
described at the beginning. We've got this hockey
puck like thing. And the string wrapped
around it pulling on it with a known force. In this problem, they
call it 150 newtons. The mass of this
thing is 75 kilograms. It's on a frictionless surface. And we want to find
its acceleration of the center of
mass and the rotation around the center of mass. So find theta double dot and
find the linear acceleration. That's basically the
name of the problem. And they give you that it's
75 kilograms, 150 newtons, and kappa, the radius of
gyration, is 0.15 meters. This is defined as the
radius of gyration. So what's that? It's a radius of gyration. It's really appropriate,
really only useful, for mostly 2D
rotational problems around an axis of rotation. And what it means is this is the
distance away from the center of rotation at which you could
concentrate all of the mass and have the same mass
moment of inertia. So this has-- here's G here
at the center, uniform disk. We know that there is an Izz
about G for this problem. And I need to pick a
coordinate system so we can talk about things here. Let me get M out of the center. So I'm going to
let z be upwards. And because I'm
looking ahead and want to keep the equation simple,
I'm going to make my x-axis here parallel to f so I only
have to deal with one vector component equation. And then that makes
the y-axis this way. So I'm interested in Izz because
I'm spinning around the z-axis. I know that for a uniform
disk, that's a principal axis, is the vertical one. And what I'm saying is
that you can then set find. There's an IzzG that can be
expressed as M kappa squared. So kappa, in effect, is just
IzzG over M square root. Now, why do we use
that kind of thing? Well, the way this
problem was set up-- I actually took it out of a book. It wasn't a uniform disk. It's a pulley
wheel or something. And it's got spokes
in here and a rim. It's still axially-- it still
has some axial symmetries. But it's getting a little messy. It's hard for you--
you can't just say that's MR squared over 2. It's got some other mass moment
of inertia about the center. But it's got holes and stuff
in it because of the spokes. So oftentimes, you'll be
given the radius of gyration because it's a little
difficult to give you a mathematical description
of what the actual Izz is. That's often why you do it. And the thing is,
you can measure. Rather than try
to calculate, you can actually just
measure the mass moment of inertia of something. So how would I measure-- let's
say I didn't know any formulas, but how would I measure
the mass moment of inertia of this in the z direction? What experiment would you do? AUDIENCE: [INAUDIBLE]
angular acceleration. PROFESSOR: She
says apply torque. Measure the angular
acceleration. Hang a weight off
of it, known weight. Wrap a string around it. Known mass. Known G. Known torque
around the center. Measure the angular
acceleration. I theta double dot
equals the torque. You know the torque,
you know the measure of the theta double dot. Calculate I. I equals M kappa
squared, and you could just say, well, the kappa
for this system is. That's how you use it. I'll give you a very
common example, really hard to calculate mass
moment of inertia. A marine propeller. You actually do want to know
the mass moment of inertia about its center for purposes
of torsional oscillations on the shaft, et cetera. Hard to calculate. Pretty easy to measure. So how many degrees of
freedom does this problem has? Well, when we say it's
2D, it's a rigid body. But it's 2D, which means
it's lying on a plane. It's a planar motion problem. Only allowed to rotate in z. Not allowed to rotate in around
the x-axis or the y-axis. So for the rigid body, six
degrees of freedom possible. There's two immediately that
you said it can't rotate. So we're down to four. It cannot translate
in the z direction. We're down to three. So that leaves us what? So the degrees of freedom
for this problem is 6 minus 3 constraints is 3. That means we have to have
three equations of motion. And they would account for
what are the possible-- in other words,
saying this, what are the possible motions
now of this problem? AUDIENCE: [INAUDIBLE]. PROFESSOR: x, y, and z. Or rotation in z. So notice we've
set the problem up. Now to go about solving it,
we need a free body diagram. So here's my disk. Here's the force. Any other-- and there certainly
has weight in the z direction, but there's no z acceleration. So in the plane of
the board, that's the only external
forces acting on this. And there's our G. Now this problem-- and
I'll say generalize on this in a few minutes. This problem can always
be restated as-- recast, let me put it that way--
as, here's your point G with a force acting
on this center of mass. See, this force doesn't go
through the center of mass. This force goes through
the center of mass. I'm going to replace that
problem with this problem. A pure moment acting
about the center of mass. You can always make
this transition. And I'll do the general case
for you in just a minute. But you can always do this. So that's kind of my
second point here. Third point, we need then--
this is our free body diagram. We need apply our
laws of motion. So sum of the forces in
the y are-- now remember, this is z coming
out of the board. y this way, x this way. Sum of the forces in the y? Zero, M acceleration of G
in the y direction, zero. So you know there's no
acceleration in the y. So it has three
degrees of freedom. That's the first
equation of motion. It gives you a trivial result.
So y double dot is zero. That's your first
equation of motion. Then you have the second
equation of motion, sum of the forces
in the x direction equals just our F i hat. It's positive x
direction because we were clever in how we set
up the coordinate system. And that's got to be Mx
double dot, i hat direction. So we know right away that
x double dot is the force F divided by M. And that's
150 newtons over 75 kilograms, or 2 meters per second squared. All right. The third one,
then, is the moment of inertia with respect
to G in this problem-- excuse me, the angular momentum
in some IzzG times omega z. And that's what
we're looking for. So this is another way of--
IzzG theta dot k hat direction. And we're going to apply that
the external torques, some of the torques, is
dH respect to G dt. And that's going to give
us IzzG theta double dot. So this third class
of problems you are best just working with
respect to the center of mass. That's kind of the point here. There's no points of contact. There's just known
external forces. You have to deal with them. Do your work with respect
to the center of mass. So we have force equations. We have moment equations. And basically you know
Izz for this problem is kappa squared M.
And you're given M, and you're given kappa. So you can now-- and what
we have to-- actually, the last thing left here is
to figure out the torque. What's the torque? Well, it's R in this
direction crossed with F in that direction. So it's Rj cross with
Fi, j cross i, minus RF, minus RF, k hat. So you can now solve for theta
double dot as RF over Izz, or RF over M kappa squared. And that says-- minus says
it's rotating this way, which is what you'd expect. Right? Now I meant to ask you a
question before we started. But think about this. If I had, right at the
beginning, had said, OK, this is a problem. If I grab this string and
pull in this direction, will there be any motion
in the y direction? I meant to start that way. I'm really kicking myself
for not doing that. Because a lot of people
think that there's possible, that it could move
off in the y direction because it's not being
loaded symmetrically. You're pulling on a side. Some people think it'll kind
of try to move away like that. It doesn't, does it? So an important generalization. We've got a rigid body. You have a force acting on it. Has a mass center here. So perpendicular to that
force is some distance. We'll call it d. You can always equate this
problem to-- and set it up as-- a force. Conceptually, you
can think of it as equal and opposite forces. But it's cancel one another. It's just like I've done
nothing to this problem. And then a force acting
at this distance F. This is our distance d. So this problem is
identical to that problem. I've just added and taken
away two more forces. So the total forces
on the system are still F. And there's still
an F operating at a lever arm d. But now, if I had put
these two together, they are equal and opposite. And they form a couple, a
moment, acting like that. So this is equivalent to--
there's G with an F on it and a moment M0. And this M0 is my D cross F. So that's the generalization
for what I did up here. I went from that to that. And this is why you can do that. And so now if you have an object
and lots of different forces acting on it, and this is
Fi down here and here's G, you can draw a radius
from G to this point. So I'll call that Ri
with respect to G. Then the way to generalize this is
that this is equal to some F total and a moment acting at G. And all that you have
to do there is F total is the summation of
the Fi's, vector sum. And the MG, the M with
respect to G here, is the summation of the Ri
with respect to G cross Fi. So that's the generalization
for multiple forces on a body. So you are making an equivalent
force acting at G and a moment acting at G. I shouldn't
call this little g. That's really confusing. So that's the
generalization when you need to do problems like this. Catch your breath while
I scrub a board here. Now we've got to move on to
these class four problems. Moving points. An example is the truck
problem that you had. Known acceleration. So there's two common ways
of doing this problem. You can do this problem
by summing forces at the center of
mass of that pipe and summing moments around it. But the moment around this
comes from a friction force here, which you don't know. So that introduces
an unknown that you have to then solve for. So if you work around G for
this pipe, you can do it. You can work around G. You
could say sum of the moments around G, sum of the
force with respect to G. But you have to deal
with unknown forces. So you're working with respect
to G implies unknown forces, e, for example, friction. So you'd really maybe rather
around the point of contact, A. Because if you sum
your moments about A, the friction force
has no moment arm, and it doesn't
appear in the answer. But this gets trickier. This is a little more
sophisticated, I'll just call it, step. And you need-- to do this,
you need a little theorem. So to work with
respect to A, you need to be able to say that the
angular momentum with respect to A-- you now are working
around a moving point, maybe accelerating. It's very handy
to be able to say it's the angular momentum
around G, which is easier to calculate, plus
RGA, the distance from the center of mass to
the point you're working on, cross the linear momentum
of the system with respect to the inertial frame. We need this. This is a formula we need. And see why this is true? This is sort of thing--
this is a formula that's come out of the blue here. And why is it true? So I don't usually like to do
proofs, but the proofs of this [INAUDIBLE] on the board. But the proof of this
is really quite simple. Here's a little mass point Mi. And this radius is R
of i with respect to A. This is R of G
with respect to A. And therefore, this is R of i
with respect to G is this one. And we know that this
plus this gives us that. We can say Ri particle i with
respect to A is RG with respect to A plus Ri with respect
to G, all vectors. So the angular momentum of that
body from the basic definition of angular momentum
is the summation of all the little mass bits
of the Ri with respect to A cross the linear momentum
of each little mass bit. But we can expand
that with that sum. So this is the summation
of my RG with respect to A plus Ri with respect
to G cross Pio, summation over all the mass bits. I'm going to expand this. And I can expand this into
this times that, summations of this times that,
and this times that. So that becomes a RGA cross
summation of-- what's Pio? This is a little Mi's,
Vi with respect to o. Each one has velocity. Each one has a mass. So this is Mi Vi with respect
to o plus the summation RiG's cross Mi Vio's. Now, notice I pulled this
one outside the summation. That's because
this is a single-- this is a fixed number. It doesn't change
in the summation. It's just the distance from
my starting point to G. So I can pull it out
and do the summation and then do the cross product. What is the summation of
all the MVi's for the body? That's the momentum of
each little particle. Add them all up,
what do you get? This is RGA cross P
with respect to o. That's this term. And this is all the
little distances from the center of mass to
the cross with the momentum of each little one. What's that? Well, this looks like a
definition of angular momentum. This is the angular momentum
of every little mass particle with respect to G added up. This is H with
respect to G, which is what we set out to prove. So the H with respect to A is
H with respect to G plus RGA cross the linear
momentum of the body. Yeah. AUDIENCE: Can you also do this
by writing Pi with respect to o, I mean the
velocity part of that, as the sum of the velocities? PROFESSOR: You got to
keep the M's in there. AUDIENCE: Well,
yeah. [INAUDIBLE]. But instead of writing
out R as a sum, you can write out the velocity
as the sum of the velocities with respect to the origin. PROFESSOR: Are you talking
about this term here? AUDIENCE: No, the definition
of angular momentum. Yeah. The velocity-- PROFESSOR: If you
could figure out-- if you had the angular
momentum [INAUDIBLE] and multiplied by Ri, that
is the angular momentum with respect to
A. But I broke it apart so that I could show you
that this formula I want to use has two pieces. I want to use that. So that if I can use that-- it's
easy to get H with respect to G sometimes. It's really hard to know
what to do with things that happen around this point A. So now let's go back. I think, to understand
this, we need to go back to our truck problem. We now have a formula that
you know where it comes from. We have our truck that is
accelerating at x1 double dot. And we want to find out what's
theta double dot for that pipe. What's it doing? So first we needed
some kinematics. And in particular, what is
the x2-- did I label this very well? I didn't. So here's my pipe. Here's my truck bed that
it's in contact with. Truck's moving at x1
double dot, we know. In an inertial
frame, the movement of the center of mass
of my pipe is x2. And it has some angular
rotation I'll call theta. So the movement
of this guy I need to be able to express in
terms of x1 and theta. Well, if this is fixed to
the truck and the truck move forward, then x2
would be equal to x1. But in fact, it rolls
back a little bit. And the distance
this point moves if it rolls through
an angle theta is it rolls backwards
an amount R theta. And we're going to need to
take two derivatives of that. x2 double dot equals x1 double
dot minus r theta double dot. So that's a little
kinematic relationship we need to do this problem. Next we need to apply
a physical law, which is the one I've derived. So this is our physics
here now, our physical law. And that's the external
torques, dH with respect to A dt, plus VAo cross Po. And in this case, is VAo 0? No, in fact, it's x1 dot, right? It's in the i direction. How about what direction is
P, the momentum of the pipe? Does it move in the y direction? Up, down? No. It only moves in the x. So this velocity
is only in the x. This is in the x, or
the i hat direction. This cross product is zero. So it just happens
that they're parallel. So this thing goes to zero. We don't have to deal with it. That's because these
guys are parallel. Parallel motion. But you did have to consider it. You did have to think about it. It's not just trivially 0. OK, so that means that the
torques about A is just dHA dt. And HA for this
problem is i-- well, it's HG, which is
IzzG theta double dot. But I have to put in this--
theta dot, excuse me. This is just the
H. I have to put in the second term,
RG with respect to A cross P with respect to o. So here's my HA. I'll write it again up here. This is IzzG theta dot in the k. And now this second term. R is Rj. My y-axis is up. The radius of this thing
is R. Here's the radius. So the moment arm is Rj crossed
with the linear momentum of that piece of pipe, which
is the mass of the pipe times x2 dot in the i direction. j cross i is minus k. So Izz with respect to G
theta double dot-- theta dot. I keep taking the derivative
a little too soon. k minus RM x2 dot k. And now some of the torques, d
by dt of H with respect to A. Take the derivatives. IzzG theta double dot k
minus RM x2 double dot k. And fortunately, none of these
unit vectors are rotating. So we don't have to
deal with any of that. And I'm almost to
the end, but now I have to go back to that original
kinematic relationship, which allows me to express x2 in
terms of x1 and R theta. And if I substitute that
in here and solve for it, I get theta double dot equals
MR over IzzG plus MR squared. x double dot whole
thing, x1 double dot. So you've accelerated
x1 double dot. The thing starts rolling. It's actually rolling
backwards, which is a plus theta direction, at this rate. We did this using
this formula for HA. Now, the beauty
of this formula is that it works for
any points that are moving, even
can be accelerating, all sorts of nasty
conditions, it's true. It's based on the fundamental
definition of angular momentum. So the book-- yeah. AUDIENCE: [INAUDIBLE]
final point here. I just want to make
sure i understood. In this first line
here, [INAUDIBLE]. That stroke doesn't
mean that V sub A was 0. It meant that that term is 0. PROFESSOR: This term's
zero because these happen to be parallel, but the
product is zero in this case. If it had not turned out to be--
they were different directions. It had to be a non-zero term. And you would have to bring it
along and take its derivative along with the other stuff. But this is a really
powerful method. Now, the book is reasonably
good in lots of points. But in chapter 17,
when it does problems that are kind of like this,
it introduces something that I find it hard to
digest what they're doing. There's particularly--
where's Matt? There's equation 1715. When you get to that bit of
the book, just ignore it. Use this method. The author tries to give you a
little trick that you can use. But the problem with tricks
is you have to memorize them. So what I've shown you today
is based on basic definition of angular momentum. That expression at the top
is always usable, not just special little conditions, which
is what the formula in the book is generated for. We've got a couple of minutes. Let you ask questions,
and then I'll just pose a conceptual
problem or two and ask you what
method you'd use. Yeah. AUDIENCE: So [INAUDIBLE]
when you [INAUDIBLE] dHA dt, is that equal to 0 because there
is no torque in the system? PROFESSOR: Oh, you know, boy,
I'm glad you caught that. Yeah, in order to
be able to do this, we've got to know
something here, right? Important catch. Thank you. Why is this true? AUDIENCE: [INAUDIBLE]. PROFESSOR: So we
chose-- that's why we chose to work around point
A. With respect to that point, there are no
external forces that create moments on that pipe
with respect to that point. And that's why you
can say that the time derivative of this
angular momentum is equal to zero because
there are no external torques. If you had picked G
to do this problem, would the sum of the
torques about G be zero? No, you'd have to put that
friction force in there and have RF and figure out F is. We completely avoided having to
calculate the friction force. That's the point of being able
to use techniques like this and make your computations
around points of contact. So textbooks have lots
of problems like this. You've got a box on a cart. And your kid's pushing it,
and he gets a little exuberant and pushes a little too
hard, accelerates the cart a little too fast, and the
box falls over and breaks the lamp or whatever's in it. And if it's this way,
and I accelerate it, it's much more tolerant. Falls over easier this way. But if I asked you,
gave you a problem and said, calculate the
maximum acceleration that I can put on this
object such that it just barely-- just right at
the edge of tipping over, but doesn't tip it over. What's that maximum acceleration
that you don't tip it over? What method would you use? I gave you four classes
of approaches to problems. AUDIENCE: [INAUDIBLE]. PROFESSOR: I hear a four. Anybody else want to bid here? More fours. Would three work? AUDIENCE: [INAUDIBLE]. PROFESSOR: Why? He says three would
be complicated. Three means taking moments
and forces with respect to the center of mass. If you do that, what do you have
to deal with in this problem? AUDIENCE: [INAUDIBLE]. PROFESSOR: You have to, then--
around the center of mass there's a friction force. There's a normal
force pushing up here. The way you do this problem
is where it's just barely about to go, all of the force
is pushing on this corner. Think of it. It's just lifting up a fraction. All of that contact point
moves to right here. You have an upward force
and a friction force, and they create
a moment about G. But if you do the
forces around G, you have to solve for those two. You do the forces around A.
The trick here is figuring out where's A. But if you
realize A is right here, and you do what we just
did, this problem's just a piece of cake.
