Download Physics 211 4-6-09 Monday My name is Dave. Dr. Barnes is out of

Physics 211
4-6-09
Monday
My name is Dave. Dr. Barnes is out of town today.
Chapter 10, chapters 5 and 6
So you've been talking about rotational Kinematics. With that you have the familiar equations
that used to be just linear:
Now in rotation speed, now you have
Then you started looking at dynamics and what caused acceleration.
We are going to translate this to rotation. We are going to find the rotational equivalent of
force, which is torque. And then acceleration, you've seen what that is. Also we have
something in place of mass. That's what I'm going to start the lecture one. This symbol i
stands for rotational inertia or moment of inertia.
Suppose I have a wheel where all the mass is in the tire with a radius (R) and a
mass is M. Let's compare that to a small disc whose mass is also M. One
has a smaller radius, both have the same mass. If they rotate at the same
angular velocity, which has more energy in it? The bigger radius has more
energy. Its got a greater tangential velocity where all the mass is. When you
talk about rotation you need a special variable that represents the geometry of what you are
working on. This depends on geometry.
The simplest case is where there isn't much geometry to speak of.
Rotational inertia: The summation of all the masses in a body times their radius
squared. If you have a bunch of masses that are connected and they are all
rotating around the center here, and they all have their own radii, you take each
mass, multiply by its radius, sum them up and you get the moment of inertia relative to this axis
of rotation. That's kinda nice when you are talking about a collection of dots.
We are interested when you take a body like a plate or a
solid object. I'm going to start with a hoop rotating
around an axis. The hoop has a radius, and the
thickness is much smaller than the radius. the hoop has
mass m. You could say that my moment of inertia is the
summation of the little masses times the radius squared.
But we have to convert this to integral language if we are going to talk about a continuous body.
imagine breaking this hoop into little pieces with a radius ri. I can pull out an r squared, and
now I have this. And you ask, what does dmi mean? You use the density of the object.
Everything we work with, you are going to do dm= density dV. Density times a small piece of
volume. You may not have seen volume integrals. Even if you haven't, I'm going to try to
keep this simple. This is constant density of a homogenous object. Its the same density no
matter where you go in that hoop.
In your book you'll see three versions of density:
• There's the usual one, that's mass per length cubed. (mass per volume).
• Then there's the aerial density (mass per length squared).
• Then there is linear density. (mass per length).
In a problem like this with a thin hoop, its a linear kind of an animal. I could actually use linear
density for this example. And then I can say that this linear density is going to equal the total
mass divided by the total length of this hoop.
I'm going to say the whole length of the hoop. I'm
changing the volume into an L. I'm kinda changing this
into a linear problem. You can look at this and say, well,
its 2 pi R and it is going to cancel. I can integrate from
zero to 2pi. I'm trying to add up the whole length of the
circle, I have to add up arc lengths, this is how I do it in an
integral.
Let's move on to a more complicated example.
Student: Is there a reason you use a big R?
When you are talking about a dimension, you capitalize it.
So what if you have a rectangular plate that rotates? I am
interested in it rotating about an axis perpendicular to its surface that goes through the center.
Let's give it a dimension of A in the X direction, a dimension of d in the Y direction, and C in the
Z direction. I'm going to start writing my integral. My moment of inertia is equal to the integral
of the volume.
Have you all seen this notation? (Mixed
response)
I'm interested in r squared, r is the distance
from the axis to whatever little piece I'm
integrating. r is the distance from the
axis. It doesn't depend on z. I can also
look at the density, mass divided by
volume.
Let's do the z integral first. Its from 0 to C.
I'm going to rush through the derivation. Taking notes is optional.
That's a boring derivation, but it takes into account the shape of an object. In a circular disc,
you'd get a totally different answer. What if I move the axis of rotation? Would the moment of
inertia be greater? Yes. Now there is stuff that is twice as far away. Its going to take more
effort to get this thing rotating.
There is a handy theorem in your book. I don't have time to derive it now, but its done in your
book. If you have an object and you find the moment of inertia around the center of mass, if
you find the moment around the axis of the center of mass, you can calculate the moment
around the new axis:
Problem #5 in your homework is a difficult one. It is related to this. I
spent 45 minutes playing around with it, I still have not figured out how
to answer it. I promise on Wednesday I'll have a good set of
directions for you to follow. It involves having a circle shape cut out
of it. So heads up on that one.
I talked about the parallel between the linear case and the rotational
case.
Have you heard of the vector cross product? You've got torque equals r
cross force. That's probably unfamiliar language. Let me demonstrate
what this means 3 dimensionally. Let's say you have a point and a force
being applied at another point, and then we've got r going through it like
this. What this cross product does on a set of vectors, you get a resulting
vector perpendicular to the plane defined by the other two vectors, and
that would be the torque vector in our case. You multiply each of these
vectors by two components of the other vectors. It would take a whole
class period to explain how that works. But this is the general definition
of the cross product. In this class you are using a special case where you
have your axis here and then both the force and the r are perpendicular to
the axis of rotation.
If I pull on a lever, would it turn around its axis? If this were a wrench and I
pulled out on it, would it spin that bolt? No. You want to have 90 degrees
here meaning your torque is maximized when you angle is 90 degrees, so
you use sine. This has the same units as work.
When you give answers to torque questions, never use Joules. I see them using Newton
meters in the book. The correct units are meter Newtons. If you have to give torque to the
computer, stick with Newton meters.