0:04

Hi, this is module 33 of Two Dimensional Dynamics and so

for today we're going to go ahead and solve an engineering problem

or engineering problems using the equations of motion that we developed

last time for a body in two dimensional planar rigid body motion.

0:24

And so again I'm going to use this graphical tool on the left hand side.

I'm going to do my free body diagrams to show my,

my forces and my moments and on the right hand

side I'm going to use these motion vectors or effective vectors

m x double dot, m y double dot and i alpha.

And so here's a, a body and so this body

again would be restricted to only rotating about the z axis.

That's, that's the only motion that I'm looking at right now.

On the left hand side I put on my forces acting on the body

and any effective moments are, excuse me, any moments that are applied to the body.

And then on my kinetic diagram I have my, two orthogonal M A vectors.

In this case I've chosen M X double dot and M Y double dot, and I also have my

rotational vector, i alpha. And so, you'll also recall from a

knowledge of equivalent force systems, that I can sum

my moments about any point on the body, and

I'll take into effect any applied moments, and the

moments due to the forces acting on the body.

And then on the right hand side, if I sum

moments about P instead of C, I can still do that.

I get the effective moment, i alpha, plus these effective forces with their moments

arms, so R from p to c crossed with these effective

forces which are m x double dot, and m y double dot.

And so that will allow us not only to sum

moments about C, but about any point on the body.

2:02

Okay, so let's go ahead and look at an example.

This is a cylinder, and it's got a mass M, a uniform cylinder, mass M.

Radius R, it's released from rest.

You have a coefficient of friction on the surface that's mu.

And we want to determine the motion of the mass center for this body, and we want to

assume that the coefficient of friction is large enough

to prevent slipping of this, this cylinder or wheel.

And so let's go ahead and look at a demo. And here's my, my wheel, and the

first thing I want to ask you is which direction is the wheel going to go?

Up or down the plane?

And so go ahead and answer that, and if I let go.

Whoops. It went uphill.

That's kind of weird.

That's my magic trick for the day. So let's now take.

You can think about that.

Why did that happen?

You saw it it went uphill and also it even started to

slide, so that's not good, because we said we would have no,

no no slipping. So here's here's a better wheel.

When I let it go, it's going to roll down the plane, no slipping.

And I want to find the motion of the mass center, and predict that.

3:21

And so what you should say is, I want to apply these equations to this problem.

And so I'm

going to use my graphical tool of FBD equals KD.

And I'd like you to begin by drawing the free body diagram.

And this is the result you should've come with.

You have your, your weight.

You have your friction force that's going to oppose impending motion

to keep it from slipping and we have our normal force.

And then on the right hand side we're going to have

our kinetic diagram that's going to display these effective forces and

effective motion or effective moments, or these motion vectors, if you will.

I've chosen x down the plane and y into the plane as my orthogonal coordinates.

So, I'm going to have an m x sub c double dot down the plane

and m y sub c double dot into the plane. And by the right

hand rule, I'm going to have an I sub zzc

alpha counter-clockwise. But since I'm only going to have

planar motion, planar rotation, I'm just going to abbreviate that as i sub c alpha.

4:34

Now, what you should realize is that I can

have an acceleration of the mass center down the plane.

But that mass center is not going to have any acceleration

into the plane,' because it's going to go parallel to the slope.

And so this kinetic vector will zero out.

4:52

And so what you should say is, we're going to go ahead

and apply these eq, equations and I'll start with summing forces in

the x direction on my free body diagram so, and set it

equal to the forces on my kinetic diagram in the x direction.

So I will have the

mg force. The x component is the sine component.

So I've got mg sine of beta, and then I've got my friction force.

It's going to be negative in accordance with my sign convention.

The normal force is only in the y direction and over

here, the only effective force I have is the mx sub

C double dot down the plane. So it's in the positive direction.

And we're going to call that equation one. Now the next step is what?

And what you should say is, okay apply the next equation.

So we're going to sum forces in the y direction.

I'd like you to go ahead and do that on your own and come on back.

6:01

Notice that I chose down and to the right positive so this would have been minus

n plus mg cosine beta

equals zero, because there's no acceleration

in the y direction, but that's the same as n minus m g cosine beta equals zero.

So it

ends up the same.

We'll call that equation two and now we have our body.

We have one more equation that we're going

to work with, and that's the, the moment equation.

So, I'm going to go ahead and sum moments about point c on the free body

diagram, set it equal to the sum of the moments about c on the kinetic diagram.

I'll, I'll call counter clockwise, positive.

And so I get, if I sum moments about C, the line of

action of the weight force, and the normal force goes through point C.

So I just get F times its moment arm, which is R.

It's going to tend to cause a counter-clockwise

rotation, so that's positive, equals my I alpha.

Notice the effective force of m x double

dot goes, it's line of action goes through c

so it's not going to cause a moment on the right hand side.

And then I substitute in i sub c around this zz axes.

That's the mass moment inertia.

I would look it up in the textbook reference, or

online, or I could do actually the triple integer again.

But this is the result.

The i, the mass moment inertia of a cylinder

about the mass center through the z axis for

a uniform cylinder is one half m r squared.

And that's theta sub double dot, theta double dot sub c.

We'll call that equation three.

8:12

but I would like you on a, as an exercise

on your own to go ahead and sum moments about

another point and see if you can get the same results.

So, you might choose p as the point of contact, and sum moments about that.

See if you can get the same result. But again, I'm going to leave that as an

exercise on your own. We're going to continue on.

I now have my x equation, my y equation, and my moment equation

and so I can sub three into one. The result I get is, is is shown here.

You can see now that the masses cancel out.

I have a mass in every term.

8:52

I would know

what sine beta is. I know what beta is.

It would be a given.

I know what the acceleration due to gravity is.

And so I end up with two unknowns, theta sub c double dot and x sub c double dot.

So I have one equation and, and, and two unknowns.

So I'm going to need some more information.

And my question to you is what other information

do I know about this problem to finish it up?

9:17

And what you should say is that you're going to have to use kinematics to

relate theta double dot sub c and x double dot sub c, which are, are two unknowns.

So, the kinematics we can use are for a no-slip wheel.

If you go back to the module where I did rolling wheels,

we can use those, those kinematic relationships, and

for a no-slip wheel we found that x sub c double dot, the linear

acceleration of the mass center, is equal to r times theta sub c double dot.

R times the angular acceleration of the mass center.

So I can now substitute, we'll call this equation four if you will.

And sub this is equation four, but we'll substitute this into equation four.

And I get

10:13

m sub g sine beta minus one half m r theta double dot

sub c squared and m r theta sub c double dot on the right hand

side since I, I substituted this in. And so I can cancel the Ms

again, and the result I have is shown here.

10:37

Okay, so continuing on now I'll solve for beta double dot c.

I've got g sine of beta.

If I carry this term to the right hand side, I get

equals three halves or three halves r, theta sub c double dot.

And I can solve for theta sub d, c double dot.

That equals 2g sine beta over 3r.

11:05

But in the problem, I want to find the motion of the mass center, so

I've got to put this in terms of x for the motion of the mass center.

I can use my kinematic relationship again.

I've got theta sub c double dot equals x sub c double dot over r.

If I substitute that in, I get x sub c double dot equals 2g sine beta over

r, or excuse me, over 3.

11:34

Now before I continue on to integrate this and find the motion

of the mass center, let's go ahead and look at the, the friction.

This was equation three if you'll recall. I can now, I can now cancel

the Rs and I end up with F equals one half MR theta

sub c double dot.

Let's go ahead and substitute in for theta sub c double dot.

And I get f equals one half m r

times 2g sine beta over 3r.

12:42

So I've got the integral would be x sub c dot, taking

acceleration to velocity equals 2g, is a constant, sine of beta

is a constant, over 3, so that's going to be multiplied by time.

Plus when I integrate,

I get a constant of integration.

However, since this, a body is released from rest, that goes to zero.

[BLANK_AUDIO]

I can integrate again to get displacement, so I've got x sub c.

Let's go ahead and do that in black, x sub c, equals.

Now I've got 2 g sine beta over 3.

When I integrate t, I get t squared over 2.

And I'm going to get another constant of integration, c2.

But, for the,

it's, again, since I'm going to call my datum when

I release it, 0, this C sub 2 goes to 0.

And, so I end up with the answer to my

problem, which is what is the motion of the mass center.

X sub c is going to be equal to g times t squared over

3 sine of beta. Again,

because these twos cancelled out.

14:30

How much friction am I going to need for the no-slip condition?

Well let's look at that.

The maximum friction

I will be able to to get is mu times n, for the Coulomb friction.

So, I know that the friction has to be less

than or equal to that max friction I'm able to get.

14:51

Okay, so that's going to be less than or equal to mu sub n.

And so, I'll substitute in for f.

That's mg sine of beta over 3 is less than or equal to mu sub

n. I'll substitute in for n as mg

cosine beta, and that leads me to mu

has to be greater than or equal to tangent beta over 3.

15:29

And that makes physical sense, because as beta gets larger, that means

I'm going to have to have a

higher coefficient of friction to prevent slipping.

Okay, so these are rather involved problems.

But you get better and better by practicing.

And so I've got a worksheet for you to do. Here's a, a crank arm,

a mechanically engineering type mechanism. So go ahead and practice this on your own.

I'd also ask you to look for other references for problems like this.

The more you practice these problems the better you'll get.

And so I'll see you next time.