This course is an introduction to the study of bodies in motion as applied to engineering systems and structures. We will study the dynamics of particle motion and bodies in rigid planar (2D) motion. This will consist of both the kinematics and kinetics of motion. Kinematics deals with the geometrical aspects of motion describing position, velocity, and acceleration, all as a function of time. Kinetics is the study of forces acting on these bodies and how it affects their motion. --------------------------- Recommended Background: To be successful in the course you will need to have mastered basic engineering mechanics concepts and to have successfully completed my courses en titled an “Introduction to Engineering Mechanics” and “Applications in Engineering Mechanics.” We will apply many of the engineering fundamentals learned in those classes and you will need those skills before attempting this course. --------------------------- Suggested Readings: While no specific textbook is required, this course is designed to be compatible with any standard engineering dynamics textbook. You will find a book like this useful as a reference and for completing additional practice problems to enhance your learning of the material. --------------------------- The copyright of all content and materials in this course are owned by either the Georgia Tech Research Corporation or Dr. Wayne Whiteman. By participating in the course or using the content or materials, whether in whole or in part, you agree that you may download and use any content and/or material in this course for your own personal, non-commercial use only in a manner consistent with a student of any academic course. Any other use of the content and materials, including use by other academic universities or entities, is prohibited without express written permission of the Georgia Tech Research Corporation. Interested parties may contact Dr. Wayne Whiteman directly for information regarding the procedure to obtain a non-exclusive license.
Module 12: Define Coefficient of Restitution; Solve an Impact Problem
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Из курса от партнера Georgia Institute of Technology
Engineering Systems in Motion: Dynamics of Particles and Bodies in 2D Motion
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Georgia Institute of Technology
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Work-Energy Principle for Particles/Systems of Particles
In this section students will learn the work-energy principle for particles/systems of particles, impulse and momentum, impact, conservation of momentum and Euler's 2nd Law - Moment of momentum.
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Dr. Wayne Whiteman, PESenior Academic Professional
Woodruff School of Mechanical Engineering
Welcome to module 12 of two dimensional dynamics.
Our learning outcomes are to define the Coefficient of Res,
Restitution and give examples, and to solve an Impact problem.
So we left off last time where we had solved a conservation
of momentum problem for two blocks that stuck and moved off together.
And so let's go back over and look at this situation.
We had one block coming in with a velocity
of v when they moved off together.
If they stuck together, their total velocity for
both blocks after the collision was v over two.
But I said last time that in most collisions we have some rebound
if you will, and so they don't move off at necessarily the same velocity.
And in that case we have to have another piece
of information which we're going to call the Coefficient of Restitution.
And here is the definition
of the Coefficient of Restitution.
It is that the relative velocity of the bodies after impact
is divided by the relative velocity of the bodies before impact.
You can find the development in any standard dynamics text,
but this is what it ends up coming out to be.
You'll notice that the B's and the A's are reversed but you
can see that in the derivation when you when, when you research that.
[COUGH] so we can have a range
of values for the Coefficient of Restitution.
If e is equal to 0, we call that a perfectly plastic collision.
And if e is equal to 1, we call that a perfectly elastic collision.
For a perfectly plastic collision, let's give it an example.
Let's say that body A is the earth, or the ground
if you will, and we're going to give it a velocity of 0.
Body B is my other body.
And so if I have a body coming in with
a certain velocity, v B1, and after it
smacks the other body, the Earth in this case.
If it has zero velocity then it's called a
perfectly plastic collision, and e is equal to 0.
And so, if we slide back over here, that was actually
the case we had with the two blocks because they moved together.
They didn't separate.
And so the velocity was the same after the collision.
Here's another
example, I have silly playdoh ball here.
If it comes down and drops and hits the ground and does
not bounce back up, then that would be a perfectly plastic collision.
Now in reality, this does bounce up a little bit so
it does have a little bit of a Coefficient of Restitution.
Let's go back over here. If the velocity after the collision is
equal to the, the relative velocity after the input, is equal to the
relative velocity before the impact, then we have a Coefficient of Restitution of 1.
So, in that case, let's go back over here.
Here I have a superball.
If the velocity before impact is equal to the velocity after impact, and we
call that a, a, a perfectly elastic collision and e is equal to 1,
so the ball would come back to ideally the
ball would come back to its the original height.
We know that's not physically possible.
But you can see that this has a
rather high Coefficient of Restitution, maybe around 90%.
Here are some other examples, this ball maybe
has a Coefficient of Restitution of less than 0.5.
You saw on the intro video
that I'm a golfer. And so here's a golf ball.
Its Coefficient of Restitution is somewhere around 0.8.
I think the regulation is that a golf ball
cannot have a Coefficient of Restitution higher than 0.83.
Okay so that's a feel for the Coefficient of Restitution.
Let's look at impact.
One type of impact we call Direct Central Impact.
First of all, we define the line of action.
That's a common normal to the perpendicular
common normal or perpendicular to the impacting surfaces.
Here's the impacting surface.
A common normal is called the line of action.
We have central impact if both mass centers lie on the line of action.
And we call it a direct impact if both velocities, as I've shown here,
also lie in the line of action. We may also have Oblique Central Impact.
Central being the mass centers continuing to lie on the line of action, but one
or both of the velocities do not have to lie on the line of action.
And we'll see that in a worksheet that we're going to do on the next slide.
One assumption, or a couple assumptions that
we're going to make, are that we have smooth
non-spinning bodies, and that the velocity components
perpendicular to the line of action are unchanged.
So we're only going to get changes in the velocity along the line of action.
So here's our, our worksheet.
We've got two identical hockey pucks. They're going to collide.
There's a Coefficient of Restitution of 0.8
and we want to find the velocity of the,
the, the pucks after collision.
And so first of all, we can apply Conservation of Momentum, Linear Momentum.
I'd like you to do that then come on back,
write the, the equation for the Conservation of Linear Momentum.
So this is what you should've come up with.
Since these are identical pucks, we're going to let m A equal m B equal m.
And so there's the information that we have.
That's not going to be enough to solve the equation alone.
But let's get started and at least look at it in the i direction.
So in the i direction, I have velocity of block A, so I've
got m times the velocity of block A.
The i direction is the sine component of 4 feet per second.
So that's going to be times 4 sine of 30 degrees.
And then I've got plus m B initial, m B initial is going to
be the fourth, fifth component of the 5 feet per second.
So it's going to be 4 5ths, and the 5 feet per second is in the negative i direction.
So that's going to be minus 5 equals m times v A final in
the i direction, and I don't know that. That's what I'm trying
to solve for. Plus m times v B.
This is final, in the i direction. I'm sorry.
That's correct. And v B final in the i direction.
I can cancel the i's and I can simplify the equation so that I get
v A final in the i direction plus v B final in the i direction equals minus 2.
And we'll call that equation 1.
We're going to also use the piece of information of the Coefficient of
Restitution, and so I've got 0.8 equals v B
final minus v A final, both unknowns, what we're trying to find in the problem.
v A initial was 4 sine of 30 in the i direction.
And v B initial
was minus a 4 5ths, and v B was in
the minus i direction. And if I
simplify that, I get v B final in the
i direction minus v A final in the i direction equals 4.8.
We'll call that equation 2. I now have two equations, two
unknowns.
And so I can solve them by combining these equations.
And so when I combine them, let's take this equation up here.
1 we've got v A final the i direction, plus I'll substitute in for
v B final from equation 2, so I've got 4.8 plus v A final in the
i direction equals 0. Oh, that doesn't
equal 0. I'm sorry, it equals minus 0.2, right?
Up here. And so I get v A final in the i direction
as a vector equals minus 3.4 i.
I can then substitute that result back into
this equation down here to find v B final in the i
direction. So I've got v B final in the
i direction equals 4.8 plus v A
final which is minus 3.4. Or v B
final in the i direction
equals 1.4 i.
Okay, so we're good as far as the i direction is concerned.
We now need to do the j direction components.
I want you to recall that I said that the, we're assuming
that the, the perpendicular velocity components
to the line of action are unchanged.
So the velocity component changed in the line of impact direction,
but not in the the off the line of impact direction, or the j direction in this
case. And so I have, in the
j direction, v A initial the j
direction, equals v A final in the j
direction. Or the j component is the
cosine of 30 degree component for body A,
so I've got 4 times cosine of 30 degrees,
or that is 3.46. So v A final as a
vector is equal to 3.46 j.
I can do v B initial now, equal to v B final.
So I've got v B initial in the j direction equals
v B final in the j direction. Or, the v
component, j component is the 3 5ths component along this slope.
So I've got 3 5ths, it's in the minus j direction.
So, it's going to
be times minus 5 or minus 3. So v B
final in the j direction, initiative in the j direction as well, is equal
to minus 3 j. I put all those results together.
I have the total velocity of the pucks, after the collision.
And so I've solved the problem.
And that's a good stopping point for this module.
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