Module 11: Impulse-Momentum Relationship; Define Impact

Loading...

Из курса от партнера Georgia Institute of Technology

Engineering Systems in Motion: Dynamics of Particles and Bodies in 2D Motion

170 оценки

Georgia Institute of Technology

Engineering Systems in Motion: Dynamics of Particles and Bodies in 2D Motion

170 оценки

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.

Из урока

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.

Module 9: Work and Kinetic Energy Principle for Particles/System of Particles; Work of a Linear Spring6:21

Module 10: Work Done by Gravity; Work Done by Friction; Solve Work-Energy Problems for Particles/System of Particles8:41

Module 11: Impulse-Momentum Relationship; Define Impact6:30

Module 12: Define Coefficient of Restitution; Solve an Impact Problem11:03

Module 13: Define Angular Momentum; Euler’s 2nd Law (The Moment Equation)7:12

Познакомьтесь с преподавателями

Dr. Wayne Whiteman, PE
Senior Academic Professional
Woodruff School of Mechanical Engineering

Hi, this is module 11 of Two Dimensional Dynamics.

Today's learning outcomes are to derive the

impulse momentum relationship, and to define impact.

So let's start off with the momentum form of Euler' 1st law.

Let's look at the definition of momentum for a system of particles.

So I have a system of particles here, each particle

has a mass m, and so this is m sub i,

this is m sub j.

Number of particles say N, capital N particles.

Each one of these particles has a momentum, a linear momentum, and

the momentum is defined as mass times velocity, and I show that there.

So, the summation of the momentum for the system of particles is each one of

these masses, times their velocities summed up

over a total number of capital M particles.

And,

I can then take the position vector from some fixed

origin in an inertial frame form this O to each of the particles.

And I can take the dot, time derivative of

that position vector, and it's the same as the velocity.

So I've replaced it there.

And then we saw in a previous module, that the summation of mR over

a system of particles is equal to the total mass times a position

vector from O, to the mass center by the definition of mass center.

And so now we can extend that to, we can say we have

an infinite number of particles, and we

extend that to a continuously distributed mass.

Now if I take the linear momentum again,

linear momentum's given the symbol capital L.

The mass is constant in this course, and so we have the derivative of

the position vector from O, to the mass center is the velocity of C.

And then I can take the, the derivative of that linear momentum for the body.

It's equal to mdvcdt since mass again is constant.

That becomes ma and we're back to the sum

of the forces equals ma of the mass center, which is Euler's 1st law.

And so what we find, is that the summation

of the forces, on a continuously distributed mass is

equal to the time derivative of the linear momentum,

just the same as it was for a particle itself.

So here's the momentum form of Euler's 1st law.

And now let's come

up with what we'll call the impulse-momentum relationship.

And so we're going to integrate these momentum form of Euler's 1st law.

And we have from some time, t1, to some time t2, the sum of the forces acting

on the body, is equal to the change in the linear momentum.

And so this left hand side, the intergal over some

period of time, the force is just called an impulse.

And the change in capital L is the change

in momentum or in this case specifically the linear momentum.

So, we now have this impulse-momentum relationship, a

particular kind of impulse is called an impact.

And an impact occurs when we have a collision of bodies over a

very short period of time and so this delta t goes to 0.

And then if the sum of the external forces in the direction of

the momentum vanishes, if this left side

vanishes, what we have is conservation of momentum.

And typically in these types of problems where we have a very large

impact force, we ignore relatively small forces,

oftentimes ignoring gravity and other negligible forces.

So let's look at an example.

I have a block at rest on a smooth horizontal surface being struck

by, so this is the block at rest here, being struck by

an identical block sliding with a velocity, or a speed actually of v.

And I want to find the velocity of the two

blocks after the collision assuming that the blocks stick together.

And so let's look at a little demo of this situation.

So, I have a block that's sitting at rest. I have another block.

We're going to assume this is a smooth surface, so it's friction-less.

And as this block comes along with a velocity v,

it sticks to the other block, and they go off together.

And so we're going to neglect friction.

We don't have any forces in the direction of motion, and so we can use conservation

of momentum. And if I go back here now, by conservation

of momentum, the total momentum before is equal to the total momentum afterward.

The only momentum I had before the collision

is the left block times the initial velocity V.

The right block has no velocity.

After the collisions just they both move together.

The m left and the m right are going to have the velocity

V after, which is what we're seeking to find for this problem.

And so I know that V after, since the blocks are sliding

together is going to be the block left and the block right's velocity.

I can just do the algebra here.

I've got M left times left divided by M left plus M right.

We said that the blocks are identical, so they have the same mass.

So V after becomes vm left over 2m left,

or v over 2.

And so, as these blocks slide

together after they've impacted, they stick together.

The original block has a velocity V.

Both blocks sliding off together by the

conservation of momentum have half that velocity.

So, I would like you to know that if

the blocks don't stick together, then conservation of momentum,

linear momentum is not enough to find the

subsequent velocities because I only have one equation.

And we're going to have two unknowns for the velocity of

each block and so we'll need another piece of information.

So let's slide back over here and look at our situation.

In the problem we just solved, the blocks go

off together in the same velocity, but in reality

in most cases, when the blocks hit each other,

they bounce off and we have two different velocities.

And so, we're going to need another piece of information

and we'll pick that up at the next module.

Ознакомьтесь с нашим каталогом

Присоединяйтесь бесплатно и получайте персонализированные рекомендации, обновления и предложения.

Coursera делает лучшее в мире образование доступным каждому, предлагая онлайн-курсы от ведущих университетов и организаций.

© Coursera Inc., 2018 Все права защищены.

Загрузить из App Store

Загрузить в Google Play