Module 17: Solve an Instantaneous Center of Zero Velocity (IC) Problem

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Do curso por Georgia Institute of Technology

Engenharia de Sistema em Movimento: A dinâmica de partículas e corpos em movimento 2D

165 classificações

Georgia Institute of Technology

Engenharia de Sistema em Movimento: A dinâmica de partículas e corpos em movimento 2D

165 classificações

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.

Na lição

Planar (2D) Rigid Body Kinematics I

In this section students will learn about planar (2D) rigid body kinematics, relative velocity equation, rotation about a fixed axis, instantaneous center of zero velocity, and relative acceleration equations.

Module 14: Define Rigid Body Kinematics; Identify three types of Planar Rigid Body Motion; Derive Relative Velocity Equation11:12

Module 15: Solve a Relative Velocity problem9:20

Module 16: Define and Locate the Instantaneous Center of Zero Velocity (IC)10:21

Module 17: Solve an Instantaneous Center of Zero Velocity (IC) Problem9:30

Module 18: Define Angular Acceleration; Derive the Relative Acceleration Equation7:54

Module 19: Solve a Relative Acceleration Problem9:39

Conheça os instrutores

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

[BLANK_AUDIO]

Hi, this is Module 17 of Two Dimensional Dynamics.

Our learning outcome for today is to

use the instantaneous center of zero velocity, which

we discussed last module, to find of velocity

of bodies in planar motion, two dimensional motion.

And so, here was the information that we, we came up with last time for the IC.

We said that if we can locate the IC, any point on a body is going to be

equal to a mega of that body crossed with a position vector from the IC to Q.

And so, let's go back and look at the crank arm piston

cylinder problem that we looked at

previously using the relative velocity equation.

This just, the IC method just give us another tool

for solving these equations, which goes from knowns to unknowns.

We know the angular velocity of body B1, we know

that point O is a fixed point. So that's the IC for

body B1. And, so I can then write an equation

using the IC. I can say that the velocity of A is equal

to, theta dot body B1 in the K direction,

crossed with R from the IC of body B1

to point A. Or the velocity of A is equal to again

theta dot B1 is 10 clockwise which is going to be in the minus

K direction, so I've got minus 10 K crossed with I.

I go minus three in the I direction, plus four in the J direction.

And, I get the velocity of A then, is equal to 40 I

plus 30 J inches per second.

Okay, so we keep going towards unknowns again.

To go from A to B, I'm going to use the IC method so I need to

locate the IC for body B2 or

the bar from A to B. We know that the velocity of A is up

and to the right. 40 I plus 30 J.

I can drop. We also know that the velocity of B

then is going to have to be to the

right because it's a slider mechanism or a piston.

So this is VB.

I can now drop perpendiculars from both of those velocity vectors.

Look where they intersect and that will be my IC.

So if I drop a perpendicular from this velocity vector and

a perpendicular from this velocity vector. This is perpendicular.

This is perpendicular.

Where they cross, will be the IC for body B2.

And

we know that this is on a four

on three slope, okay. From our dimensions

here we also know that

this distance is 12.4. Lets do this in a different color here.

So this is 12.4 inches from

B over to

A. And

I, therefore know that this distance from O to B is 9.4 inches,

because that's minus 3. And

I want to locate the location of the I for, body B2.

I need to know this distance or this height,

which I'll call H, by similar triangles I know

that H is to 4 as

9.4 is to 3 and so H

ends up being 12.53 inches, and that's the distance from here to here.

So I've got the location of the IC for body B2.

I can now use it to finish up the problem.

So here again, here's the location of the IC,

it's 12.53 inches below B for body, bar A B.

I, I have already found the velocity of A. It's 40 I plus 30 J or 50 inches

per second total, on, at this direction. So I've

got the velocity of A, by my

instantaneous center equation

is equal to theta dot B2.

Body B2 in the K direction. That's omega.

Crossed with R from, the I for body B2, to point A.

Okay.

And when I do that I get velocity of A was found to be 40 I plus 30 J.

The angular velocity of bar B2 is unknown, so that's theta

dot B2 K. Crossed with and going

from I to A, the position

vector is minus 12.4 in the I direction, and plus 16.53 in the J

direction. So this is going to be minus

12.4 I plus 16.53 J.

And I can now do my cross products so I get 40

I plus 30 J equals let's see, K cross

I is minus J so that, excuse me, K cross I is

plus J. But I have a minus 12.4 times

theta dot.B2 so that's going to be

minus 12.4 theta dot B2 in the J direction

and then I've got K crossed J is a minus I

16.53 theta dot B2 in the I direction.

Now I can match components solve for the unknown which is theta dot B2.

I'll match I components.

[BLANK_AUDIO]

When I do that, I've got 40 Equals minus 16.53 theta dot B2.

Or theta dot B2 [COUGH]

equals minus 2.42.

Just as we found with the relative velocity equation.

So we should get the result, if we don't we know we'd done something wrong.

It's interesting to note that if you match

J components, you'll get the same result for theta

dot B2 which you should. Finally we, we found the velocity of A.

We found the angular velocity of bar B2.

Now using that angular velocity in the IC I

can find the velocity for any point on bar AB.

So, the one I want to find is the velocity, the piston pin B.

And so I'm going to write my relative velocity equation again.

And I'll get the velocity of B is equal to

theta dot B2 in the K direction crossed with a position vector from

the IC for body B2 to point B.

And we know that theta dot B2 is minus 2.42.

We solved for that

R from I of B2 to B is just 12.53 in the J direction.

And if I do that cross products, I've got K cross J is minus I.

But I've got a minus 2.42 times 12.53 so minus a minus is a plus.

I get 30.3 and it's in the I direction

inches per second for the velocity of bar B.

And so we get the same answer as we did when we

used the relative velocity equation. Okay, that's solving a, a relative

velocity velocities on a planar rigid two dimensional body

and I want to give you a chance to do it again using the IC method.

So here's another worksheet for you, very

similar but it's great practice for you and

once you finish this, I have the solution in the, in the module handouts, and

I'll see you next time.

[BLANK_AUDIO]

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