That''s certainly true. But the question is how do you apply it to this ploblem?

aha! I think i''ve got it.

Muse:
the force applied by engine #3 acts upon the center of mass for translational acceleration, right?

but, it also creates an angular acceleration, so the linear acceleration produced by the engine is always moving in a circle.

interesting.



A man said to the universe:
"Sir I exist!"
"However," replied the universe,
"The fact has not created in me
A sense of obligation."

Well, physics is not that difficult. You have a limited set of short equations but the trick is to understand and apply them correctly =)


As some guys here said, a rigid body problem like this is more easily handled if you separate the motion of the center of mass and the rotational motion of the body.

Each motion follow a particular law.

1)The motion of the center of mass follows a sort of Newton's law.
The sum of the forces acting on a body is equal to the mass of the body multiplied by the acceleration of the center of mass.

F= m*A

2)For the rotational motion:

The change of rate of the angular momentum is equal to the torque.

d/dt(L)=M

Since the angular momentum is a function of the angular velocity, you can easily get an expression for the rotational motion.

I know, it looks a bit complicated and my lousy english doesnt help for sure =)


I've just solved the problem that way, I'm going to write up the solution and post a graph of the motion. (Btw how do you post a pix on this messageboard?)

lcf




Edited by - SuperBlaireau on 2/20/00 2:11:15 PM

yeah, i understand everything now, thanks people.

angular acceleration is almost exactly like Newtons second law

linear
Force = Mass * Acceleration

angular
Torque = I * Acceleration

where I is a variable called the Rotational Intertia. This describes the mass and wieght distribution of all particles in the object, since rotating a particle further away requires more work.

the I for the tube i mentioned rotated perpendicularly about its center is:
I = 1/12 * Mass * Length^2

so, if I apply 10 newtons of force to a 10 meter long rod, with a mass of 50kg, its acceleration is:

10 = 1/12 * 50 * 100 * A
A = 0.024 r/s^2


thats about 1.3 degree;s per second acceleration.
If the bar were shorter in length, the acceleration would be higher.

Rotational physics are neat


A man said to the universe:
"Sir I exist!"
"However," replied the universe,
"The fact has not created in me
A sense of obligation."

Well, it's not complicated but it's not that simple!

You have to use use the first equations in space fixed reference frame. Since the rod rotate, the force perpendicular to the rod will rotate as well..

Here is a pic to explain this:

black arrow: the force vector
stripped arrow: the y-component of the force in a space fixed reference frame..(xy)



The x and y components of the force will change=>change of the x and y acceleration of the center of mass..

I hope it helps..
If I can get this damn scanner to work you'll have a pix of this fine movement...

In facts here are the differential equations of the motiont:

d/dt(angular velocity)=L*F/(4*I)*t^2
a_x=-F*sin(L*F/(4*I)*t^2)/M
a_y=-F*cos(L*F/(4*I)*t^2)/M

where
a_x, a_y are the components of the acceleration
I the "rotational inertia"
L the lenght of the rod
t, time

I'll post how to get this equations later =)



Edited by - SuperBlaireau on 2/21/00 2:12:52 AM

Edited by - SuperBlaireau on 2/21/00 4:04:44 AM

Well, since it''s impossible to get a clean mathematical expression of the motion of the center of mass (the diff. equations are virtually impossible to solve) I''ve used a numerical integration..

Here is the graph..



To get a good idea of the real motion of the rod, just imagine the center of the rod (=center of mass in this case) following the red path and the rod rotating itself around the center of mass faster and faster...

I''m gonna write a small opengl programm to make the things clear =)

well, the scanner works now..
here is my solution:



what do you think of it?

lcf

very interesting.

probably best solved in a real time demo though, because i am horrible at advanced math.

===============================================
I saw a man pursuing the horizon;
Round and round they sped. I was disturbed at this; I accosted the man.
"It is futile," I said, You can never -- "

"You lie," he cried, And ran on.

So you need a demo?

Np, here is one, just coded one hour ago!

(thx Nehe for the source code frame work)

rod demo

Instruction:
1.you need an openGl enabled graphic card
2.use the arrow keys to move the view and "page up" and "page down" keys to zoom in/zoom out!


Note:

1)The pix of the trajectory above is in facts the velocity, not the position! What you see is that the velocity tend to have a fixed value when the time "grows"
So the rod will move in a straight line very quickly cause there are no more acceleration!!!
("the rod rotates so fast that the force cancels out!")


2)The diff. equations are highly unstable when t is huge (try to plot cos(t^2) to know what I mean)!! That''s why the numerical solution (the demo) is not very accurate if you let the programm run a long time. For instance, in the demo, the rod will change direction suddenly! It''s not true, the rod must follow a straight line when t is important)