linear force vs torque proportions

In working on coding a video game, I am becoming a little confused..

Suppose there are two identical bars at rest in space.  There are two forces of same magnitude acting on these bars.  See bad artwork:

Bar 1                                       Bar 2

---------------                         ---------------
         ^                                                 ^
         ^                                                 ^

In the case of Bar 1,  the bar will not rotate at all and will receive a certain amount of linear force, which is correct.

In the case of Bar 2,  while the same linear force is applied to the center of mass, a rotational force is also being applied.

In cases 1 and 2, an equal amount of energy is entering the system, yet clearly in case 2 more work is being done.  Rotating the bar takes energy, thus less energy must be going into moving the bar linearly?? At least, that's how it seems to make sense to me.  yet, resources I've read (http://www.cs.cmu.edu/~baraff/sigcourse/notesd1.pdf       - pg 29 thru 32) suggest that they each get the same linear velocity.

Has bar 2 traveled less distance because some of the energy from the force went into causing it to rotate. And if so, how do you determine the proportions?

LVL 3
OutsideTheBoxAsked:
Who is Participating?
I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

ozoCommented:
work is force * distance
the force moves further in case 2, because the bar will rotate in addition to moving lineaely
0
ozoCommented:
Impulse = force * time
impulse causes change in linear momentum
in case 2, the same force would either act for less time or more distance, since the end of the bar is moving.
0
OutsideTheBoxAuthor Commented:
@ ozo;
I understand that from an arithmetic perspective. But what I don't understand (from a conservation of energy point of view) is how more work can be done when the same mount of energy has been put into both cases. Why isn't the same amount of work being done when the same amount of energy is put into both systems?  

If someone were to say "bar 1  ctr of mass moves a distance of x in t time, bar 2's ctr of mass moves .8x in t time because some energy went into making it spin such that the point of contact with the force  traveled .2x distance about the center of mass (or whatever amount of distance the inertial tensor allowed to equate to the same amount of energy ." , I'd say" that make sense". But that's not what I'm interpreting.

What is wrong with my understanding?
0
Rowby Goren Makes an Impact on Screen and Online

Learn about longtime user Rowby Goren and his great contributions to the site. We explore his method for posing questions that are likely to yield a solution, and take a look at how his career transformed from a Hollywood writer to a website entrepreneur.

OutsideTheBoxAuthor Commented:
prev post meant for your 1st post

0
ozoCommented:
This was explained on page 32 of your resource
0
ozoCommented:
the same mount of energy has not been put into both cases
It takes more energy to move the same force through a larger distance
0
OutsideTheBoxAuthor Commented:
>the same mount of energy has not been put into both cases
 but my force vectors had the same magnitude & the time was the same.  Maybe that's when I'm confused. How did case 2 get more energy? Where did the "more" come from. In both cases the force vector's magnitude was the same.
 Am I wrong to thing of my force vector magnitude as proportional to the amount of energy introduced into the system?
0
ozoCommented:
the amount of energy introduced into the system is proportional to (the force vector) · (the distance vector that the force moves through)
0

Experts Exchange Solution brought to you by

Your issues matter to us.

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Start your 7-day free trial
OutsideTheBoxAuthor Commented:
ok, I'm going to chew on all this a while & try to grasp it. I'll post back later.
0
NovaDenizenCommented:
I'm confused about the force acting on Bar 2.  At t=0, the force will be upwards on that particular point of the bar.  After t=0, the bar will be moving and rotating.  Where will the force be applied then?  At the same point?  At the same x-coordinate?  Will the force remain constant in direction and magnitude?  
0
sublimationCommented:
Strange question. :-)

If you know the mathematics then what else do you need to know.

Rotating a bar takes enery, so does pushing the bar in a straight line.

Maybe the fact that friction acts in differnt ways for different movements is confusing you... If you roll a bar along a smoth service, it will encounter less friction than spinning the bar on a smooth surface. So although you apply the same amount of energy, one may seem to travel further.
0
mrider01Commented:
Have you considered this?  On Bar A the force is constantly being applied to the center or mass.  Also the surface on which the force is getting applied is perpendicular to the force.  Therefore, all of the force on Bar A is going into moving Bar A.  So Bar A moves d distance, and rotates 0 degrees, over time t.
Now lets look at Bar B.  Once  Bar B starts rotating, the force is no longer perpendicular to the surface.  Then less of the force is being applied to the center of mass.  So less of the force goes to moving the bar forward, and less of the force goes into rotating the bar.  Think about what would happen if the force on Bar B would be applied infinitely.  It would eventually rotate enough that the force would not even be in the bar anymore.  The force would simply push the bar out of its way.  Note my equally lame pictures:

Bar B, t=0                                   Bar B t=infinity
-------------                                           /^                
            ^                                             /  ^              
            ^                                           /    ^          

Think of how the force is being applied with respect to the center of mass.  In Bar A, the force is always directly on the center of mass.  On Bar B, once the rotation begins, the force is acting less and less upon the center of gravity, until the bar turns so much that the force is not effecting the bar's center or gravity at all.

So this is where the disappearing force goes.  Let's say the force is created by a car driving into the bar.  The car pushing Bar A might travel 10 feet over 2 seconds, due to the resistance of the bar.  The same car, pushing with the same force may travel 11 feet over 2 seconds, while only pushing the center of mass 9 feet, because there is less resistance force being supplied by Bar B.  

I hope this helps, and that I'm not as unclear as I fear I might be.
0
It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
Game Programming

From novice to tech pro — start learning today.