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Physics Simulation

  1. Jul 27, 2009 #1
    I'm currently writing my own physics simulation engine for some practice with physics since I plan on taking it as my major along with computer science and would like to be ahead of the curve. Plus I'm just plain interested in it. I've read through some papers such as http://chrishecker.com/images/c/c2/Gdmphys2.pdf" [Broken] paper from an online SIGGRAPH course.

    I've come across a major problem in the Hecker article however.... unless I read it wrong, he doesn't take it account where the force is applied when determining the linear force on an object. My intuition says that if I hit an object square in the middle, all of the force will be directed into the linear acceleration of the object. If I hit it on the corner however, more of the force will be put into the angular momentum and less in the linear direction. Is this the right way to think of it?

    This is an excerpt from the article on 2D rotation:
    1. Calculate the CM and the moment of inertia at the CM.
    2. Set the body’s initial position, orientation, and linear and angular velocities.
    3. Figure out all of the forces on the body, including their points of application.
    4. Sum all the forces and divide by the total mass to find the CM’s linear acceleration
    5. For each force, form the perpdot product from the CM to the point of force application and add the value into the total torque at the CM (Eq. 11).

    Step 4 seems completely wrong to me. You should see where the linear force acts on the object relative to the center of mass and then go from there.

    So my question is how exactly are linear and angular force related. Do they both add up to equal the total force imparted on the object?

    P.S. Easy on the notation, I'm not too knowledgeable on all of the symbols :uhh:
     
    Last edited by a moderator: May 4, 2017
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  3. Jul 27, 2009 #2

    diazona

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    It actually doesn't matter where on the object the force is applied. A force that's applied off-center doesn't somehow get "split" between linear and angular motion; it contributes fully to both linear and angular acceleration. The linear part of this is just Newton's law
    [tex]\vec{F}=m\vec{a}[/tex]
    Note that that equation doesn't include anything about where the force is applied; it only includes the total force.

    Note that there is actually no such thing as "linear force" or "angular force" - there's just force. What you're probably thinking of when you say "angular force" is torque, which is force multiplied by the perpendicular distance between a chosen pivot point (often the center of mass) and the line along which the force is applied. There is something analogous to Newton's law for torque and angular acceleration:
    [tex]\vec{\tau} = \mathbf{I}\vec{\alpha}[/tex]
    Here I is the moment of inertia, which you can sort of think of as "angular mass". (It's the thing that plays the same role in the angular motion equation that mass does in the linear motion equation)
     
  4. Jul 27, 2009 #3
    So lets say I push a 40kg cube across a frictionless surface with a 10 N force.

    F = m * a
    10N = 40kg * a
    a = .25 m/s[tex]^{2}[/tex]

    Now if I push at the very corner of the cube, it will still have that same acceleration? But at the same time, it will also have a torque applied to it so it will be rotating.

    I'm just thinking that in the second scenario there is more energy in the system because there is rotational energy too. It just doesn't seem intuitive to me.
     
  5. Jul 27, 2009 #4

    A.T.

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    Step 4 was about computing the instantaneous linear acceleration, not energy.
     
  6. Jul 27, 2009 #5
    Well, I apologize for my lack of enlightenment but I'm still not convinced from any of the explanations. I think it has to do with MY explanation... I'll break it down to its very basics. If you guys could solve these two simple scenarios with explanations as to why they are the way they are, I think that would do wonders for my understanding.

    I have a rectangular box ( 1m wide and 10m long ) which weighs 10kg that I'm pushing along a frictionless surface.

    Code (Text):

     ----------------------------
    |              *             |     * = center of mass ( centered in the box )
     ----------------------------
                        ^       ^
                        |       |
                        A       B      Force A = Force B = 10 N
    Scenario A - Force A is pushing on the box at the center of mass. No rotation should be imparted.

    Scenario B - Force B is pushing on the edge of the box. Rotation should result.

    From what I understand, you guys are telling me that the acceleration forward will be the same in both cases. It doesn't matter where the force is applied. This doesn't make sense to me though...
     
  7. Jul 27, 2009 #6

    diazona

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    Well, for one thing, if the box is sitting on a surface, it's probably not going to rotate. It might fall over, but that's about all. There are forces applied by surface that you'd have to take into account.

    But let's say the box is sitting in empty space, not touching anything. In that case, the acceleration forward will be the same in both cases. It really doesn't matter where the force is applied. Sorry if it doesn't make sense, but that's the way it is... I'm not sure what else to tell you about it.
     
  8. Jul 27, 2009 #7
    Could you or someone provide some example to show that the states of the two objects somehow contain the same energy/force/whatever is equal between them?

    Or somehow rework my understanding of the conservation of energy? Maybe energy isn't even related in this problem.

    If the box A is translating with an energy of 100 joules and rotating with an energy of 20 joules, and box B is just translating with an energy of 100 joules, you're telling me that both of those scenarios can be caused just by a force being applied right at T(0), just in different places? I don't know if this is a point of misunderstanding either but the force I apply isn't a continuous force. Imagine I'm punching the box, so the force is applied at an instant, not pushing it. I'm mainly modeling collisions right now, so it all happens in zero time pretty much.

    Sorry if I'm being a hound but I really feel like I should have a firm grasp on this before I continue further in my program. I really appreciate the help so far.
     
  9. Jul 27, 2009 #8
    1) Look up "center of percussion" in your physics books, or in the school library. The basis of center of percussion is where the force or impulse is applied to the object relative to the center of mass.
    2) Lay a hammer on the floor. Kick the handle end so it spins. How much did the hammer's center of mass move. Not much. Now kick the hammer near the head. Did it spin as much? No. Did the center of mass move? Yes. Can you explain the difference?
     
  10. Jul 27, 2009 #9
    I looked it up and it just seemed to further the idea that the further you hit an object from its center of mass, the slower it will translate because more energy is put into the rotation of the object... Which is the complete opposite of what diazona said... I am thoroughly confused now haha oh dear. I'm not sure if I'll ever fully understand what is going on here. :frown:
     
  11. Jul 28, 2009 #10

    A.T.

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    Again: the instantaneous linear acceleration has little to do with the total energy or linear momentum you will put into the object. If you kick something at the edge it will quickly rotate away from your foot, and you wont be able to transfer linear momentum to it. If you kick it in the middle it will stay in front of your foot and you can apply the force for a longer time period to it, thus transferring more linear momentum.

    But the instantaneous linear acceleration depends only on the forces acting in a single time point, not a time period. If you apply a very big force for a very short time, then this:

    ...|----->
    ...|
    ...O
    ...|
    ...|

    is the same as this

    ...|-->-->
    ...|
    ...O
    ...|
    ...|

    which is the same as this:

    ...|-->-->
    ...|
    ...O
    ...|
    <--|-->

    which is the same as this:

    ...|-->
    ...|
    ...O-->-->
    ...|
    <--|

    which is the same as this:

    ...|-->
    ...|
    ...O----->
    ...|
    <--|

    where you can see that the force applied to the center is the same as the one applied to the top in the initial case. You also see that there is half of the initial force acting at each end in opposite directions. They do not contribute to linear acceleration but they create the same angular acceleration as did the full initial force at one end only.
     
    Last edited: Jul 28, 2009
  12. Jul 28, 2009 #11

    rcgldr

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    The linear and angular reactions to a force are independent. The linear reaction is an acceleration, = force / mass, no matter where the force is applied. However, the rate of acceleration at the point of application of that force will be greater if the force also results in angular acceleration.

    For example, say you have a 1kg block resting on a high friction but massless treadmill. The treadmill applies a 1 N (Newton) horizontal force to the block. The block's linear acceleration is 1 m / s2. The blocks angular acceleration is 0. The treadmill accelerates at 1 m / s2.

    Replace the block with a 1 kg uniform solid cylinder that is free to roll with it's axis horiztontal and perpendicular to the direction of treadmill movement. The treadmill applies a 1 N horizontal force to the bottom surface of the cylinder. The cylinder's linear acceleration is 1 m / s2. The cylinder's angular acceleration = torque / inertia, α = (1 N r) / (1 kg r2 / 2) = (2 N) / (kg r) = 2 m / (s2 r). The related linear surface acceleration = α r = 2 m / s2. Total surface acceleration = 3 m / s2. The treadmill accelerates at 3 m / s2.

    In both cases, the treadmill applies the same force, but in the case of the cylinder it has to accelerate 3 times as fast to accomplish this. From a standing start, the work done by the treadmill (and the energy gained by the accelerated object) in time t (seconds) for the block = 1/2 J (t/s)2, for the cylinder = 3/2 J (t/s)2 (where J = Joule = N m).
     
    Last edited: Jul 28, 2009
  13. Jul 28, 2009 #12
    This makes sense. So I'm probably just getting the illusion that when I push something at the end, less of the force is going into the linear component because it rotates, therefore decreasing the amount of force going into translating the object because the normal is point away from my finger.

    So depending on where I apply the force, it can end up doing less work on the object? I think I'm finally starting to get an understanding of it. Thanks guys.

    One more example and then I'll drop this thread: if I have a rocket engine aimed directly away from the center of mass, it should push the object perfectly forward with no torque. However, if I put the rocket engine off center, it will go into a wild spin and it will barely make any progress. Is this just because the tangential velocity of the rocket changes, constantly canceling out any forces acting upon it? And is this problem even related to what we're talking about?
     
  14. Jul 28, 2009 #13

    A.T.

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    Or you are even applying less force, because it resists less to rotation. It is hard to control the amount of force you apply to an object which is not fixed.


    The engine rotates with the rocket so the direction of the linear acceleration changes too, if you fire the engine continuously. But you can still get somewhere, when you fire the rocket in short bursts, just when the engine points in the desired direction. If you had two opposing off center engines you could fire them alternating after half rotation avoiding spinning up more and more. Since you don't put any energy into rotation, this should be quite efficient in terms of terminal velocity / fuel spent.
     
    Last edited: Jul 28, 2009
  15. Jul 28, 2009 #14
    Alright I think I finally understand what's going on here... thanks for bearing with me guys I really appreciate the help. It's important I get these basics down so I have a good foundation of knowledge to build off of.
     
  16. Jul 28, 2009 #15

    rcgldr

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    It would still spin up the same with respect to rocket duration time (times two if both motors on at the same time), but the linear acceleration component would improve. If the offcenter motors were on the same side of the rocked, pointed in the same direction, you'd just fire both at the same time to avoid any angular acceleration.
     
  17. Jul 28, 2009 #16
    A good example of this problem is to consider pulling two spools of thread along with equal force. On one spool, the thread is not allowed to unwind (thus the spool does not spin). On the other spool, the thread is allowed to unwind.

    If you pull both spools at the same time (i.e., if you race them) you'll see that they both cross the finish line at the same time. However, one spool was given rotational motion along with its linear motion, so it has more energy than the other. The key is that the hand pulling the spool that is rotating also has to move farther than the hand on the other spool--that is, the work done on the rotating spool is greater.
     
  18. Jul 28, 2009 #17

    diazona

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    Yeah, that's worth thinking about. I was also going to mention that when you hit, say, a block off center, making it rotate, your hand (or whatever you're hitting with) moves along with the end of the block as it rotates, so the force acts over a larger distance than it would if you were hitting it on center. Which means more work is done, and that accounts for the extra kinetic energy of rotation.

    dgreenheck, I'm glad you're getting this figured out.

    P.S. Jeff, nice going on the quantitative calculation :smile:
     
  19. Jul 28, 2009 #18
    Ah, perfect examples. I think the problem was I never really relayed what my misunderstanding was. You can consider this case solved :)

    Thanks very much guys I really appreciate it.
     
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