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Dominoes: Translation and Rotation

  1. Apr 24, 2005 #1
    Hi,

    I have a few questions about rigid body dynamics. I'm planning on making a 3D domino simulation similar to those domino rally games we all played as kids. I understand that the domino effect can get pretty complicated because it creates a chain of forces, but right now I'm having trouble simply understanding how a single domino acts.

    Without much luck, I've searched all over google and this message board for several different phrases about rigid body dynamics, eccentric forces, advanced dynamics, linear and rotational motion, and so on. I found an answer to one of my questions in this thread, but I would like to ask it anyway so that I can clearly understand why that answer is right and how it applies to my problem.

    As far as my physics education goes, I'm currently wrapping up a freshman physics course specifically about classical mechanics and specifically for scientists and engineers, so I'm not advanced but I should be able to understand direct explanations and equations. I've taken math through Cal II, Discrete Math, and Linear Algebra.

    Here is my problem:

    [​IMG]

    I suppose you all would have preferred that I call the force, F, an impulse. It's an instantaneous force. Imagine thumping the domino with your middle finger.

    Is my free-body diagram correct? I figured the static frictional force works in the same horizontal direction as F when F is above B, otherwise the domino would spin in place about the CoM.

    Problem 1: [​IMG]

    Is [tex]R_2[/tex] equal to [tex]R_1[/tex]?

    Why?

    My intution tells me that [tex]R_2[/tex] should be larger because more of the force goes into creating torque when F is applied at A than it does when F is applied near B, so less goes into creating linear acceleration. I'm guessing my big error here is thinking about the force as going into anything. Why does the force create the same linear acceleration no matter how far it is applied from the CoM? Will you explain it to me in more detail than a single sentence that says something like "conservation of momentum"? I have a feeling I'm missing some simple, fundamental idea about how forces relate to torque and linear acceleration.

    If [tex]R_2 = R_1[/tex], then consider the following image:

    [​IMG]

    Assume there is no frictional or gravitational force in this example. Is the horizontal acceleration of the CoM equal to 0 regardless of where I put [tex]F_2[/tex] between B and C?

    Problem 2: [​IMG]

    Here is my biggest problem, the one I cannot find an answer to anywhere.

    By small F I mean small relative to the mass. Imagine tapping a domino softly so that it falls over without point D sliding. Or imagine barely tapping a domino so that it starts to lean, but returns to its original vertical position. That is a small force here.

    By large F I mean large relative to the mass. Imagine thumping a domino so hard that the number of revolutions it completes appears to be a larger number than the horizontal displacement in centimeters. That's not a very relevant ratio, I know, but that is what I mean by a large force here.

    Did I state that incorrectly? Should I talk about F being small and large relative to [tex]f_s[/tex]?

    What I want to know is: does the axis of rotation change depending on the magnitude of F?

    My intuition tells me that the domino rotates about D when the force is small, changing the moment arm to be the full height of the domino instead of half of the height and thus changing torque differently.

    My intuition also tells me that the domino rotates about the CoM when the force is large.

    How does the frictional force affect the axis of rotation?

    Once again, I have a feeling that I'm missing some simple, fundamental idea.

    Is it always rotating about the CoM and I'm confusing the CoM's linear acceleration with angular acceleration around another point? Once the angle from the vertical is larger than 0, does the gravitational force along with the rigid body's points of contact cause the CoM to accelerate towards the ground, giving me a false view of rotation? Does that mean that simulating it on a computer is more of an issue of collision detection than of another torque equation?

    Since I gave these problems in two dimensions, I'm okay with an answer involving only two dimensions. I would like to fully understand how this all works in two dimensions before moving on to three dimensions anyway.

    I know there are physics SDKs out there for programmers, but I'm more interested in learning than simply getting it done. This fall I will be turning this into a directed study project to go towards my undergraduate computer science degree, and I'm doing that so that I can learn as much physics, graphics programming, audio programming, and UI programming as I can before I finish my degree next year.

    I will sincerely appreciate your effort if you give me reasonably detailed answers to my questions. Let me know if I forgot to mention any important information.
     
  2. jcsd
  3. Apr 24, 2005 #2

    Andrew Mason

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    Just a suggestion: you should try to make your questions more succinct.

    The second impulse (which creates no rotation) provides horizontal motion to the centre of mass:

    [tex]F\Delta t = mv_{0x}[/tex]

    The distance that the domino moves horizontally before stopping is:

    [tex]s = v_{0x}\Delta t[/tex]

    and [tex]\Delta t = \sqrt{2gh}[/tex] where h is the height that the centre of mass falls.

    [tex]s = v_{0x}\sqrt{2gh}[/tex]

    So s is proportional to initial horizontal speed of the centre of mass. Since in the first instance, the impulse produces both rotational and translational motion, the horizontal speed of the centre of mass will be less than in the second. On the other hand, in the second instance it falls farther - by (L1-L2)/2 - so the time of fall is longer. So I don't think you can tell just by looking at it. You have to work it out.

    Can you work out the horizontal speed of the centre of mass in the first instance?

    The domino starts rotating about the bottom right corner. The axis of rotation changes when the domino is no longer pivoting about the contact point with the floor. Up to this point, there is a gravitational torque acting on it (non-constant due to changing angle of [itex]\vec g[/itex] to the radius of turning) It will then rotate about its centre of mass with constant angular speed.

    The magnitude of F will simply make the change in axis of rotation occur sooner unless it is sufficient to overcome the static friction force. If F is large enough to make the bottom slide, the pivot point will move along the bottom of the domino (to the left).

    AM
     
  4. Apr 24, 2005 #3

    Hurkyl

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    Some complications:

    The frictional force doesn't always point in the direction of applied force. For example, I can push on a domino so lightly it won't move... in that case, the frictional force points opposite the direction of applied force. (And the normal force would be applied at the corner, rather than at the center of length of the base)

    The corner can also slip. It's probably too hard to see with a domino, but I did an experiment with a thin book, and when I gently nudged it so it would tip over, I could hear it sliding across my desk shortly before it finished tipping.
     
  5. Apr 24, 2005 #4
    Oh, trust me, I normally would; I am a technical writing minor and a ruthless editor. In this case I wanted to:

    • introduce myself
    • convince anyone who is willing to help me that I am partially literate and worth the effort spent typing out descriptive answers
    • clearly identify what I don't understand

    That's why I asked mostly yes or no questions. I was hoping for a direct yes or no followed by descriptions. :cool:


    Does that mean [tex]a = F/m[/tex] is usually false for rigid bodies?

    Must I derive a special equation to calculate the horizontal speed of the center of mass for this specific rigid body?

    I think I understand what you're saying about the pivot position of the domino, but I'm still a little confused.

    First, that seems incredibly complicated to implement in a software algorithm.

    Second, if the domino is rotating about the bottom right corner, what makes the domino stop pivoting about that contact point? Is it simultaneously pivoting about the center of mass?

    That question might not make sense since you already understand what's going on, but the way I currently see it, if the domino only pivots about the right contact point at first, then the impulse would create all of the torque about that point, causing the domino to slam into the ground as if it were bolted down with a hinge at that point.

    If it is not rotating about the center of mass from the very beginning, then what causes the pivot point to eventually leave the ground and change the axis of rotation to the center of mass?

    Thanks for your help.

    Anyone else who is willing to pitch in on any of my other questions will make my day.
     
  6. Apr 24, 2005 #5

    Andrew Mason

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    Welcome to PF!

    No. F=dp/dt = ma for the domino. The problem is that this rigid body is not free. It is connected, by friction, to the surface it is resting on. The domino applies force for a certain time to the surface it is pivoting on and this reduces the momentum gain of the domino [itex]m\Delta v = F\Delta t - f_s\Delta t_s[/itex]

    Lets take the case where the domino pivots on the lower right corner (static friction is sufficient to hold the corner). When it receives the impulse at the top, the direction of the centre of mass is radial from the pivot point, so it is partly horizontal and partly vertical.

    I should say that I neglected the fact that, in the second instance, the kinetic friction also reduces the momentum of the domino in a similar way.

    It stops pivoting about that point when there is no longer a horizontal force at that point ie. when the domino has fallen over completely. At this point, there is no external torque. It is then rotating only about its centre of mass. And, yes, it is rotating about its centre of mass as it is pivoting. It is synchronous rotation (synchronous with the rotation about the pivot).

    As it falls, so long as the pivot point holds, gravity is providing torque as well.

    AM
     
  7. Apr 25, 2005 #6
    I dont mean to interfere.. but how can one approximate the time the force is applied in the impulse equation? Also, doesnt that assume that all the of the momentum is transferred into the next domino,which isnt the case, since the domino is still moving after impact. Does it stop momentarily and then gets reaccelerated by gravity?
     
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