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Homework Help: Energy of Falling Dominoes - Rotational and Gravitational

  1. Jun 17, 2011 #1
    First of all, hi! I'm new here.

    1. The problem statement, all variables and given/known data
    Rather than a specific problem, my friend and I are doing an Extended Experimental Investigation on the energy transfer of dominoes for our grade 12 assignment.

    2. The attempt at a solution

    We have approached it using loss of gravitational potential energy (Ug) and then comparing that with rotational kinetic energy determined by the theta: Theta is the angle from the vertical to the point as where it has fallen.

    Gravitational energy becomes rotational energy.
    The domino's leading edge that it rotates around does not slide
    The domino has a toppling point, and its velocity at this toppling point is 0m/s

    We then aim to compare these two values with experimental data gathered with a camera.

    Here's what we have done (equation number):

    To find the gravitational potential energy, we calculated the height of the center of gravity at the toppling point* (1), found Ug at this point, then calculated Ug at the end point in the same method (2).
    The difference was found, and hence the loss of gravitational energy, which should have been converted into rotational kinetic energy.

    *The toppling point is when the point of the center of gravity of the domino is outside the vertical of the axis of rotation. I.e, draw a diagonal from the edge the domino is rotating on to the opposite corner (this line will pass through the center of gravity) and when that diagonal line passes vertical, the domino starts to fall due to gravity. If the diagonal is not past the vertical, it will fall back to an upright position.

    To find the rotational kinetic energy theoretically we did the following:

    The Domino's velocity is 0 at the toppling point, at which the angle is dependent on the height (3).
    The acceleration of the domino is not constant, however is can be modeled. As the angle of fall increases (theta), the component of gravity actually affecting the domino increases, hence the acceleration increases(4). Thus by integrating the acceleration, one can find the velocity (i say velocity here, but it is more the speed of the domino) of the domino due to the falling angle (5).

    Once this velocity is acquired, it was used to find the rotational velocity (omega) (6). Using the mass moment of inertia (7) and the rotational velocity, the rotational energy was found (8).

    3. Relevant equations

    The equations jpeg.


    Basically, the calculated change in Ug and Ekr does not match and i want to know why!!; they are out by about a factor of 7. I have attached my spreadsheets if you want to look at exact calculations, however, because the formulas derived are dependant on height as well as theta, in the rotational energy spreadsheet columns E, F, G, H, and K are all for the height of 0.05m. The rotational energy calculated is at a theta value of 1.42 i think - Pretty much the last cell in the bottom right.

    Attached Files:

    Last edited: Jun 18, 2011
  2. jcsd
  3. Jun 18, 2011 #2
    I'm a little confused on why you chose to integrate for the rotational energy. The conservation of energy should easily give the rotational energy at each point.
  4. Jun 19, 2011 #3


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    Welcome to PF SaminS!

    Sounds reasonable

    It's not clear to me why the velocity should be 0 at the tipping point in a typical scenario. I mean, if another adjacent domino falls into the one being considered and pushes on it, then one would think the velocity would be non-zero from the moment of contact onwards. However, if the problem you're considering is one in which a lone domino is initially at rest and is perched in unstable equilibrium on its pivot point, and then begins to topple, then that's fine. I just don't know if that scenario matches the experiment you did.

    I'm not sure about either of these. At the tipping point, the diagonal of the rectangular face is vertical, and so the distance of the centre of mass above the pivot point is just d/2, where the diagonal, d, is given by:

    [tex] d = \sqrt{h^2 + w^2} [/tex]

    What you have in your equations is a little different. Similarly, once the domino falls down and is horizontal, then the height of the centre of mass above the surface on which the domino is resting is just equal to one half of the domino thickness: w/2.

    They say a picture is worth a thousand words. If what you're saying corresponds to the first image attached below, then I agree with you, and I think we're on the same page. It's once that diagonal goes past vertical that the direction of the torque around the pivot point switches in such a way as to try to make the domino fall over.

    I didn't look too closely at this part here, but basically my approach to the problem would be using the concept of torque. It is conceptually clearer, and you can use rotational dynamics for everything. The torque, [itex] \tau = I\alpha [/itex] where [itex] \alpha = d\omega /dt [/itex] is the angular acceleration. But you also know that the magnitude of the torque is equal to the force (in this case the weight mg) multiplied by the perpendicular distance to the axis of rotation. This perpendicular distance is a function of theta, so one can write it as [itex] d_\perp (\theta) [/itex]. The second diagram I've attached (and in particular the red triangle in it) should help you with the geometry of the situation so that you can find [itex] d_\perp (\theta) [/itex]. (In the second diagram, I relabelled what I was calling [itex] \theta_{\textrm{tip}}[/itex] as [itex]\phi [/itex] in order to match your notation). Once you have that, you can equate expressions for torque

    [tex] I\frac{d\omega}{dt} = mgd_\perp (\theta) [/tex]

    Now you have to do some manipulating using the rules of differential calculus in order to get it so that you can solve for the final [itex] \omega [/itex] by integrating both sides w.r.t. [itex] \theta [/itex] over the range of angles spanning the motion.

    I managed to get the same answer using both methods with the procedure I outlined above. (I didn't bother to go quite so far as to get a numerical answer, but my expressions for omega were the same in both cases).

    Yeah, I see this as being done just as an exercise, sort of the rotational equivalent of deriving the work-energy theorem from [itex] W = \int F \,dx [/itex]

    Attached Files:

  5. Jun 22, 2011 #4
    Yea, I assume that its zero in the case of one domino.

    Sorry about this. It's a bit of a misunderstanding. You're perfectly correct about the gravitational energy of the tipping point being greater than when it is vertical. I just had a typo in my formula (my factor of 2 inside the square root)

    About the end point: I forgot to say how we assume that the domino finishes supported by the next domino, which is supported by the next, and the next, and the next etc etc. Hence, it still has a slight bit more gravitational energy than its height of the center of mass when it is completely horizontal, as it is propped up as if after falling in a real domino chain. The formula on my sheet is actually a calculation of [itex] d_\perp (\theta) [/itex], though in a slightly different way than your diagram recommends. Rather than combining the two angles like your diagram suggests, we used two different triangles - your way is heaps simpler, but nevertheless my way still works :).

    Exactly what I was talking about. :)

    Thanks very much, this makes sense. I'll give it a go and see how it works.

    Yea, exactly.

    Again, thanks very much.
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