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Finding Optimal Axis of Rotation with Constraints

  1. Nov 7, 2015 #1
    I really don't have much experience with calculus ( :sarcasm: Hooray! ), but earlier I was reading an introductory article explaining the usefulness of the Lagrange multiplier in dimensional optimization,


    and it reminded me that another curiosity of mine is trying to find the optimal point along which one could pivot a drumstick, whose shape I will define as:

    a relatively thin cylinder, with a half-sphere on the 'butt end', and on the other a paraboloid, whose tip is partially overlapped by a smaller shape- be it an ellipsoid, a barrel-esque shape (haha, burlesque), or a sphere. For the sake of argument it is made of an isotropic material.

    ImageUploadedByPhysics Forums1446959249.812896.jpg

    (This picture obviously does not use the correct proportions of scale, but I hope it gets the general point across)

    I really don't have much experience with adding and subtracting volumes for computing that of a... object of 'composite shape'(?) (my words), but while the cylinder and half-sphere are quite straightforward, I'm pretty sure that in order to find the total volume for such an object I'm going to need to 1) subtract a spherical/ellipsoidal cap from the paraboloid, and 2) subtract the tip of the paraboloid from the sphere/ellipsoid.

    I went ahead and wrote the volumes of each shape below:

    Vcylinder = πrc2hc

    Vsphere = \frac{4}{3}πrs3

    Vparaboloid = \frac{π}{2}hprp2

    Vellipsoid = \frac{4}{3}πrMrm2

    Vs,cap = \frac{πh2}{3}(3rs-h)

    Ve,cap = πrm2(\frac{2}{3}rM - h + \frac{h3}{3rM2})

    (It occurred to me to do it this way by intuition, visualizing how the boundaries of the paraboloid and sphere/ellipsoid intersect each other. I also realize that the volume of a shape could also be calculated in two dimensions, of which the Cartesian integral is taken from 0 to Lnet , followed by the polar integral from 0 to 2π.)

    Note: The phrases "Cartesian integral" and "polar integral" are my own, simply because I either do not know or do not recall the 'official' terminology for these methods.

    Anyway, I know that there need to be some particular constraints: to minimize the moment of inertia, yet maximize the potential angular momentum when moving through an arc of

    s = \frac{πra}{2}


    ra = \frac{Lstick - CoM}{2}+d

    CoM is the center of mass along the stick's length, and d is the distance between the axis of rotation and the stick's center of mass along its length, and the axis of rotation must also lie along the stick's length. (I apologize for the repetition of phrase- I'm not sure how else to word it at the moment)

    Now the parallel axis theorem states:

    I = ICoM + Md2 ,

    so the 'optimal' moment of inertia will lie somewhere between \frac{1}{12}ML2 and \frac{1}{3}ML2, with a corresponding distance d .

    Could one not use the idea of the Lagrange multiplier to find the answer for a question such as this?

    In the past I had considered the trebuchet as being a good study for optimizing the location of the pivot, but even then the best I could find online was an approximation which assumed a beam with unchanging dimensions throughout its length.
    Last edited: Nov 8, 2015
  2. jcsd
  3. Nov 8, 2015 #2


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    It is not that complicated . Look up how to calculate volumes of solids of revolution .
  4. Nov 8, 2015 #3


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  5. Nov 8, 2015 #4


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    nb : Divide solids of revolution with varying profile into short sections for easier calculations .
  6. Nov 8, 2015 #5
    I understand. I could write a piecewise-defined 2-D equation r(x) describing the changing radius from 0 to L, take the indefinite integral, then multiply it by 2π.

    Vstick = 2π * [integral;0,L] r(x) dx

    However, I already have this information, and the integral would just result in the exact same ratios I'm about to add/subtract.

    Here are some clarifying points about how each volume relates to each other:

    rp = rc
    rs,b = rc
    hp = \frac{1}{2}hc

    I have two problems though, the first is something geometric which I don't understand algebraically. The second is a simple decision I'll have to make.

    1) If the tip of the paraboloid stops in the dead center of the sphere/ellipsoid, what is h (for the purpose of calculating the volume of the spherical/ellipsoidal and paraboloidal caps)?

    2) Some times it seems as though the radius of the 'bead' (tip) of the stick is a bit smaller than that of the shaft (the main portion), and some times it's not that much smaller, so I would be tempted to say:

    \frac{1}{2}rc =< rs,t =< \frac{2}{3}rc
    Last edited: Nov 8, 2015
  7. Nov 8, 2015 #6
    But how do you know you need these volumes to start with?
    Your problem does not seem very specifically stated. What do you mean by "optimal" point to pivot a drumstick? What quantity will be maximized (or minimized) to make it optimal?

    Are you thinking about the center of percussion, maybe?
  8. Nov 8, 2015 #7
    In order to calculate the center of mass of this non-uniform cylinder it becomes necessary to find its volume, followed by its mass (assuming it's isotropic)

    I honestly had never heard of that term until now, but after reading the cursory explanation on Wikipedia that may be at least close...
  9. Nov 8, 2015 #8
    When playing a percussion instrument with drumsticks there are two main gripping types: "Matched" (where the grip of one hand is the mirror image of the other), and "Traditional" (where the grip, and consequently the motion, of one hand, usually the 'weaker' hand, is transformed from flexion/extension to pronation/supination by flipping the hand over (supination) and turning the stick to where the tip is now pointing across to the other side of the body).

    There is a third grip type (a subtype of matched grip) is called "French", where the movement is adduction/abduction. Some percussionists may even, depending on dynamic needs of the piece, use combinations of grips, changing them from one moment to the next as the need arises, such as from Matched grip to Traditional grip with one hand and French grip with the other.

    Note: There are also subtypes of Traditional grip, but enough of that...

    Last edited: Nov 8, 2015
  10. Nov 8, 2015 #9
  11. Nov 8, 2015 #10
    I'm hoping to derive a formula which provides the distance from the stick's center of gravity to the pivot point along its length that displays a balance between minimizing the torque needed to achieve a particular angular momentum when traveling through an angle of π/2, and yet also minimizes the moment of inertia of the stick.
  12. Nov 16, 2015 #11
    What I think I'm trying to find I believe has to do with the 'center of percussion', but it's almost like it's other way around...

    It's not the point where a force doesn't result in a rotation, but more like a point where a torque doesn't result in a translation (which I'm guessing is the center of mass...).
    It's a point between the center of mass and the shortest end of a non-uniform rod that results in the least amount of 'shock' (perhaps vibration is a better word to use?), whenever the longest end strikes an object (such as a circular membrane).
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