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Minimizing Moment of Inertia, keeping Moment constant

  1. Jul 18, 2013 #1
    Minimizing the Moment of Inertia while keeping the Moment constant

    Hi there. I am dealing with a mathematical problem which seems to be much harder than I initially expected:

    Minimize the functional

    [itex]J(\Omega) = \frac{1}{\rho} I_{z} = \int \!\! \int \!\! \int_\Omega \left( x^{2} + y^{2} \right) dx dy dz[/itex]

    subject to

    [itex] W(\Omega) = \frac{1}{\rho} x_c m(\Omega) = x_c \int \!\! \int \!\! \int_\Omega dx dy dz = \int \!\! \int \!\! \int_\Omega x dx dy dz = C = constant[/itex]

    i.e. the unknown to be optimized for is the domain of integration [itex]\Omega[/itex]. How to solve this problem as generally as possible? Shall one assume that: a) [itex]\Omega[/itex] is continuous? b) [itex]\Omega[/itex] is differentiable, and (if yes) in which sense?

    Those who are familiar with mechanics immediately notice that the problem in fact is: Assuming constant density [itex]\rho[/itex] throughout the body, minimize the Moment of Inertia [itex]I_{z}[/itex] around the z-axis while keeping the Moment [itex]x_c m[/itex] around the same axis constant. Anyway, the problem as it stands is of purely mathematical nature so I think it belongs to this section.

    This is what I tried so far,

    1. Introducing a Lagrange multiplier [itex]\lambda[/itex]

      [itex]\bar{J}(\Omega) = J(\Omega) + 2 \lambda W(\Omega) = \int \!\! \int \!\! \int_\Omega \left( x^{2} + y^{2} + 2 \lambda x \right) dx dy dz = \int \!\! \int \!\! \int_\Omega \left( (x + \lambda)^{2} + y^{2} - \lambda^{2} \right) dx dy dz[/itex]

    2. Deciding what [itex] \Omega[/itex] should look like

      [itex]\bar{J}(\Omega) = \int_{z_l}^{z_u} \!\! \int \!\! \int_{\Omega_z} \left( (x + \lambda)^{2} + y^{2} - \lambda^{2} \right) dx dy dz = \int_{z_l}^{z_u} \!\! \int_{y_l(z)}^{y_u(z)} \!\! \int_{x_l(y,z)}^{x_u(y,z)} \left( (x + \lambda)^{2} + y^{2} - \lambda^{2} \right) dx dy dz[/itex]

    3. Deriving equations for the extremals of [itex]\bar{J}(\Omega)[/itex]

      [itex] \bar{J}_{\! x_l} = - \int_{z_l}^{z_u} \!\! \int_{y_l(z)}^{y_u(z)} \left( (x_l(y,z) + \lambda)^{2} + y^{2} - \lambda^{2} \right) dy dz = 0[/itex]
      [itex] \bar{J}_{\! x_u} = \int_{z_l}^{z_u} \!\! \int_{y_l(z)}^{y_u(z)} \left( (x_u(y,z) + \lambda)^{2} + y^{2} - \lambda^{2} \right) dy dz = 0[/itex]
      [itex] \bar{J}_{\! y_l} = - \int_{z_l}^{z_u} \!\! \int_{x_l(y_l(z),z)}^{x_u(y_l(z),z)} \left( (x + \lambda)^{2} + y_l(z)^{2} - \lambda^{2} \right) dx dz = 0[/itex]
      [itex] \bar{J}_{\! y_u} = \int_{z_l}^{z_u} \!\! \int_{x_l(y_u(z),z)}^{x_u(y_u(z),z)} \left( (x + \lambda)^{2} + y_u(z)^{2} - \lambda^{2} \right) dx dz = 0 [/itex]
      [itex] \bar{J}_{\! z_l} = - \int_{y_l(z_l)}^{y_u(z_l)} \!\! \int_{x_l(y,z_l)}^{x_u(y,z_l)} \left( (x + \lambda)^{2} + y^{2} - \lambda^{2} \right) dx dy = 0[/itex]
      [itex] \bar{J}_{\! z_u} = \int_{y_l(z_u)}^{y_u(z_u)} \!\! \int_{x_l(y,z_u)}^{x_u(y,z_u)} \left( (x + \lambda)^{2} + y^{2} - \lambda^{2} \right) dx dy = 0[/itex]

      where [itex] \bar{J}_{\! \cdot} = \frac{\partial \bar{J}}{\partial \ \cdot} [/itex].

    After this point I am kind of stuck. I am not even sure that the expressions in 3. are correct, but I believe so. I have not been able to fully evaluate any of the integrals in 3. or even analyze them in any other meaningful way. Obviously, by looking at the integrand of the triple-integral, a coordinate transformation along the x-axis is possible, followed by a transformation to polar coordinates. The integrand then becomes [itex] ( r^{2} - \lambda^{2}) r[/itex] which doesn't seem to be any simpler. After the transformation the domain of integration is still unknown so nothing has been gained.


    PS. I believe that problems like this one must have been solved ages ago, but I couldn't find anything. If anybody knows the solution to the problem and the proof thereof, please let me know. A conjecture from my side is that if the object is finite in the z-direction, i.e. [itex]z_{u}-z_{l}[/itex] is finite, then the optimal shape is a half-disk with thickness [itex]z_{u}-z_{l}[/itex] with obvious orientation. If the thickness of the disk is allowed to increase, it extends along the z-axis and becomes more and more "slender". When [itex]z_{u}-z_{l} \rightarrow \infty[/itex] the radius of this half-disk goes to zero. This conjecture that goes along well with intuition is what I am trying to prove, however.

    EDIT: I think I had some of the equations for the extremals in 3. wrong. They are now corrected. I would be happy if someone would check them for me.

    EDIT 2: I think I had the equations correct the first time. They are now again corrected! Please still check them for me.
    Last edited: Jul 18, 2013
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  3. Jul 18, 2013 #2


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    A cuboid [0,x0] x [-e,e] x [0,f] has ##J \propto x_0^3 e^3 f## with ##W \propto x_0^2 e f##. If you increase f and decrease x0 and e, J gets arbitrarily small while W can stay constant. Therefore, there is no minimum, and the lower bound is zero.
  4. Jul 18, 2013 #3
    Thank you! You provided an example for the case when there are no bounds on the domain of integration in any direction. Then it turns out that the lower bound on [itex]I_z[/itex] is zero. I actually have found such examples myself, namely a cylindrical wedge, i.e. "a piece from a round cake". BUT, what happens if [itex]z_u - z_l[/itex] is fixed and finite? Can somebody help me with that one? Naturally, that case is more interesting from a practical standpoint.

    When minimizing the Moment of Inertia for a cylindrical wedge where the thickness is fixed, it turns out that the optimal piece is a half-disk with thickness [itex]z_u - z_l[/itex]. Intuition tells me that it should also be the optimal solution among ALL possible shapes, but intuition has been wrong before, :)
    Last edited: Jul 18, 2013
  5. Jul 18, 2013 #4
    Where did you come across this problem, if you don't mind me asking?
  6. Jul 18, 2013 #5
    I am glad you asked! Well, it's all about balancing crankshafts for combustion engines :) It's amazing how often I run into mathematical problems when trying to do things in practice. I am probably thinking a bit too much at times.
    Last edited: Jul 18, 2013
  7. Jul 18, 2013 #6


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    Most folks usually drill out the bottom of one or more of the counterweights. Seems to work without exploding your head.
  8. Jul 19, 2013 #7
    Yes, but I love math too so I cannot resist doing this, :)

    Further progress! I have been calculating like crazy yesterday and this is what I have been able to find out. It seems like the optimal solution is a solid cylinder of length [itex]L = z_u-z_l [/itex] with its center-line parallel to the z-axis. The center of the cylinder is at [itex](x_c,y_c,z_c)=(-\lambda,0,z_c)[/itex] and radius [itex]|\lambda|[/itex], where [itex]z_c[/itex] is arbitrary and [itex]\lambda[/itex] depends on the Moment [itex] x_c m = \rho C[/itex] and the length [itex]L[/itex]. I am not completely done with the math though so don't quote me just yet!

    This is actually a result I almost "guesstimated" a few weeks ago but discarded it for some reasons I cannot remember. However, it makes somewhat sense since a cylinder is the solid that "packs the mass" most tightly around its z-axis and hence rotates about it the easiest, i.e. it has the lowest Moment of Inertia. Moving the center of rotation to some other axis parallel to its z-axis (i.e. our z-axis) just adds another component to that Inertia. But as already stated, I am not done yet so my conclusions might be wrong.
    Last edited: Jul 19, 2013
  9. Jul 19, 2013 #8


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    If we have bounds in z:
    1) the optimal shape will be symmetric along the whole range of the z-axis, so your problem is just 2-dimensional.
    2) the shape has to be convex (otherwise moving masses towards the missing piece would reduce J while keeping W).
    3) the whole border of the shape has to have the same J/W ratio

    (3) will give lines of equal J/W ratio. Look for a line of sufficient area inside to get the required W (all lines outside of this have a worse ratio, all inside have a better ratio). (1) and (2) guarantee that this shape will be optimal.
  10. Jul 19, 2013 #9
    Ok, I think I got it! Here is the result:

    Given a fixed maximal length [itex]L = z_u-z_l[/itex] of the sought solid the solution that minimizes [itex]J(\Omega)[/itex] subject to the constraint [itex]W(\Omega)=C[/itex] is a solid cylinder with its center-line parallel to the z-axis, center at [itex](x_c, 0, z_c)[/itex], length [itex]L[/itex], and radius [itex]|x_c|[/itex]. The parameter [itex]x_c = \sqrt[3]{\frac{C}{\pi L}}[/itex] and [itex]z_c = \frac{1}{2}(z_l+z_u)[/itex]. Furthermore, [itex]\min\left\{J(\Omega)\right\} = \frac{3}{2} \pi L x_c^{4} = \frac{3}{2} \sqrt[3]{\frac{C^{4}}{\pi L}}[/itex].

    It has been some good days! Don't spoil it by telling me the solution is all wrong :P

    EDIT: The equations for the extremals in the OP are completely wrong! Unfortunately I cannot change them anymore.
    Last edited: Jul 19, 2013
  11. Jul 19, 2013 #10
    Thank you for your help in particular. It seems like you have an amazing insight into physics and mathematics. Me myself, I was just using brute force here, :)
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