Minimizing Moment of Inertia, keeping Moment constant

[KingBongo]

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

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

subject to

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

i.e. the unknown to be optimized for is the domain of integration \Omega. How to solve this problem as generally as possible? Shall one assume that: a) \Omega is continuous? b) \Omega 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 \rho throughout the body, minimize the Moment of Inertia I_{z} around the z-axis while keeping the Moment x_c m 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 \lambda
   
   \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
2. Deciding what \Omega should look like
   
   \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
3. Deriving equations for the extremals of \bar{J}(\Omega)
   
   \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
   \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
   \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
   \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
   \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
   \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
   
   where \bar{J}_{\! \cdot} = \frac{\partial \bar{J}}{\partial \ \cdot}.

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 ( r^{2} - \lambda^{2}) r which doesn't seem to be any simpler. After the transformation the domain of integration is still unknown so nothing has been gained.

HELP?

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. z_{u}-z_{l} is finite, then the optimal shape is a half-disk with thickness z_{u}-z_{l} 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 z_{u}-z_{l} \rightarrow \infty 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.

 

[mfb]

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.

 

[KingBongo]

mfb:
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 I_z 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 z_u - z_l 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 z_u - z_l. Intuition tells me that it should also be the optimal solution among ALL possible shapes, but intuition has been wrong before, :)

 

[bolbteppa]

Where did you come across this problem, if you don't mind me asking?

 

[KingBongo]

bolbteppa:
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.

 

[SteamKing]

Most folks usually drill out the bottom of one or more of the counterweights. Seems to work without exploding your head.

 

[KingBongo]

SteamKing:
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 L = z_u-z_l with its center-line parallel to the z-axis. The center of the cylinder is at (x_c,y_c,z_c)=(-\lambda,0,z_c) and radius |\lambda|, where z_c is arbitrary and \lambda depends on the Moment x_c m = \rho C and the length L. 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.

 

[mfb]

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.

 

[KingBongo]

Ok, I think I got it! Here is the result:

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

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.

 

[KingBongo]

mfb:
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, :)