4
$\begingroup$

I was taught that perpendicular axis theorem is valid only for laminar objects and not for 3-D objects.I have difficulty in understanding this intuitively.

I mean why would such a condition even exist?

We can consider a 3-D object to be a collection of infinitely divided elements which are stacked one upon on the other to form a 3-D object.Now,we know that for a 2-D object the theorem is valid and we can find moment of inertia of each element about an axis passing through its geometrical centre and perpendicular to its plane and now we can add up moment of inertia of each element using simple addition to get moment of inertia of the 3-D body(because all elements are symmetrical).

This is the same as using perpendicular axis theorem for 3-D object right?

Improve this question
$\endgroup$
7
  • $\begingroup$ lets say the object sits on the x-y plane and you wish to calculate $I_z$ for sake of simplicity. Your infinitely thin slices also have tiny offsets from the x-y plane that need to be taken into account via the parallel axis theorem. This leads to a quadratic term in your infinitesimal which only goes away if the object is 2D $\endgroup$
    – Triatticus
    Commented Jul 9, 2018 at 13:32
  • 1
    $\begingroup$ To make a specific example from @Triatticus's post: consider a sphere or a cube. Obviously $I_x = I_y = I_z$, because of symmetry. So the "2d" perpendicular axis theorem $I_z = I_x+ I_y$ clearly gives the wrong answer! $\endgroup$
    – alephzero
    Commented Jul 9, 2018 at 14:30
  • $\begingroup$ @ Triatticus If i'm finding moment of inertia of each slice about Iz ,then I can simply add up moment of inertia of each slice because my axis is along y direction only for each slice.we would have to use parallel axis theorem only if my axis (axis about which i'm finding moment of inertia) was along x or y direction. $\endgroup$
    – Banchin
    Commented Jul 9, 2018 at 14:30
  • $\begingroup$ @alephzero I get it,totally,but could you explain me where I was wrong or confused in my way of thinking $\endgroup$
    – Banchin
    Commented Jul 9, 2018 at 14:33
  • $\begingroup$ What I mean is that you have to integrate along the z direction to add up your infinite slices, but each slice perpendicular axis theorem brings in a parallel axis theorem for each x and y moment which depend on $(dz)^2$ $\endgroup$
    – Triatticus
    Commented Jul 9, 2018 at 15:11

2 Answers 2

Reset to default
5
$\begingroup$

For a 3D object although the slices are all laminar, and the Parallel Axis Theorem applies independently for each lamina, it does not apply for all of them together, ie for the 3D object. The reason is that although the laminas have the same z axis, they do not have the same x and y axes. So you cannot simply add the moments of inertia (MOI) for the x and y axes for each lamina. The x and y axes for each lamina are parallel but they are off-set from each other and from the x, y axes you are using for the 3D object. As @Triatticus points out, you need to apply the Parallel Axis Theorem to each lamina.

You can add the values $I_z$ for each lamina to get the value of $I_z$ for the 3D body, provided that all the z axes coincide. The lamina do not have to be symmetrical to do this, and the z axis does not even have to pass through the centre of mass of each lamina. However, this is not an application of the Perpendicular Axis Theorem - there are no perpendicular axes here.

Suppose your 3D object consists of two co-axial disks each of mass $m=\frac12 M$ and radius $a$ separated by a distance $2b$.

The MOI for this 3D object about the common z axis is $I_z=2\times \frac12 ma^2=\frac12 Ma^2$. It equals the sum of the $I_z$ for the 2 disks because the z axes coincide. If the 3D object satisfied the Perpendicular Axis Theorem the MOIs about perpendicular axes would be $I_x=I_y=\frac14 Ma^2$. However, where is the origin for these x, y axes located? When you apply the Perpendicular Axis Theorem for each disk you are using a different origin for each, usually the centre of each disk. When you check the Perpendicular Axis Theorem for the 3D object you are using the geometrical centre of the 2 disks as origin, which is the midpoint of the centres of the 2 disks.

The MOIs for each disk about x and y axes through their own centres of mass are $I_x=I_y=\frac14 ma^2$ which satisfies the Perpendicular Axis Theorem. But these axes do not coincide with the x and y axes through the COM for the 3D object. The disk x, y axes are each offset by a distance $b$ from the 3D object x, y axes. Using the Parallel Axis Theorem the MOIs about x and y axes through the COMs of the 3D object are $$I_x=I_y=2(\frac14 ma^2+mb^2)=\frac14 Ma^2+Mb^2 \ne \frac12 I_z$$ Equality only applies when $b=0$ - ie when the disks coincide.

Generally the moments of inertia for a 3D object are defined by distance from the relevant axis : $$I_x=\int (y^2+z^2)dm$$ $$I_y=\int (x^2+z^2)dm$$ $$I_z=\int (x^2+y^2)dm$$ Note that these definitons use the same co-ordinate system throughout, including the same origin. Then $$I_x+I_y=\int (x^2+y^2+2z^2)dm=I_z+2\int z^2 dm$$ The Perpendicular Axis Theorem $I_x+I_y=I_z$ only applies if $z=0$ for all points of the object - ie if the object is confined to the $xy$ plane.

Improve this answer
$\endgroup$
7
  • $\begingroup$ I think the OP is asking about finding the $I_z$ of each section independently first by using the perpendicular axis theorem on each section, then adding them up to get the entire $I_z$. Kind of like taking a bunch of sheets, separately finding $I_z$ for each, stacking them, and then saying the $I_z$ of the composite body is the sum of each separate $I_z$ $\endgroup$
    – BioPhysicist
    Commented Jul 9, 2018 at 17:02
  • $\begingroup$ Yes I agree. That procedure gives the correct answer if the z axes for each sheet coincide. The point is that this procedure does not work for the x and y axes, which do not coincide. So whereas $I_z$ for the composite 3D object is the algebraic sum of individual $I_z$ values for each lamina, the values of $I_x, I_y$ for the composite 3D object is not the algebraic sum of the individual $I_x, I_y$ values for each lamina. $\endgroup$
    – sammy gerbil
    Commented Jul 9, 2018 at 17:17
  • $\begingroup$ I see what you are saying. I guess it comes down to whether the OP is interested in just $I_z$ or in applying it to the other axes. It seems like he does not to me. The method is valid for $I_z$, but I would not call it the 3D version of the perpendicular axis theorem. $\endgroup$
    – BioPhysicist
    Commented Jul 9, 2018 at 17:19
  • $\begingroup$ The Perpendicular Axis Theorem relates 3 mutually perpendicular moments of inertia $I_z=I_x+I_y$. I think the OP is asking why this formula (which applies for laminas) does not apply for a 3D object (which can be divided into laminas). $\endgroup$
    – sammy gerbil
    Commented Jul 9, 2018 at 17:29
  • $\begingroup$ I guess the intent of the OP is not clear then. I completely see where you are coming from though. $\endgroup$
    – BioPhysicist
    Commented Jul 9, 2018 at 17:32
1
$\begingroup$

Your method is correct, but I wouldn't call it the same thing as the perpendicular axis theorem. This is because of the points others have already brought up. For a 2D object in the x-y plane we know that $I_z=I_x+I_y$. Objects in 3D can also be rotated about these same axis, and it is not true in general that $I_z=I_x+I_y$ (as has already been pointed out). I would instead say you are just applying the perpendicular axis theorem to each slice so that you can add up each $I_z$ to find $I_z$ of the composite body. But I guess at this point it all just depends on what you mean when you "define" the "3D perpendicular axis theorem".

Addition:

As pointed out, you also have to keep in mind while the composite $I_z$ is the sum over each $I_z$, the same cannot be said of the moments of inertia about the other axes. $\Sigma I_x$ is not equal to $I_x$ for example.

Improve this answer
$\endgroup$
3
  • $\begingroup$ @sammygerbil I don't think this is the case. He could use the parallel axis theorem if he knew $I_z$ for each sheet but did not align each sheet when stacked. Then $I_z=\Sigma (I_n+md_n^2)$. But if we are stacking each sheet with everything aligned I do not see how he must be using the parallel axis theorem. He is just using the fact that we can add each separate $I_z$ together since translation in the z-direction does not change $I_z$. $\endgroup$
    – BioPhysicist
    Commented Jul 9, 2018 at 18:50
  • $\begingroup$ @sammygerbil the parallel axis theorem is for finding $I$ given a known $I$ for the same body. In this case we are finding $I$ for a composite body by adding up all of the $I$'s from each separate body. This is why I do not think the parallel axis theorem is applied here. As for the perpendicular axis theorem, it is being applied separately to each separate body (thin sheet) in order to find $I$ of the whole body. Adding up multiple $I$'s due to composing multiple bodies is not the parallel axis theorem. $\endgroup$
    – BioPhysicist
    Commented Jul 9, 2018 at 22:12
  • $\begingroup$ @sammygerbil We take each lamina and apply the perpendicular axis theorem to each of them separately to get a $I_z$ for each. That is the end of using the perpendicular axis theorem here. Then we stack the lamina such that their z-axes are aligned. Now we have a 3D object. What is $I_z$ of this new object? Well it is just the sum of all of the $I_z$'s we found previously. Performing this sum is not using the parallel axis theorem. It is just from the definition of $I_z$ and how the integral is a linear operation. $\endgroup$
    – BioPhysicist
    Commented Jul 10, 2018 at 0:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.