Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Principal axes of a rigid body

  1. Apr 19, 2010 #1
    It is a known fact that every rigid body has a set of orthogonal axes called principle axes about which when rotated the angular momentum vector and the angular velocity vector point in the same direction. Inside the body we can take any point as the origin and find a combination such that the products of inertia vanish with respect to that point. So, in that sense this set of principle axes is not unique, isn’t it? Feynman in his lectures has stated that the principal axes always pass through the centre of mass. Is this a necessary condition satisfied by the principal axes?
     
  2. jcsd
  3. Apr 19, 2010 #2

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    That a rigid body has principal axes is a consequence of the nature of the inertia tensor. The inertia tensor is a real symmetric positive definite matrix. Any real symmetric matrix can be diagonalized by some orthogonal matrix. That orthogonal matrix embodies the principal axes. Positive definite matrices make the diagonalization even nicer. It's just math in other words.
     
  4. Apr 19, 2010 #3

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi Amith2006! :smile:

    "The" principal axes are defined as going through the centre of mass.

    But if a body is constrained to pivot about a fixed point other than the centre of mass, then its angular momentum will still be parallel to its angular velocity if the angular velocity is parallel to a principal axis.

    So in that sense, only the directions of the principal axes matter. :smile:

    (if you suddenly remove the pivot, of course the rotation will stay the same, but the whole body will start moving with the instantaneous velocity of the centre of mass)
     
  5. Apr 19, 2010 #4
    Hello Amith,

    You are correct in stating that the products of inertia reduce to zero for some set of axes, at any point.

    These are called the principal axes for that point.

    Since the centroid is a point there must be a set of such axes at the centroid.

    These are called the centroidal principal axes.

    So Feynman was correct, there are principal axes through the centroid.

    However there is more since the parallel axes theorems can be applied to products of inertia.

    This connects the centroidal axes (principal or not) to axes through some general point but parallel to the ones through the centroid.

    In this way the centroidal axes are more 'special' than any other.

    Incidentally the points don't have to be inside the body and often are not with irregular shapes, eg an L shape.

    Do you know how to prove your statement about products of inertia being zero ?
     
  6. Apr 20, 2010 #5
    Now, I am getting a sense of it. Here I go!
    We can find a co-ordinate system with any point inside the body as origin, but for it to be termed as a principal co-ordinate system the axes should be parallel to the principal axes of the rigid body under consideration. Though the origin may differ but all principal co-ordinate systems have their axes pointing in the same direction. In fact, any 2 principal co-ordinate systems can be related by a translation or rotation or both. The center of mass may or may not be the origin of the principal co-ordinate system though having the center of mass as the origin greatly simplifies the problem of determining the components of the inertia tensor wherein the off diagonal elements vanish due to symmetry. But in the case of irregular bodies with no symmetry at all, it is difficult to identify the principal axes and as a result the products of inertia are non zero and they have to be determined from first principles. Right?
     
  7. Apr 20, 2010 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi Amith2006! :wink:

    Fine :smile: down to here …
    Symmetry is unnecessary …

    any rigid body has three principal axes, which are the eigenvectors of the inertia tensor, and so the off-diagonal elements of the tensor (and the products of inertia) must vanish relative to those axes. :smile:
     
  8. Apr 20, 2010 #7
    I suggest you review your understanding of both moments of inertia and products of inertia.

    Moments of inertia are always positive because they contain a square term, relative to a single axis, eg x2 .


    Products of inertia can be positive or negative because they contain a 'cross product' (not in the vector sense), relative to two axes, eg xy. Since either x or y can be negative.
     
    Last edited: Apr 20, 2010
  9. Apr 20, 2010 #8
    Tiny tim and Studiot, please don’t think I am challenging you; I am just trying to find the flaws in my understanding and would like to work on it. Forgive me if I have said something stupid. If I have rightly understood your question, I would like to defend my first statement that, “having the center of mass(CM) as the origin greatly simplifies the problem of determining the components of the inertia tensor wherein the off diagonal elements vanish due to symmetry”. I don’t mean to say that if the origin is at the CM, then the products of inertia vanish. What I meant is that if the co-ordinate system has its origin at the CM, the products of inertia can be evaluated easily. The following is from the book on Classical mechanics by Gregory which says” In almost all problems, the products of inertia can be inferred from symmetry and known results. Only a body with virtually no symmetry would require products of inertia to be calculated from first principles”. This has been substantiated in the following problem. This statement has been substantiated through the following problem from the same book-Consider the case of a uniform rectangular plate placed in the X-Y plane. You are asked to find the inertia tensor at 2 points- one at the center of mass and the other at one of the corners of the plate. You will find the products of inertia vanish when the origin is at the center of mass due to symmetry but when the origin is at one of the corners of the plate, all except one vanish and for that the definite integral of mxy needs to be evaluated to get I_xy( = I_yx)
    Now the second statement I made,“But in the case of irregular bodies with no symmetry at all, it is difficult to identify the principal axes and as a result the products of inertia are non zero “. I do understand that any rigid body has three principal axes. In the case of symmetric bodies, the symmetry axis of the body obviously is also a principal axis and hence inertia tensor when evaluated is a diagonal one whereas in the case of asymmetric bodies you cannot straightaway identify the principal axis and you will have to start with an arbitrary co-ordinate system which may or may not be the principal co-ordinate system as a result of which the products of inertia do not vanish. Later on you can find the eigen values of the inertia tensor, which will be the principal moments of inertia and its corresponding eigen vectors will give directions of the principal axes.
     
  10. Apr 20, 2010 #9
    No problem, we are all trying to help here. It's good that you are thinking about this, it can be rather daunting to get your head around, especially in 3D.

    Tidying up the English can come later, but what about the question I asked in my first reply post? I gave you a hint in the second one.

    I also mentioned the transfer theorem and wonder if you are totally happy with this statement?

    Sorry I don't have time for more now but will return, matrices and all.
     
  11. Apr 20, 2010 #10
    Hello Studiot,
    Now, coming to the question you asked, I would like to take a specific case rather than a general one. For example, consider the case of a rectangular plate of mass m lying in the X-Y plane whose principal co-ordinate axes is located at the center of mass of the plate. If you try to calculate products of inertia we have,
    I_xy = summation of mxy
    Here for every point in the X-Y plane with a given value of y, there is a point with a value +x and -x and vice-versa. So, when the sum is taken over all such values, I_xy becomes zero. Since the plate is in the X-Y plane,
    z=0
    Consequently, I_yz and I_xz become zero. The same thing happens in a rigid body of any shape. I hope this is what you were expecting!
     
  12. Apr 20, 2010 #11
    Well sort of.

    It's actually neater than that. As you point out

    This means that the product xy must take on some negative values and some positive value
    and all values in between since the product is continuous between its greatest negative value and its greates positive one.

    Obviously it take on zero at some point.

    This argument, when put in mathematical terms, constitutes a valid mathmatical proof using the intermediate value theorem from analysis.

    Now about your other statement

    You can reasily perform the rotations for concurrent systems by (pre)applying a rotation matrix.
    How would you apply a translation, say [tex]x \mapsto x - a[/tex] ?

    You may like to know that of all the possible values of moment of inertia for a body the smallest possible I belongs to an axis through the centroid.

    This is clearly not true for products of inertia which can be zero or negative.

    I don't know if you have come to Euler's formulae yet, but when you do there is a most interesting example of real life chaotic behaviour to be had by spinning a brick about each of the three main axes.
     
  13. Apr 21, 2010 #12
    Google "Polhode Rolls without Slipping on the Herpolhode"

    Bob S
     
  14. Apr 22, 2010 #13
    Thanx to all of you.

    -- The more you learn, the more you realize how little you know…
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook