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Moments of Inertia

  1. Apr 16, 2003 #1


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    This may be a sort of odd question.

    I know what a moment of inertia is. I know what they represent, I know how to use them, I even know how to calculate them from scratch if need be.

    But I don't know why they are what they are.

    Why is it that this wierd value with seemingly random d^2 terms provides the mass distribution about an axis?

    Probably more to the point:

    What, conceptually, is a tensor?
  2. jcsd
  3. Apr 16, 2003 #2

    Tom Mattson

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    The definition from which you calculate them from scratch tells you a lot. The moment of inertia about an axis due to a point mass is (dm)r2. Those factors for different shapes arise out of the simple adding up of all the point contributions.

    A second rank tensor (which the moment of inertia is) is, for these purposes, that which can be represented as a square matrix (3x3, in this case). It needs to be 3x3 so that it can be contracted with two 3-vectors, as in the expression for rotational kinetic energy, like so:


    where ω is a column vector representing the angular velocity and ωT is its transpose.
    Last edited: Apr 16, 2003
  4. Apr 16, 2003 #3


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    Re: Re: Moments of Inertia

    But why r^2 (other than to get the units to come out right...)

    Again, this is a stupid -mybraincan'tcomprehendsomethingentirelyunlessIgetitentirely- type question.

    I know what it does, I just don't know why it does what it does.
  5. Apr 16, 2003 #4
    Okay well here goes an attempt to explain it:
    • So lets start with what we know. Torque by definition is force times the lever arm (or rxF to be more precise about it). This definition makes sense so we will start from here.
    • Given this definition of torque we can say that Στ = (ΣF)r. Here we'll assume, (just becuase I don't know how to do so otherwise) that all the forces are acting at the same distance and are acting perpindicularly.
    • Okay but the second term looks familiar to us. Newtons law ΣF = ma, so we'll substitute and get Στ = mar.
    • But now were dealing with rotations not linear motion so we'll remember that a = αr. So substituting again we find Στ = mαrr. Simplifying we get Στ = mr2α.
    • Since we know in rotations that Στ = Iα it follows that from above I = mr2.
    Okay I know that's a little presumptious but does it help at all? I'll see if I can't find a better way.

  6. Apr 16, 2003 #5


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    THATS what I needed.

    Thanks climbhi!
  7. Apr 16, 2003 #6

    Tom Mattson

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    Oh, I thought you wanted to know why the coefficients to (mr2) were all different!

    Very nice, climbhi.
  8. Apr 16, 2003 #7
    Thanks, I'm glad that helped. So Tom do you have any ideas on how you could tweak with what I had above so that you don't have to make that assumption that I made. In other words can you generalize it to a system of n particles where the forces aren't neccesarily acting perpindicularly on these particle and get the result to come out I = ∫r2dm?
  9. Apr 16, 2003 #8

    Tom Mattson

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    I'll do the first one now (nonperpendicular forces), and the second one tomorrow (continuum of mass).

    Here's the "vector version":


    Now note that rr=r2r,

    where the italicized vector is the unit vector in the radial direction. So...


    That the product αxr gives the right direction is left as an exercise for the reader.

    (That means I gotta run!)
  10. Apr 18, 2003 #9
    No argument with your approach to tensors, Tom. For my (engineering) students, I write:

    A scalar has magnitude and no direction
    A vector has magnitude and one direction
    A (2nd order) tensor has magnitude and two directions
    (one for the traction vector and one for the plane of action)


    scalar = 0th order tensor has 3^0 components (in 3-space)
    vector = 1st order tensor has 3^1 components
    2nd order tensor has 3^2 components

    and so on to higher order tensors.

    A (2nd order) tensor is then a kind of supervector.


  11. Apr 19, 2003 #10
    Can't the form for the moment of inertia be derived from conservation laws?
  12. Apr 19, 2003 #11
    Probably, I remember deriving it from first principles in my physics class but when i went back to check my notes I couldn't find it. I think we started by throwing a whole series of dots on the board and looking at what happens on each particle and then applying it to the whole system. I can't remember how to do it though. Anyone wanna take a stab at it?
  13. Apr 20, 2003 #12
    aXr doesn't give the right direction


  14. Apr 20, 2003 #13
    The parallel axis theorem helps to extend calculating the moment of inertia to arbitrary 3-dimensional bodies.
  15. Apr 21, 2003 #14
    Yes, parallel axis theorem = most convenient thing ever!
  16. Apr 21, 2003 #15


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    The inertia tensor originates from the study of (discrete or continuous) rigid systems, which are systems that preserve relative distances among their parts.
    The inertia tensor does not involve any dynamical consideration (forces and torques) in its definition. It is simply a powerful tool to calculate both kinetic energy and angular momentum of a rigid system.
    This tensor extends the concept of mass to rotating systems; mass determines how much speed you will gain when you transfer a certain (linear) momentum, inertia tensor determines what angular speed you will attain by trasfering a certain angular momentum:
    (Consider anyway that angular momentum depends on a prefixed pole and so will the inertia tensor; the problem can be simplified by breaking the motion into two parts one relative to the the center of mass the other treating the whole system as concentrated in the center of mass)
    What characterizes a (3D real symmetric) tensor is that there are three direction of space mutually orthogonal that make the tensor diagonal. Hence a (3D real symmetric) tensor can always be thought as the rotation of these diagonal tensor. The way the diagonal tensor acts on the components of a vector is really simple since it simply multiplies each of the components for a simple number. The three numbers (principal inertia moments in our case) usually differ from each other; a scalar multiplying a vector can be simply thought as a tensor with three identical principal components.
    To go back to the comparison with mass and velocity/momentum, there are conditions in which the (effective) mass will be different dipending on the direction of momentum (e.g. in many crystals) and these can be represented by an (effective) mass tensor. The same happens with many physical objects usually considered as scalars such as conductivity, index of refraction, etc.

    The simplest way to see how each component contributes to the inertia tensor is by using a system constituted by a single mass rotating around a prefixed axis with (variable) angular velocity ω
    Let us consider the origin as the pole for momentum calculation and a rotation axis passing through the origin/pole
    Now if you want to extend the concept to a system of masses you need to have a system in which the velocity of each mass point can be written as a vector product of the SAME angular velocity and position.
    In general this is possible for rigid sytems only after separating the contribution of the center of mass.

    Mathworld link for angular momentum (http://scienceworld.wolfram.com/physics/AngularMomentum.html) could help but beware that it is full of little mistakes and it does not even mention that its calculation of the inertia tensor works only for rigid systems.

    Hope this helps, Dario
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