1
$\begingroup$

Perhaps someone can suggest the right terms for the following mathematical objects related to moment of inertia?

  1. A inertia tensor $I$. $$I \equiv \begin{bmatrix} I_{1,1} & I_{1,2} & I_{1,3} \\ I_{2,1} & I_{2,2} & I_{2,3} \\ I_{3,1} & I_{3,2} & I_{3,3} \\\end{bmatrix}$$
  2. A product of inertia is an off-diagonal entry in the tensor: $I_{1,2} = I_{2,1}$, $I_{1,3} = I_{3,1}$, or $I_{2,3} = I_{3,2}$.
  3. A principal moment of inertia is a diagonal entry in the tensor: $I_{1,1}$, $I_{2,2}$, or $I_{3,3}$. This is the semantic of moment of inertia discussed in elementary treatment of Physics.
  4. What is the term for $I_2$ and $I_3$ in the last line below? $$\begin{align*} I_{1,1} &= \sum_{j} m_j\;\left(r^2_{j,2} + r^2_{j,3}\right) \\ &= \sum_{j} m_j\,r^2_{j,2} + \sum_{j} m_j\,r^2_{j,3} \\ I_{1,1} &= I_2 + I_3 \end{align*}$$
Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ I have a hard time to see what the question could possibly be. You seem to be asking about the right terminology but the question itself seems to contain the answers. $\endgroup$
    – Luboš Motl
    Jun 23, 2014 at 17:59
  • $\begingroup$ I don't know the term for $I_2$ and $I_3$ in number 4. $\endgroup$
    – Tadeus Prastowo
    Jun 23, 2014 at 18:33
  • $\begingroup$ The semantic of $I_2$ (or $I_3$) is the perpendicular distance of a mass to $x_2$-axis (or $x_3$-axis) after the position $r$ of the mass has been projected onto the $x_2$$x_3$-plane. $\endgroup$
    – Tadeus Prastowo
    Jun 23, 2014 at 18:55
  • $\begingroup$ Mathematically, $I_2 \neq I_{2,2}$ and $I_3 \neq I_{3,3}$ because by definition, $I_{1,1} = \sum_j m_j\,\left(r^2_{j,2} + r^2_{j,3}\right)$, $I_{2,2} = \sum_j m_j\,\left(r^2_{j,1} + r^2_{j,3}\right)$, and $I_{3,3} = \sum_j m_j\,\left(r^2_{j,1} + r^2_{j,2}\right)$, and therefore, $I_{1,1} \neq I_{2,2} + I_{3,3}$. $\endgroup$
    – Tadeus Prastowo
    Jun 23, 2014 at 19:09
  • $\begingroup$ @TadeusPrastowo You said "The semanatic ... plane". This definition is not conistent because $I_2$ would have dimensions of length. Are you reading any book in particular? $\endgroup$
    – jinawee
    Jun 23, 2014 at 19:43

2 Answers 2

Reset to default
2
$\begingroup$

A inertia tensor $I$. $$I \equiv \begin{bmatrix} I_{1,1} & I_{1,2} & I_{1,3} \\ I_{2,1} & I_{2,2} & I_{2,3} \\ I_{3,1} & I_{3,2} & I_{3,3} \\\end{bmatrix}$$ A product of inertia is an off-diagonal entry in the tensor: $I_{1,2} = I_{2,1}$, $I_{1,3} = I_{3,1}$, or $I_{2,3} = I_{3,2}$.

True.

A principal moment of inertia is a diagonal entry in the tensor: $I_{1,1}$, $I_{2,2}$, or $I_{3,3}$. This is the semantic of moment of inertia discussed in elementary treatment of Physics.

This is not true, the principal moments of inertia are the diagonal elements (eigenvalues), only if you have diagonalized the inertia matrix (which you can always do).

What is the term for $I_2$ and $I_3$ in the last line below? $$I_{1,1} = I_2 + I_3 $$

If $ I_{2}=I_{2,2}$ and $ I_{3}=I_{3,3}$, once you have diagonalized $I$. In general, we have (dropping the indices):

$$I = m \begin{bmatrix}y^2+z^2 & I_{1,2} & I_{1,3} \\ I_{2,1} & x^2+z^2 & I_{2,3} \\ I_{3,1} & I_{3,2} & x^2+y^2\\\end{bmatrix}$$

But in a planar body one of the coordinates is zero, in the example below (continuos body but same concept), $y=0$. Then you can see that:

$$I = \int d\mathbf{m} \begin{bmatrix}z^2 & I_{1,2} & I_{1,3} \\ I_{2,1} & x^2+z^2 & I_{2,3} \\ I_{3,1} & I_{3,2} & x^2\\\end{bmatrix}$$

So the moment of inertia associated to the z axis is the sum of the other two.

enter image description here

Improve this answer
$\endgroup$
13
  • $\begingroup$ Is there any term for a diagonal element in an inertia tensor when the matrix is not yet diagonalized? $\endgroup$
    – Tadeus Prastowo
    Jun 23, 2014 at 19:02
  • $\begingroup$ @TadeusPrastowo By term do you mean expression? Do you know how to calculate eigenvalues? $\endgroup$
    – jinawee
    Jun 23, 2014 at 19:03
  • $\begingroup$ No, I mean the terminology. When I communicate with someone else and I need to say the noun for a diagonal element in the moment of inertia tensor matrix that is not yet diagonalized, what is the noun? $\endgroup$
    – Tadeus Prastowo
    Jun 23, 2014 at 19:05
  • $\begingroup$ $I_2 \neq I_{2,2}$ and $I_3 \neq I_{3,3}$ because by definition, $I_{1,1} = \sum_j m_j\,\left(r^2_{j,2} + r^2_{j,3}\right)$, $I_{2,2} = \sum_j m_j\,\left(r^2_{j,1} + r^2_{j,3}\right)$, and $I_{3,3} = \sum_j m_j\,\left(r^2_{j,1} + r^2_{j,2}\right)$, and therefore, $I_{1,1} \neq I_{2,2} + I_{3,3}$. $\endgroup$
    – Tadeus Prastowo
    Jun 23, 2014 at 19:08
  • $\begingroup$ That is why I ask whether there is any name for $I_2$ and $I_3$ because they are not equal to $I_{2,2}$ and $I_{3,3}$, respectively. $I_{2,2}$ and $I_{3,3}$ are called moments of inertia about an axis. $\endgroup$
    – Tadeus Prastowo
    Jun 23, 2014 at 19:11
0
$\begingroup$

They are just the equation above rewritten to a shorter format using the definition of the moment of inertia about an axis:

$$ I = \sum_{i=1}^N{m_ir_i^2}. $$

The equation above uses $r_{j,2}$ and $r_{j,3}$, which correspond to $I_2$ and $I_3$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ But, $I_2$ (or $I_3$) is not a perpendicular distance to $x_2$-axis (or $x_3$-axis) while the Wikipedia article defines $r$ in the "moment of inertia about an axis" to be the perpendicular distance of a mass to an axis. The semantic of $I_2$ is the perpendicular distance of a mass to $x_2$-axis after the position of the mass has been projected onto the $x_2$$x_3$-plane. $\endgroup$
    – Tadeus Prastowo
    Jun 23, 2014 at 18:53
  • $\begingroup$ Mathematically, $I_2 \neq I_{2,2}$ and $I_3 \neq I_{3,3}$ because by definition, $I_{1,1} = \sum_j m_j\,\left(r^2_{j,2} + r^2_{j,3}\right)$, $I_{2,2} = \sum_j m_j\,\left(r^2_{j,1} + r^2_{j,3}\right)$, and $I_{3,3} = \sum_j m_j\,\left(r^2_{j,1} + r^2_{j,2}\right)$, and therefore, $I_{1,1} \neq I_{2,2} + I_{3,3}$. $\endgroup$
    – Tadeus Prastowo
    Jun 23, 2014 at 19:09

Your Answer

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

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