Moment of inertia of a cube along the diagonal.

  • Thread starter carllacan
  • Start date
  • #1
274
3

Homework Statement


Calculate the moment of inertia of a cube which rotates along an axis along its diagonal.

Homework Equations


Moment of inertia definition: [itex]I = \int \rho (\vec{r}) \vec{r} ^2 dV[/itex]
Angular velocity vector; [itex]\vec{\omega}=\omega (\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})[/itex]


The Attempt at a Solution


My biggest problem is on finding the position vector for the volume element. My try at it goes like this:
[itex]
\vec{r} = \vec{p}- \vec{p}·\vec{\omega} \frac{\vec{\omega}}{\omega}
[/itex]
Where [itex]\vec{p}[/itex] is the point of interest. With this we calculate the difference between the point's vector and a vector with the direction of [itex]\vec{\omega}[/itex] and magnitude equal to that of the component of [itex]\vec{p}[/itex] along [itex]\vec{\omega}[/itex]. This should be the shortest vector from the axis of rotation to the point. Am I right?

I think, however, that there is an easier way to solve this. Isn't there any shortcut similar to the Steiner Theorem, only for rotates axes?

Thank you.
 

Answers and Replies

  • #2
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,796
1,668
The moment of inertia values for a body form a tensor. Instead of trying to calculate the moment of inertia directly using the diagonal as an axis, calculate the moments using an axis aligned with the edges of the cube passing through the centroid and rotate them to the diagonal.

This discussion is relevant:

http://www.eng.auburn.edu/~marghitu/MECH2110/staticsC4.pdf [Broken]

See Eq. 4.8 on page 5
 
Last edited by a moderator:
  • #3
274
3
The moment of inertia values for a body form a tensor. Instead of trying to calculate the moment of inertia directly using the diagonal as an axis, calculate the moments using an axis aligned with the edges of the cube passing through the centroid and rotate them to the diagonal.

This discussion is relevant:

http://www.eng.auburn.edu/~marghitu/MECH2110/staticsC4.pdf [Broken]

See Eq. 4.8 on page 5
The section you refer to talks about translation of axis. Isn't this problem more about a rotation of axes? Anyway, thanks for the link. I used the results on section 4.2 for a general axis and I obtained this:
[itex] I = \frac{\rho}{\sqrt{3}}(-\frac{1}{2} ab^2c^2+\frac{1}{2} a^2bc^2-\frac{1}{2} a^2b^2c + \frac{2}{3}b^3)[/itex]
Does it make any sense? It doesn't for me. Why is the horizontal dimension given appear on its own? Shouldn't it be symmetrical?
 
Last edited by a moderator:
  • #4
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,796
1,668
Eq. 4.8 (and Sec. 4.2) come before Sec. 4.3, Translation of Coordinate Axes.

As to the validity of your calculations, I don't know. Why don't you show the full work up, starting with your calculation of the inertia tensor before rotation?
 
  • #5
274
3
Ah, right, I misread you answer.

My calculations look like this:
[itex]
I = \int dV| \vec{u} x \vec{r}|^2 = \int dV \left | \begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\
x & y & z \\ \end{array} \right | = \int dV \frac{1}{\sqrt{3}} (z - y, -z + x, y - x)^2 =
[/itex]
[itex]
\int dV \frac{1}{\sqrt{3}} (z^2 -2zy + y^2 -z^2 + 2zx -x^2 + y^2 - 2yx + x^2) = \frac{1}{\sqrt{3}} \int dV ( -2zy + y^2 + 2zx + y^2 - 2yx) =[/itex]

[itex] \frac{1}{\sqrt{3}} ( -\frac{ab^2c^2}{2} + \frac{a^2bc^2}{2} - -\frac{a^2b^2c}{2}+ \frac{2}{3}b^3 )
[/itex]
 
Last edited:
  • #6
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,796
1,668
It's not clear what the length of the edge of the cube is.

You also have 2 i's in your integral determinant.

If the edge length of the cube is say 'a', then the inertia tensor of the cube of uniform density about its centroidal axes will have entries only on the main diagonal (the products of inertia are identically zero because of symmetry). These values on the main diagonal will be

I[itex]_{xx}[/itex] = I[itex]_{yy}[/itex] = I[itex]_{zz}[/itex] = [itex]\frac{m}{12}[/itex]*(a[itex]^{2}[/itex]+a[itex]^{2}[/itex]) = [itex]\frac{m}{6}[/itex]*a[itex]^{2}[/itex]

m = ρ a[itex]^{3}[/itex], where ρ is the density

and off diagonal

I[itex]_{xy}[/itex] = I[itex]_{yz}[/itex] = I[itex]_{xz}[/itex] = 0

You can apply Eq. 4.8 to this tensor without evaluating the integral.
 
  • #7
AlephZero
Science Advisor
Homework Helper
6,994
292
Something went wrong in post #5, because the last term in your answer has the different dimensions (length^3) compared with the others (length^5).

And if this is a cube, why are there three different dimensions a b and c?

The really neat way to do this is not Eq 4.8, but look further down at sections 4.4 and 4.5. For a cube, the "ellipsoid of inertia" is not a general ellipsoid. The three axes are the sane length, so it is a sphere. That means the inertia is the same about any line through the centroid. The inertia tensor about any set of axes through the centroid is a diagonal matrix with the three equal terms on the diagonal.
 

Related Threads on Moment of inertia of a cube along the diagonal.

  • Last Post
Replies
1
Views
4K
Replies
6
Views
2K
  • Last Post
Replies
13
Views
48K
  • Last Post
Replies
4
Views
890
  • Last Post
Replies
1
Views
9K
  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
2
Views
2K
Replies
2
Views
8K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
1
Views
1K
Top