Calculating Moment of Inertia Tensor for Rod Along x Axis

w3390
Messages
341
Reaction score
0

Homework Statement



A thin rod has mass M and length L. What is the moment of inertia tensor about the center of mass if placed along the x axis.

Homework Equations



I would write the inertia tensor in component notation, but I don't know how to use Latex.

The Attempt at a Solution



I am getting lost in the inertia tensor equation. For example, when I try to find the I_11 component, plugging in I get:

I_11 = int[rho((x_2^2 -x_1^2)+(x_3^2-x_1^2))dV]

I_11 = iiint[rho(-2x_1^2+x_2^2+x_3^2)dx_1dx_2dx_3]

This is where I am stuck. I do not understand what to use as my various bounds or if this equation is even correct.

Any help would be much appreciated.
 
Physics news on Phys.org
Okay, so I think I figured out my problem. I was doing something wrong with the component form.

Here is what I got for my inertia tensor:

I_11 = 0
I_22 = (1/3)ml^2
I_33 = (1/3)ml^2
I_12 = I_21 = 0
I_13 = I_31 = 0
I_23 = I_32 = 0So it looks something like this:

[..0...0.....0...]
[..0...(1/3)ml^2...0...]
[..0...0...(1/3)ml^2]

Does this look correct?
 
it should be 1/12 ml^2 you are doing it about the center
 
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...

Similar threads

Replies
15
Views
2K
Replies
3
Views
4K
Replies
1
Views
3K
Replies
4
Views
5K
Replies
5
Views
4K
Replies
12
Views
1K
Back
Top