Moment of inertia of two bars at right angle using Steiner's theorem

[SchroedingersLion]

Homework Statement

Assume we have two thin bars of length L and mass M. The moment of inertia of one bar with respect to axis through center of gravity of said bar is $I=\frac {mL^2}{12}$. The two bars stick together at their ends, forming a right angle.
What's the moment of inertia of the whole system with respect to that point?

Homework Equations

I saw two different approaches, but I don't know, why one of them is wrong.

The Attempt at a Solution

First approach:
I can write the total $I_{tot}$ with respect to a given axis as the sum of the two moments of inertias of the separate bars with respect to said axis. So with respect to an axis going through the point where they stick together, let's call it 'O', I can write:
$I^O_{tot}=I^O_{1}+I^O_{2}$
For each $I^O_{i}$ I can write $I^O_{i}=I + Md^2$ with d being the distance between center of gravity of one bar to the point O, i.e. $d=\frac {L}{2}$.
Inserting everything, I get $I^O_{tot}=2I+ 2Md^2 =...=\frac {7}{6}ML^2$

Now, the second approach.
In an earlier exercise, one was to calculate the center of gravity coordinates of the whole system. Using a coordinate system whose axes follow the bars from point 'O', the coordinates of the center of gravity should be $\vec r_{cog}=L * (\frac{1}{2}, \frac{1}{2})$.

Now I want to use Steiner's theorem to first calculate the moment of inertia of the whole system with respect to an axis through $\vec r_{cog}$, i.e. $I^{cog}_{tot}$, and then use it again to calculate $I^O_{tot}$ again from there.

$I^{cog}_{tot}=I^{cog}_1 + I^{cog}_2 =I_1 + Md^2 + I_2 + Md^2$
Here, d is the distance of the center of gravity of one bar to the center of gravity of the whole system. But this should again simply be $\frac {L}{2}$, just like earlier!
But then I would get $I^{cog}_{tot}=2I+ 2Md^2 (=I^O_{tot}$ c.f. above).
If I used Steiner's theorem now to get $I^{O}_{tot}$ from here, I would get a larger value than in my first approach.

Does anyone see my error?SL

 

[BvU]

the coordinates of the center of gravity should be $\vec r_{cog}=L * (\frac{1}{2}, \frac{1}{2})$.

Really ?

 

[SchroedingersLion]

Really ?

Hmm... I would write the mass density as $\rho(\vec r)= \frac M L ( \delta(x) 1_{[0,L]}(y) + \delta(y) 1_{[0,L]}(x))$

Then the center of gravity is given by $\vec r_{cog}=\frac 1 M \int_{x=0}^L \int_{y=0}^L \vec r \rho(\vec r) \, dx \, dy$.
Do you agree up until here?

 

[BvU]

I agree. But I simply take the two centers of mass and find the 'average'

 

[SchroedingersLion]

I agree. But I simply take the two centers of mass and find the 'average'

Makes sense, but then I solved the integral wrong...

$$\vec r_{cog}=\frac 1 M \int_{x=0}^L \int_{y=0}^L \vec r \rho(\vec r) \, dy \, dx \\=\frac 1 L (\int_{x=0}^L \int_{y=0}^L \delta(x) (x,y) \, dy \, dx + \int_{x=0}^L \int_{y=0}^L \delta(y) (x,y) \, dy \, dx) \\= \frac 1 L (\int_{x=0}^L \delta(x) (xL, \frac {1}{2}L^2) \, dx + \int_{x=0}^L (x,0) \, dx ) \\= \frac 1 L ( (0, \frac {1}{2}L^2) + (\frac 1 2 L^2,0) ) \\= L * (\frac 1 2, \frac 1 2) \neq L * (\frac 1 4, \frac 1 4)$$

 

[BvU]

Isn't that for a plate instead of for two bars ? [edit] oops, too quick. Let me read first

 

[BvU]

Ah, you want to divide by 2M, being the mass of the whole thing.

 

[SchroedingersLion]

Right! So I get $L*(\frac 1 4, \frac 1 4)$.
So if my two approaches from above are equivalent, I should get the same result, right?

 

[BvU]

Think so, yes.

 

[SchroedingersLion]

Doesn't work for me.

First approach yields (the value from first post was wrong):
$I^O_{tot}=2I+ 2Md^2=\frac {ML^2}{6} + 2M \frac {L^2} {4}= \frac 2 3 ML^2$

Now second approach:
Total moment of inertia with respect to center of gravity at $\vec r_{cog}=L*(\frac 1 4, \frac 1 4)$:
$$I^{cog}_{tot}=2I + 2Md^2\\ = \frac {ML^2}{6} +2M |(0, \frac L 2) - (\frac L 4, \frac L 4)|^2 \\
=\frac {ML^2}{6} +2M \frac {2L^2}{16} \\
=\frac {ML^2}{6} + \frac {ML^2}{4}\\
=\frac {5} {12} ML^2 +\frac {3} {12} ML^2 = \frac 2 3 ML^2$$

Lol. I did it correctly in LaTeX but wrong on my paper. So it really is the same. Man, I am tired now...

Thank you BvU for your help!

edit: wait. something is off...
It's not the same. Now I would have to use Steiner again to get to $I^{O}_{tot}$. Than I would have different results.

edit2: Ok, found the error it is $I^{cog}_{tot}=...= \frac {5}{12} ML^2$
and then, let's see...

 

[SchroedingersLion]

Ok, with $I^{cog}_{tot}= \frac {5}{12} ML^2$ it follows via Steiner: $$I^{O}_{tot}=I^{cog}_{tot}+2Md^2\\
= \frac {5}{12} ML^2 +2M |(\frac L 4, \frac L 4)|^2 \\
=\frac {5}{12} ML^2 + 2M 2*\frac {L^2}{16}\\
=ML^2 (\frac {5}{12}+ \frac{1}{4}) = \frac 2 3 ML^2$$

Now it's ok. Sorry for the confusion, I must say that I am in a bit of a hurry today...

Thanks again!

 

[BvU]

You're welcome. Take it easy: if you have the luxury of time, laying something aside and picking it up later is often quite useful. Works like a dream for my sudoku puzzles .