Rotational inertia of square about axis perpendicular to its plane

[member 731016]

Yes, that will always be the case for a square.

Thank you for your reply @erobz !

 

[member 731016]

Won't be easier to just find the moment of inertia for a square of side L and mass m and then use the results for any value of L and for any way this side is related to another parameter in the problem, like to the radius here? You can do this in two or three lines, even by using the integral.
For one side of the square, the moment of inertia relative to the center is just $$\int_{-L/2}^{L/2} [(L/2)^2+x^2]\lambda dx =\lambda L (L/2)^2+\lambda \frac{x^3}{3}\Big|_{-L/2}^{L/2} =\lambda L (L/2)^2+\lambda \frac{2L^3}{3\times 8}$$
I considered the x axis parallel to the side I am looking at.
Using $M=\lambda L$ you get
$$I_{side} =m\frac{L^2}{4}+m \frac{L^2}{12}$$
The last term is the moment of inertia for an axis going through the center of the bar (look it up in a table) and the first term is what you add by using the parallel axis theorem. You could write this right away, without any integral.
Then the monemt of the square is just $I_{square}=4I_{side}$.
Only now is time to see what happens if $L=R\sqrt{2}$.

Thank you very much for that @nasu!

 

[member 731016]

I have done that @nasu!

$I_{side}= \frac {2mR^2}{4} + \frac{2mR^2}{12}$
$I_{side} = \frac {mR^2}{2} + \frac{mR^2}{6}$
$I_{square} = 4[\frac {mR^2}{2} + \frac{mR^2}{6}]$
$I_{square} = 2mR^2 + \frac{2mR^2}{3}$
$I_{square} = \frac{8mR^2}{3}$

Many thanks!