Clarify about the moment of inertia

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opeth_35
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hey, I want to ask you something about to calculate the moment of inertia for any shape.

My problem is that I can not calculate that moment of inertia values for any shape. Such as

rectangle or stick and the others.

I have tried to calculate for stick and I have found the solution of Ix=∫y2.dm is ∫(L/2)^2dm and after that point how can i calculate dm for this integrnt. I wrote insted of dm which is dm=M/L.dL but I cannot forward within the equation.. ı think there is a very simple thing in this equation but I cannot see that and I cannot solve thıs equatıon properly. I fell like a blind for that.

Please if you help me clarify for that solving i would be appreciate for taht.
 
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opeth_35 said:
I have tried to calculate for stick and I have found the solution of Ix=∫y2.dm is ∫(L/2)^2dm and after that point how can i calculate dm for this integrnt. I wrote insted of dm which is dm=M/L.dL but I cannot forward within the equation..
Try this:
Ix = ∫x2dm

dm = (M/L)dx

So: Ix = (M/L) ∫x2dx

You should be able to continue the calculation now.
 
i have a problem again about that solution, Could you check this out again?
 
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opeth_35 said:
i have a problem again about that solution, Could you check this out again?
Giving your solution in an attachment makes it more difficult to comment on each step.

Why do you think you need to substitute L/2 for x in the integrand? First do the integration, then substitute the range of variables.
 
because I have to calculate moment of inertia according to the central axis. that's why I have taken that L/2.

I have been trying to clarify that problem since morning but I still can not. If you say to me how to solve that. Iwill finish that things. please. by the way.. I have tried to solve without putting L/2 first. I found Ix= ML^2 / 24.

Maybe it is so simple to see that here for you, but I cannot. If I bored you sorry! I am just going to second class of my undergraduate.
 
opeth_35 said:
because I have to calculate moment of inertia according to the central axis. that's why I have taken that L/2.
What you're trying to do is evaluate the definite integral:

[tex](M/L)\int_{-L/2}^{+L/2} x^2 dx[/tex]

First find the antiderivative of x^2, then evaluate using the limits of integration.
 
that has been so clear for me, Thank you for helping me :) my problem was to understand the boundries.. okey..

have a nice days Doc Al:)