Moment of Inertia of an Infinite Rod with Non-Uniform Density?

chromium1387
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Homework Statement



A thin rod extends along the x-axis from x= +b to infinity. It has a non-uniform linear mass density of A/x^{4} where A is a constant with units of kg m3. Calculate the moment of inertia of the rod for rotation about the origin.

Homework Equations


The Attempt at a Solution


Well, I know that you have to put the rod in a coordinate system, break it up into small pieces, take the sum of the those, take the limit as n goes to infinity, and integrate.
So, I basically want to \intx^2dx because the word "thin" suggest that the y-coordinates do not matter. I understand that by breaking the rod into small pieces and choosing one, I have a \Deltam which I need to relate in terms of dx. I can do this using using the \frac{\Delta m}{M}=\frac{\Delta x}{A/x^{4}}, where \Delta m is my piece of mass, M is the total mass, \Delta x is my small width, and then A/x^{4} is my linear mass density. But I don't really know where this b and infinity and limits of integration come into play. :( If any can help me get started, that would be awesome!
 
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hi chromium1387! :smile:
chromium1387 said:
Well, I know that you have to put the rod in a coordinate system, break it up into small pieces, take the sum of the those, take the limit as n goes to infinity, and integrate.

So, I basically want to \intx^2dx because the word "thin" suggest that the y-coordinates do not matter. I understand that by breaking the rod into small pieces and choosing one, I have a \Deltam which I need to relate in terms of dx. I can do this using using the \frac{\Delta m}{M}=\frac{\Delta x}{A/x^{4}}, where \Delta m is my piece of mass, M is the total mass, \Delta x is my small width, and then A/x^{4} is my linear mass density.

yes :smile:

except that's not the way linear density works …

the mass of a length L is linear density time L, so the mass of [x , x+∆x] is A∆x/x4 :wink:
But I don't really know where this b and infinity and limits of integration come into play.

you're adding (integrating) the moment of inertia for every [x , x+∆x]

so you start at one end of the rod, and go to the other … ∫b
 
Ohhh.. My bad. Silly algebra mistake.
And that makes sense. I'm just used to placing one end at the origin.
Sooo, after I integrate and everything, I get \frac{MA}{b}?
Thanks for your reply!
 
chromium1387 said:
Sooo, after I integrate and everything, I get \frac{MA}{b}?
Thanks for your reply!

erm :redface: … why do you keep talking about M ? :confused:

apart from that, yes :smile:
 
I don't know.. haha. It's just incorporated into the density, right? :P
 
it doesn't exist! :biggrin:
 
okay. :)
 
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