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

In summary, Chromium1387 attempted to solve homework equation involving the moment of inertia of a thin rod. However, he was lost and could not find where M came into play. He concluded that after integrating, he would get MA/b.
  • #1
chromium1387
16
0

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[itex]^{4}[/itex] 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 [itex]\int[/itex]x^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 [itex]\Delta[/itex]m which I need to relate in terms of dx. I can do this using using the [itex]\frac{\Delta m}{M}[/itex]=[itex]\frac{\Delta x}{A/x^{4}}[/itex], where [itex]\Delta m[/itex] is my piece of mass, M is the total mass, [itex]\Delta x[/itex] is my small width, and then A/x[itex]^{4}[/itex] 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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  • #2
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 [itex]\int[/itex]x^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 [itex]\Delta[/itex]m which I need to relate in terms of dx. I can do this using using the [itex]\frac{\Delta m}{M}[/itex]=[itex]\frac{\Delta x}{A/x^{4}}[/itex], where [itex]\Delta m[/itex] is my piece of mass, M is the total mass, [itex]\Delta x[/itex] is my small width, and then A/x[itex]^{4}[/itex] 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
 
  • #3
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 [itex]\frac{MA}{b}[/itex]?
Thanks for your reply!
 
  • #4
chromium1387 said:
Sooo, after I integrate and everything, I get [itex]\frac{MA}{b}[/itex]?
Thanks for your reply!

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

apart from that, yes :smile:
 
  • #5
I don't know.. haha. It's just incorporated into the density, right? :P
 
  • #6
it doesn't exist! :biggrin:
 
  • #7
okay. :)
 

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

What is the moment of inertia of an infinite rod?

The moment of inertia of an infinite rod is a measure of its resistance to rotational motion. It is calculated by multiplying the mass of the rod by the square of its distance from the axis of rotation.

How is the moment of inertia of an infinite rod different from that of a finite rod?

The moment of inertia of an infinite rod is larger than that of a finite rod with the same mass and dimensions. This is because the infinite rod extends infinitely in both directions, while a finite rod has endpoints that contribute less to the moment of inertia.

How does the mass distribution of an infinite rod affect its moment of inertia?

The mass distribution of an infinite rod does not affect its moment of inertia, as long as the mass is evenly distributed along the length of the rod. This is because the distance from the axis of rotation remains the same for all points on the rod.

What is the formula for calculating the moment of inertia of an infinite rod?

The formula for calculating the moment of inertia of an infinite rod is I = (1/12) * m * L^2, where I is the moment of inertia, m is the mass of the rod, and L is the length of the rod.

How is the moment of inertia of an infinite rod used in physics and engineering?

The moment of inertia of an infinite rod is an important concept in rotational motion and is used in various applications, such as calculating the torque required to rotate the rod and determining the stability of structures. It is also used in designing machines and tools that involve rotational motion.

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