Deriving the moment of inertia of a thin rod.

1. Apr 15, 2014

Dethrone

1. The problem statement, all variables and given/known data

I need to derive the moment of inertia of a thin rod with its axis of rotation at the end of the rod. http://en.wikipedia.org/wiki/List_of_moments_of_inertia
The third one.

2. Relevant equations
I = mr^2

3. The attempt at a solution
I completely understand how the moment of inertia is mL^2/3, but what I don't understand is why linear density is used. M/L. Shouldn't it be volume density, or area density at the very least since a thin rod is 3D?

2. Apr 15, 2014

BruceW

Hey, welcome to physicsforums. Yes, you can use volume density. And since they are talking about a very thin rod, you would then need to take the limit that the width of the rod is very small. Continue with this idea, you should end up getting the answer mL^2/3 either way.

3. Apr 15, 2014

Dethrone

Thanks for the welcome! I guess my new question is how I can get volume density to work out to mL^2 /3. My attempt: density = M / (pi r^ w) , where w is the width of rod. Am I right so far?

I was just introduced to all this angular stuff last week, so pardon me if this seems to be a dumb question, but what is the usefulness of a "very thin rod". Would finding the moment of inertia of just a regular cylindrical rod not suffice?

4. Apr 15, 2014

BruceW

Well, the density is the mass of the rod, divided by the volume of the rod. So you've got 'M' correct. But for the volume of the rod, you have pi r^ w This definitely is not right. It should be the volume of a cylinder of length 'L' and width (i.e. diameter) 'w'.

Finding the moment of inertia of a regular cylindrical rod is more difficult than finding the moment of inertia of a 'very thin rod'. In fact, the regular cylindrical rod is number 8 on the wikipedia page you linked to. (well, the moment of inertia through it's center, rather than at the end). So you can see under what approximation number 8 will become approximately the same as number 4.

5. Apr 15, 2014

Dethrone

How about M / (pi (w/2)^2 L)?
From the volume of a cylinder, I've replaced the radius with w/2 and the height by L.

Last edited: Apr 15, 2014
6. Apr 15, 2014

BruceW

yep, that's it. So, that is the density. So next thing would be to calculate the moment of inertia, using the density and shape of the cylinder. Have you learned about these kinds of integrals yet? I guess you have done the one in 1 dimension. In 3 dimensions it's not so different, except instead of distance from the origin, you need to use distance from the closest point on the axis of rotation.

7. Apr 15, 2014

Dethrone

This is what I have so far: (assuming axis of rotation is at the end)
integral from 0 to L of (M / [ pi (w/2)^2 L ]) x^2 dx)

I think x is the distance from the axis of rotation, and dx are very small changes in Length
Where did I go wrong?

8. Apr 15, 2014

BruceW

that is (almost) the 1 dimensional equation. For the 3 dimensional equation, the distance from the axis of rotation is more complicated than just x (since we are saying the rod is not 'very thin'), and for the integral you also need to integrate over dy and dz. And the limits for the y and z integrals are essentially the equation for a circle. (Since we are talking about a cylinder here).

9. Apr 15, 2014

Dethrone

The integral would be something like (M / [ pi (w/2)^2 L ]) ( (x^2 + y^2) dx dy dz? or am I completely off.? I really have no clue.

Last edited: Apr 15, 2014
10. Apr 15, 2014

BruceW

yeah, that's right. You've implicitly chosen the z axis as the axis that the rod spins around, since x^2+y^2 is the squared distance from the z axis. The main thing you still need to do is get the limits of integration to match the shape of the cylinder.

11. Apr 15, 2014

Dethrone

I completely guessed that, because x^2 + y^2 defines a circle and x^2 + y^2 + z^2 makes no sense to me. I also find this hard to visualize. Also, does that mean I = ∑ mx^2 doesn't apply for more than two dimensions? Since it tells me I need to square the distance.

Z bound: integral from 0 to L
Y bound: integral from -pi/2 to pi/2 (no idea)
X bound: integral from -sqrt(r^2- y^2) to + sqrt(r^2- y^2) (no idea)

I haven't been taught integration yet, so I don't think I'll be able to get the bounds. Everything I know about it right now are just self-learned. And I'm not required to know how they're derived yet, but it really bugs me to not fully understand it.

Last edited: Apr 15, 2014
12. Apr 16, 2014

BruceW

mx^2 generally only applies for a one dimensional object. So, for example, the very thin rod is approximately one dimensional object, since it is very thin.

the x bound and z bound should be swapped around, since the z axis is the axis that the rod spins around. But once you've swapped around the x and z bound, you will have the correct bounds for both of them. And for y bounds, you just need the two extreme values for y. And remember to integrate over the bounds -sqrt(r^2- y^2) to + sqrt(r^2- y^2) First, since this includes the y variable.

Or, you could change to cylindrical coordinates, which might make things easier. Anyway, you're fairly close to the answer. I think the important thing is to see how number 8 on the wikipedia page will approximately be the same as number 4, under a certain limit. If you can see that, then that is all you really need, to understand how the very thin rod is derived as a special limiting case of the regular rod.

also, there is nothing wrong with just doing the calculation in 1 dimension to begin with. If we assume the rod is very thin, then we immediately know that the distribution is approximately 1 dimensional. So we don't need to worry about the other two dimensions in calculating the answer. But you do need to use mass per length, instead of mass per volume, since for the 1 dimensional case, you are no longer integrating over volume, but are integrating over a length instead.