# Moment of inertia, huh?

1. May 17, 2004

### acgold

I've been doing moment of inertia problems and haven't had any trouble so far until these:

Using integration, find the moment of inertia of a rod about an axis through its center if the mass per unit length is $$\lambda=\lambda_o x$$. The answer is supposed to come out to $$I=\frac 1 8 ML^2$$.

Using integration, find the moment of inertia of a disk of radius $$a$$, about its center if a) $$\sigma=\sigma_o r^{-1}$$ b) $$\sigma=\sigma_o r^2$$. The answers are a) $$I=\frac 1 3 Ma^2$$ b) $$I=\frac 2 3 Ma^2$$

Last edited: May 17, 2004
2. May 18, 2004

### arildno

Okay, I'll do the first one for you.

The line density is given as $$\lambda=\lambda_{0}x$$

It's no wonder that you are stuck with this exercise, because this expression is by itself meaningless.
The correct expression must be:
$$\lambda=\lambda_{0}|x|$$, -L/2<=x<=L/2
i.e the absolute value of x, rather than x itself.
(I assume you gave us ALL the information present in the exercise?)
We gain:
$$I=\int_{-\frac{L}{2}}^{\frac{L}{2}}\lambda_{0}|x|{x}^{2}dx=\lambda_{0}\frac{2}{4}(\frac{L}{2})^{4}$$

The mass M of the rod is readily calculated:
$$M=\int_{-\frac{L}{2}}^{\frac{L}{2}}\lambda_{0}|x|dx=\lambda_{0}(\frac{L}{2})^{2}$$
Combining the expressions yiels the desired result.

3. May 18, 2004

### acgold

Thank you thank you thank you...I've been working on that problem for hours. Yeah I gave you ALL the information I had. My professor is notorious for having typos. I did the first problem the way you showed me and I understand it 100% now. I'm off to work on the last two on my own...thanks for all your help. BTW, how did you know it was the absolute value of x? Do the sigma expressions require absolute values also? Thanks again.

4. May 18, 2004

### arildno

My reasoning was as follows:
1. Density has to be non-negative.
Hence, the only meaningful interpretation of the given expression as it stood, was that x is measured from one end of the rod (x=0 at one end) , but that does not give the stated expression.

2. Hence, I assumed that the density functionshould bemeasured from the point of the rotation axis, but then I had to introduce the absolute value in order to have a meaningful density function.

3. As to the area densities, I don't think so, since the radius pr. definition is non-negative (I haven't checked them yet)

Last edited: May 18, 2004
5. May 18, 2004

### acgold

Thanks, you're a much better teacher than my professor. :) I've been working on the second problem for a few minutes now and I'm stuck again. Maybe another typo...I don't know but I'll keep trying it.

6. May 18, 2004

### arildno

Well, thanks for the compliment,
(But there has to be a good reason why he's a professor, and I'm not!)
Your professor has the second problem right.
In order to solve it, you must use polar coordinates!!
Remember that the proper area element, dA, has the form: $$dA=rdrd\theta$$,
with the limits:
$$0\leq{r}\leq{a},0\leq\theta\leq2\pi$$
Good luck!