# Eliminate constants/Moment of Inertia

1. Nov 19, 2012

### aaronfue

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

Find the moment of inertia, Iy, about the y-axis using a vertical strip, width dx.
Use the given equation and eliminate constants k and c, express answer in terms of number and powers of a and b.

2. Relevant equations

I can't figure out how to eliminate the constants k and c?

3. The attempt at a solution

I believe I have the integration ready:

$\int^{a}_{0}$ x$^{2}$ (b-y)dx
where y=k(x-a)2 + c

#### Attached Files:

• ###### Moment of Inertia.JPG
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2. Nov 19, 2012

### Staff: Mentor

Pick out two points on the curve where you know the coordinates. Use these values to obtain two equations in two unknowns (c and k).

3. Nov 19, 2012

### aaronfue

Ok. So far this is what I have:

I know the point at the origin (0,0). And x=a and y=b.

0 = k(0 - a)2 + c = ka2 + c ==> k = -$\frac{c}{a^2}$

So,

y=(-$\frac{c}{a^2}$)(x-a)2 + c

I'm still not sure how to get rid of the c?

4. Nov 19, 2012

### Staff: Mentor

5. Nov 20, 2012

### aaronfue

I see..... so x=a, y=b

b=(-$\frac{c}{a^2}$)(a-a)2 + c
b=(-$\frac{c}{a^2}$)(0)2 + c
b=c

....I think??

6. Nov 20, 2012

### Staff: Mentor

Yup. So replace c in the y(x) function and then plug in the (0,0) point to see what k is all about...

7. Nov 23, 2012

### aaronfue

Thanks! Would you also check if my integration is set up correctly?

$\int^{a}_{0}$ x2 [b-($\frac{-b}{a^2}$)(x-a2) + b] dx

How would I get my answer in terms of a number and powers of constants a and b (per instructions)?

Last edited: Nov 23, 2012
8. Nov 23, 2012

### Staff: Mentor

For your integration, be sure that you're squaring the right thing in the integrand; is it just "a" that is squared?

Once you perform the integration the result will be in terms of a constant and powers of a and b.

Note that the tacit assumption is that the 'material' that the figure is made of has a uniform density of 1 mass unit per unit area. If instead you want to assume a mass M for the object, you should calculate its density from M and the total area and incorporate this into the determination of the moment of inertia.

9. Nov 23, 2012

### aaronfue

I thought that the equation for Iy was

A x2 dA

Why would the "a" be the only thing squared?

10. Nov 23, 2012

### Staff: Mentor

Isn't the function $-\frac{b}{a^2}(x - a)^2 + b$, so that it's the whole term $(x - a)$ that's squared, not just the a in the term?

11. Nov 23, 2012

### aaronfue

My answer came out to be:

$\frac{3ba^3}{10}$

I don't understand how to get a power of a constant with this problem?

12. Nov 23, 2012

### Staff: Mentor

a and b are constants. So in your answer b is to the power of 1 and a is to the power of 3.

The numerical constant 3/10 looks off to me though. Can you show more detail in your work?

13. Nov 23, 2012

### aaronfue

I've attached a scan of my work. Now that I've looked at the problem, I think I left out a "b-" for my "ydx". I think that it should be (b-y)dx. Seem right?

#### Attached Files:

• ###### MOI #1.JPG
File size:
15.5 KB
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Last edited: Nov 23, 2012
14. Nov 23, 2012

### Staff: Mentor

Yup. b-y(x) is the length of the differential strips.

15. Nov 24, 2012

### aaronfue

After including "b-" in my problem my answer came out to be:

$\frac{8ba^3}{15}$ - $\frac{2ba}{3}$,

without factoring.

Radius of Gyration: Ky = $\sqrt{\frac{(\frac{8ba^3}{15} - \frac{2ba}{3})}{(b-y)dx}}$

#### Attached Files:

• ###### Calc's.JPG
File size:
45.4 KB
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16. Nov 24, 2012

### Staff: Mentor

Your integral should look something like:

$$\int_0^a x^2\left[ b - \left[-\frac{b}{a^2}*(x - a)^2 +b\right]\right]dx$$
$$= \int_0^a x^2\left[\frac{b}{a^2}(x - a)^2\right]dx$$
$$= \int_0^a b x^2 - 2\frac{b}{a}x^3 + \frac{b}{a^2}x^4 \, dx$$

Evaluating the last form term by term should leave you with terms having b to the power of one, a to the power of three only. Try the integral again.

17. Nov 24, 2012

### aaronfue

I see what I did wrong. Instead of having (-$\frac{2b}{a}$x3), I had: (-$\frac{2b}{a^2}$x3).

I now have: ba2*($\frac{8ba}{15}$ - $\frac{2}{3}$)??

I believe I integrated correctly.

Last edited: Nov 24, 2012
18. Nov 24, 2012

### Staff: Mentor

If that's your result for the integration, it's not correct. Integrate term by term and show your work.

19. Nov 25, 2012

### aaronfue

Here is the new integration that I did. I didn't know if I should leave it as is or try to simplify it. There are no instructions that say to do so.

#### Attached Files:

• ###### Integration.JPG
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20. Nov 25, 2012

### Staff: Mentor

In the first line of your attachment, the term with the coefficient -2b/a should be an x3 term, not x2. There's one x2 term, one x3 term, and one x4 term going into the integration.