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
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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
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45.4 KB
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83
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.