Moment of inertia (double integral)

[Pascal1p]

Homework Statement

Determine the moment of inertia of the shaded area about the x axis.[/B]

Homework Equations

Ix=y^2dA

The Attempt at a Solution

Okey so I now get how to do this the standard method. But I want to know if the method I tried is correct as well or where my mistake lies.

My attempt:
I look at the portion above x-axis and then multiple it with 2.

So I tried to use the method that I have been taught, a small element that touches the graph like in the picture.
dA= (Rcos (a)- X)dady
since x^2+y^2=R^2, x= sqrt (R^2-y^2)
so I(x)= y^2(Rcos(a)-sqrt (R^2 - y^2))dady
First I integrate this to da with the limits 0 to 1/2a (since the angle goes from 0 to 1/2a)
and then that answer I integrate to dy with limits 0 to Rsin(a/2) (since y goes from 0 to Rsin(a/2)

So basically after the first integration I get:
y^2Rsin(a/2)-(a/2)(y^2)sqrt(R^2-y^2)dy

and then when I integrate this I get this very long and quite complicated equation but is it correct?
I know now the simpler way, but I find it important that I understand it.

 

[andrewkirk]

No, it's not correct. The error arises here:

dA= (Rcos (a)- X)dady
since x^2+y^2=R^2, x= sqrt (R^2-y^2)

The equation x^2+y^2=R^2 only applies to points on the curved line, and the X in your equation dA= (Rcos (a)- X)dady is the x coordinate of a point on the diagonal straight line - not on the curved line.

 

[Pascal1p]

No, it's not correct. The error arises here:

The equation x^2+y^2=R^2 only applies to points on the curved line, and the X in your equation dA= (Rcos (a)- X)dady is the x coordinate of a point on the diagonal straight line - not on the curved line.

Ah okey I think I get it. So it would have worked if I had the equation of the straight line and implented that? instead of x^2+y^2=R^2

Can I use x= y/tan(a/2) ?

Owh and by the way, I have not had double integrals before, but how do you know if you should integrate towards da or dy first? How to figure out the order of integration?

 

[andrewkirk]

Can I use x= y/tan(a/2)?

Yes, that should work.

Owh and by the way, I have not had double integrals before, but how do you know if you should integrate towards da or dy first? How to figure out the order of integration?

In most cases it is possible to switch the order of integration. You just need to be careful to adjust the integration limits accordingly when you do so. So you can integrate in either order. Choose whichever one gives the easiest looking integration to do.

 

[Pascal1p]

Yes, that should work.
In most cases it is possible to switch the order of integration. You just need to be careful to adjust the integration limits accordingly when you do so. So you can integrate in either order. Choose whichever one gives the easiest looking integration to do.

Thanks, you been a great help =D

 

[Pascal1p]

So if I get it straight the integral will be come:
dA= (Rcosα- y/tan (α/2))dy
and Rcosα= x = √(r^2-y^2)
so dA= (√(r^2-y^2) - y/tan (α/2)) dy
and Ix= y^2dA so I(x)= (y ^2√(r^2-y^2) - y^3/tan (α/2)) dy and then integrate this from 0 to Rsin(α/2).

I thought at first it would be dA= (Rcosα- y/tan (α/2))dαdy, but I had no justification why in this situation dα would be there. Because in order to get the area of the rectangle element you don't multiply anything with a small angle.
But so is: I(x)= (y^2√(r^2-y^2) - y^3/tan (α/2)) dy and then integrate this from 0 to Rsin(α/2)

When I use standard integral for (y^2√(r^2-y^2)dy, I get a fairly complicated equation.
I know the answer should be R^4/8(α-sinα), but I don't know if the answer for 2* I(x) = 2* (y^2√(r^2-y^2) - y^3/tan (α/2)) dy from 0 to Rsin(α/2) simplifies to that.

 

[Pascal1p]

Yes, that should work.
In most cases it is possible to switch the order of integration. You just need to be careful to adjust the integration limits accordingly when you do so. So you can integrate in either order. Choose whichever one gives the easiest looking integration to do.

So I think dA= (Rcosα- y/tan (α/2))dαdy is wrong aswell, the dα seems random and I believe should not be in there. But since cosα is a variable related to dy, I need to rewrite it. So the correct integral to me seems: dA= (√(r^2-y^2) - y/tan (α/2)) dy ?

 

[andrewkirk]

the correct integral to me seems: dA= (√(r^2-y^2) - y/tan (α/2)) dy ?

Yes, that looks right.