How do I find the moment of inertia for a curve?

1. Mar 27, 2009

rock.freak667

1. The problem statement, all variables and given/known data
Say I am given some curve f(x,y) (revolved around some axis), how do I find the moment of inertia about an axis?

I know how to find the moment of inertia of things like a uniform rod, ring and sphere using

$$I=\int r^2 dm$$

I believe I am supposed to to pick an elemental piece such that the revolved element is through the axis I want. But if I use I=$\int$r2 dm, I don't get anywhere.

I've various places that I am to use a double integral or even a triple integral. But I don't know how to set these up to compute the moment of inertia.

2. Mar 27, 2009

Dr.D

Have you met up with the Theorems of Pappus yet?

3. Mar 27, 2009

rock.freak667

Nope. All they did in class was how to get the moment of inertia of some figure (it escapes me what shape it was). But it was a triple integral, and they just put in the limits and some integrand (how they got it, I know not) and then just put the answer.

4. Mar 27, 2009

Dr.D

All right, then you will have to do something similar. Think in cylindrical coordinates, with the axis of revolution as the polar axis of the cylindrical coordinates. Let's suppose that your given curve is y = f(x). We want to use y as the maximum radius, and x as the z value in the cylindrical coordinate system, so the volume element is

dv = r dr dth dz
where
r is the radius to a point inside the volume, 0<=r<=f(z)
th is the angle theta that measures angle around the z axis, 0<=th<=2*pi
z is the original x value range

Then make your triple integration with all the proper limits and this should give you the volume.

5. Mar 27, 2009

rock.freak667

then I must still use the formula I=S r2 p dv (p=rho, S=integral)

what if I need to find the moment of inertia of a plane figure such as a triangle or rectangle?

6. Mar 27, 2009

Dr.D

That is a different problem, solved in a different way. Get through this one for now.

7. Mar 27, 2009

rock.freak667

I need to find the moment of inertia about the y axis and in is really cm

So my integrals would be like this:

$$I_y= \pho \int_0 ^{0.03} \int_0 ^{2 \pi} \int_0 ^{0.03} r^3 dr d\theta dz$$

8. Mar 28, 2009

Dr.D

In this problem, x is the radius, so solve the curve for x = f(y). Then use that as the upper limit of integration on r.

9. Mar 28, 2009

rock.freak667

Then it should be:

$$I_y= \pho \int_0 ^{0.03} \int_0 ^{2 \pi} \int_0 ^{\frac{y^3}{9}} r^3 dr d\theta dz$$

But I do not understand how x is the radius here if the curve shows that the x distance is not constant.Also why then would they give the distances 3cm and 3cm (vertically)?