Physics Forums (http://www.physicsforums.com/index.php)
-   Introductory Physics Homework (http://www.physicsforums.com/forumdisplay.php?f=153)
-   -   How do I find the moment of inertia for a curve? (http://www.physicsforums.com/showthread.php?t=303129)

 rock.freak667 Mar27-09 09:53 PM

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

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.

 Dr.D Mar27-09 09:58 PM

Re: how do I find the moment of inertia for a curve?

Have you met up with the Theorems of Pappus yet?

 rock.freak667 Mar27-09 09:59 PM

Re: how do I find the moment of inertia for a curve?

Quote:
 Quote by Dr.D (Post 2136519) Have you met up with the Theorems of Pappus yet?
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.

 Dr.D Mar27-09 10:44 PM

Re: how do I find the moment of inertia for a curve?

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.

 rock.freak667 Mar27-09 10:48 PM

Re: how do I find the moment of inertia for a curve?

Quote:
 Quote by Dr.D (Post 2136587) 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.
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?

 Dr.D Mar27-09 10:50 PM

Re: how do I find the moment of inertia for a curve?

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

 rock.freak667 Mar27-09 11:02 PM

Re: how do I find the moment of inertia for a curve?

http://img23.imageshack.us/img23/1364/picwmz.jpg

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

So my integrals would be like this:

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

 Dr.D Mar28-09 08:26 AM

Re: how do I find the moment of inertia for a curve?

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.

 rock.freak667 Mar28-09 01:43 PM

Re: how do I find the moment of inertia for a curve?

Quote:
 Quote by Dr.D (Post 2136977) 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.
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)?

 All times are GMT -5. The time now is 09:00 AM.