# Second moment of inertia.

• April 22nd 2006, 11:29 AM
Second moment of inertia.
Hey guys!

I came across this question and its on second moment of inertia, which I totally don't know the first thing about. So I was wondering if you can explain it process from the start so I can do other problems too.

Here it goes:

A flywheel consists of a 10mm thick disc of diameter 90mm symmetrically inside a hoop with internal diamemter 90mm, external diameter 100mm and thickness 50mm. Both components of the flywheel are made from the same material and its total mass is 1078g. Determine from first principles the flywheel's second moment of inertia about a central axis perpendicular to the face of the disc.

As the question states 'from first principles' I also want to know what the easier method for finding the solution will be.

I hope you guys can help. As it seems like a tough question!
• April 22nd 2006, 01:13 PM
topsquark
Quote:

Hey guys!

I came across this question and its on second moment of inertia, which I totally don't know the first thing about. So I was wondering if you can explain it process from the start so I can do other problems too.

Here it goes:

A flywheel consists of a 10mm thick disc of diameter 90mm symmetrically inside a hoop with internal diamemter 90mm, external diameter 100mm and thickness 50mm. Both components of the flywheel are made from the same material and its total mass is 1078g. Determine from first principles the flywheel's second moment of inertia about a central axis perpendicular to the face of the disc.

As the question states 'from first principles' I also want to know what the easier method for finding the solution will be.

I hope you guys can help. As it seems like a tough question!

IF I understand the concept (I've never heard of it before, either) it really isn't that complicated.

The formula for "second moment of inertia" ie. "second moment of area" (see Wikipedia) is $I=\int y^2dA$ where dA is an area element and y is the perpendicular distance from the axis of rotation and the area element. (Given this formula I am frankly wondering how it became known as a moment of inertia. It appears to have nothing at all to do with the concept!)

My only question is what exactly are we calculating the second moment of inertia for? The hoop, the disk, or the combination?

In any event, I can walk you through doing the integral (which is more or less trivial in any of the above cases.)

I will use area elements that are bits of hoops. So $dA = (dr)(rd\theta)$. If we are, for example, finding the moment of the disk we then integrate over all theta and all r from 0 to the radius of the disk, which I'm calling R:
$I = \int_0^R \int_0^{2 \pi} r^2 r\, dr d \theta$

So $I=2 \pi * \frac{1}{4}R^4$. To get the moments of the other objects, merely change the limits on the r integral.

-Dan
• April 24th 2006, 08:23 AM