Moment of Inertia of Plate Problem

1. Mar 21, 2008

seichan

1. The problem statement, all variables and given/known data
A uniform plate of height h= 0.69 m is cut in the form of a parabolic section. The lower boundary of the plate is defined by: y=1.10x^2. The plate has a mass of 5.62 kg. Find the moment of inertia of the plate about the y-axis.

2. Relevant equations
I=sum(m*r^2)
OR
I= int(r^2)dm

3. The attempt at a solution
Alright, so I figured I would use the first formula to figure out the moment of inertia. There is only one mass and one radius... so at first I figured it would just be 5.62*.69^2 kg m^2. I realize now that this is faulty and the idea is more of rotating the entire parabola around the axis. So, I figured it was similar to solid rotation in calculus.
So, I rotated the parabola about the y axis, resulting in two layers of cylindrical shells, with the first possessing raidus R1 and the distance to the farthest one being R2, respectively. You start with int(r^2)dm. The term dm is equal in this case to p*dv, where p is density and dv is the volume differential. The volume differential is equal to 2pi*p*h*R^3*dR. By substituting in our radiuses and integrating from R1 to R2, we are left with [(pi*p*(h)/2](R2^4-R1^4). P is constant in this case and the volume by cylindrical shells is (piR2^2-piR1^2). If we solve for the mass, we are left with M=p*v, which is equal to p*pi(R2^2-R1^2)*h. We can simplify the formula for the moment of inertia we had before, arriving at .5*M(R1^2+r2^2).
However, I have no idea how to use the lower boundary of the plate, 1.10x^2 or how to get the radiuses for the cylinders. Any help you could give would be great.

Last edited: Mar 21, 2008
2. Mar 21, 2008

Mindscrape

Why are you using cylindrical or spherical coordinates? This is a thin plate, and you can even see from symmetry that the x component will be zero (why the problem says to only do y). So just find the moment of inertia for y, and use cartesian coordinates to prevent an unnecessary mess.

Have you done vector calculus yet?

3. Mar 21, 2008

seichan

No, I don't know Vector Calculus. How would I just find it for y? I've attempted just using the original formula ( The sum of the mass times the radius squared) and this resulted in an incorrect answer. I think the most pressing problem is that I have no idea how to use the 1.10x^2=y information.
Cartesian coordinates are also not a strong suit of mine...
I tried calculating the moment of inertia for just the y- or attempted anyway. I used the h, y=.69 and plugged it into the equation that was given, 1.10x^2. I then tried to place this into the slot for radius of the moment of inertia and multiplied it by the mass but it was still wrong. Am I using the wrong idea for a radius?

Last edited: Mar 21, 2008
4. Mar 21, 2008

Mindscrape

Okay, so the whole idea behind finding a moment of inertia is that you are taking all these different point masses, each of which have their own moments of inertia, and summing up their contributions. So, for example, even though the x direction has that general shape, we should see each of the point masses contribute equal and opposite inertias to cancel each other out.

In setting up your integral what you want to do is look at the formula

$$I = \int r^2 \rho(\mathbf{r}) dr$$

and think, "Okay, so the formula is telling me that I need to integrate over all the distances along the y axis." You have a curve that is relating the y coordinates to the x coordinates. Now figure out which part of the given curve equation you want to use. In fact, maybe you want to even try out both and make sure that one of them gives you zero.

Don't worry about rho (mass/area) for now, since it is constant and comes out of the integral. Later on, to get the right value, you will need to put rho back in, but that is easy because it is just finding the area of curve, which is something every calc student does at least a gazillion times.

Sorry, I have to go, so this is all the help I can give for tonight.