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October 31, 2014

October 31, 2014

Posted by **Kara** on Friday, March 21, 2008 at 10:04pm.

Alright, I've done this so far:

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.

- Physics (Mechanics) -
**drwls**, Saturday, March 22, 2008 at 6:37amThe half-width of the parabolic section at any height y above the x axis is

x = sqrt (y/1.1)

Perform one integral for the total mass in terms of H, using the known mass. This will tell you the density x thickness. Then perform a second integral for the moment of inertia, using that density you have determined.

The mass integral is

5.62 kg = (density)(thickness)*INTEGRAL OF 2x dy for y from 0 to 0.69 m. Substitute sqrt (y/1.1) for x before integrating.

The moment of inertia integral is

I = (density)(thickness)* (2/3)INTEGRAL OF x^2 dy for y from 0 to 0.69 m

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