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

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.

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 6:37am
The 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

## Answer This Question

## Related Questions

- physics - A uniform plate of height = 1.80 m is cut in the form of a parabolic ...
- Science - According to the theory of plate tectonics, how do two plates interact...
- Physics CENTER OF MASS - A thin rectangular plate of uniform areal density σ...
- PHYSICS CENTER OF MASS - A thin rectangular plate of uniform areal density σ...
- mechanics - A roller of radius 400mm rolls without slip between two parallel ...
- Physics - A thin rectangular plate of uniform area density σ1 = 1.01 kg/m2 ...
- physics - A thin rectangular plate of uniform area density σ1 = 1.00 kg/m2 ...
- Science - What type of plate boundary is located at the Juan de Fuca plate and ...
- Physics - Two parallel plates are 4.85 cm apart. The bottom plate is charged ...
- Physics - In the figure a uniform, upward electric field of magnitude 1.90 × 103...

More Related Questions