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.
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
To find the moment of inertia of the plate about the y-axis, we can use the parallel axis theorem and break down the plate into infinitesimally small rectangular strips.
Consider an infinitesimally small strip of width dx at a distance x from the y-axis. The height of this strip can be found using the lower boundary equation y = 1.10x^2.
The mass of this strip is given by dm = density * area = density * (base * height) = density * (dx * y) = density * (dx * 1.10x^2).
The moment of inertia of this strip about the y-axis is given by dI = dm * x^2 = density * (dx * 1.10x^2) * x^2 = density * 1.10x^4 * dx.
To find the total moment of inertia about the y-axis, we integrate the expression for dI from x = 0 to x = some value x_max, which corresponds to the base of the parabolic section.
I_y = ∫(density * 1.10x^4 * dx) from 0 to x_max.
To determine x_max, we solve the equation y = 1.10x^2 for x_max. Substitute y = H to obtain x_max.
H = 1.10x_max^2
x_max^2 = H/1.10
x_max = √(H/1.10)
Now, integrate the expression for I_y:
I_y = density * 1.10 * ∫(x^4 * dx) from 0 to x_max
= density * 1.10 * [(1/5)x^5] from 0 to x_max
= density * 1.10 * (1/5)(x_max^5 - 0^5)
= density * 1.10 * (1/5)(x_max^5)
The mass of the plate is given as 5.62 kg. Hence, density = mass/volume. The volume is the area of the parabolic section multiplied by the width (dx).
Let's calculate the value of density (ρ):
Mass = density * volume
5.62 = ρ * ∫(1.10x^2 * dx) from 0 to x_max
5.62 = ρ * 1.10 * (1/3)x_max^3
Solve for ρ:
1.10 * (1/3)x_max^3 = 5.62/ρ
ρ = (5.62)/(1.10 * (1/3)x_max^3)
Now, plug in the value of ρ into the expression for I_y:
I_y = [(5.62)/(1.10 * (1/3)x_max^3)] * 1.10 * (1/5)(x_max^5)
Simplify:
I_y = (5.62 * (1/3) * (1/5) * x_max^5) / (1.10 * x_max^3)
Cancel out x_max^3:
I_y = (5.62 * (1/3) * (1/5) * x_max^5) / 1.10
Finally, substitute the value of x_max = √(H/1.10):
I_y = (5.62 * (1/3) * (1/5) * (H/1.10)^(5/2)) / 1.10
Simplify further if needed.
Note: This solution assumes that the plate is of uniform density.
To find the moment of inertia of the plate about the y-axis, you are correct in using the formula for the moment of inertia of a continuous body. However, instead of using cylindrical shells, you can use the parallel axis theorem to simplify the calculation.
The parallel axis theorem states that the moment of inertia about an axis parallel to and a distance 'd' away from an axis passing through the center of mass is equal to the moment of inertia about the center of mass plus the mass times the square of 'd'.
In this case, you can consider the plate as a collection of infinitesimally thin strips along the y-axis. Each strip has a height 'dy' and a width 'dx'.
To calculate the moment of inertia of each strip, you can use the formula for the moment of inertia of a thin rod rotating about its end, which is given by (1/3) * mass * length^2. In this case, the length of each strip is dx, and the mass of each strip is equal to its density * dy * dx.
Now, let's go step by step to find the moment of inertia.
1. Start with the equation of the lower boundary of the plate, y = 1.10x^2.
2. Since you are rotating the strip about the y-axis, the distance 'd' for each strip is equal to the x-coordinate of the strip.
3. The length of each strip is dx, and the mass of each strip is equal to its density * dy * dx.
4. The density of the plate is equal to its mass divided by its volume. The mass of the plate is given as 5.62 kg, and the volume of each strip is equal to its height (dy) multiplied by its width (dx).
5. Substitute the equation of the lower boundary, y = 1.10x^2, into the expression for the moment of inertia of each strip.
6. Perform the integration to sum up the moment of inertia of all the strips from x = 0 to x = R, where R is the x-coordinate of the farthest point on the parabolic section.
The final expression for the moment of inertia of the plate about the y-axis can be obtained by plugging in the value of R into the integral expression.