Cut Washer Moment

This problem involves a sizeable amount of work. It was quite satisfying to get the right answer at the first go, cause I thought it is easy to err with the approach I used. Here it is.

Consider the portion of the mass below the line $y = x +0.5$ . It is bounded by the line and two concentric circular segments. First, I considered the circle:

$x^2 + y^2 = 1$

To compute the moment of inertia of this circular segment about the Z-axis, it is necessary to find the points of intersection between the circle and the line $y = x +0.5$ . Solving for the points of intersection gives a quadratic equation. The solution of this quadratic equation gives the x-coordinates of the points of intersection between the circle and the line. They are:

$x_{\mathrm{LiL}} = \frac{-1-\sqrt{7}}{4}$ $x_{\mathrm{LiH}} = \frac{-1+\sqrt{7}}{4}$

The subscript has a definite meaning. In the subscript $\mathrm{LiL}$ , the first L signifies whether the segment lies below (L) or over (U) the given line. The second L signifies whether the x-coordinate is lower(L) or higher (H) in value. The 'i' signifies whether the circle considered is the inner concentric circle or the 'outer'(o) concentric circle. This notation is used throughout the solution

Knowing the coordinates, it is now possible to compute the moment of inertia of the lower-inner area by inspecting the diagram. The expression is as such:

$I_{\mathrm{Li}} = \left(\int_{x_{\mathrm{LiL}}}^{x_{\mathrm{LiH}}} \int_{-\sqrt{1-x^2}}^{x+0.5} \left(x^2+y^2\right) \ dy \ dx \right)+\left( \int_{x_{\mathrm{LiH}}}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \left(x^2+y^2\right) \ dy \ dx\right)$

Now, consider the segment of radius 2 below the line.

$x^2 + y^2 = 4$

To compute the moment of inertia of this circular segment about the Z-axis, it is necessary to find the points of intersection between the circle and the line $y = x +0.5$ . Solving for the points of intersection gives a quadratic equation. The solution of this quadratic equation gives the x-coordinates of the points of intersection between the circle and the line. They are:

$x_{\mathrm{LoL}} = \frac{-1-\sqrt{31}}{4}$ $x_{\mathrm{LoH}} = \frac{-1+\sqrt{31}}{4}$

Knowing the coordinates, it is now possible to compute the moment of inertia of the lower-outer area by inspecting the diagram. The expression is as such:

$I_{\mathrm{Lo}} = \left(\int_{x_{\mathrm{LoL}}}^{x_{\mathrm{LoH}}} \int_{-\sqrt{4-x^2}}^{x+0.5} \left(x^2+y^2\right) \ dy \ dx \right)+\left( \int_{x_{\mathrm{LoH}}}^{1} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \left(x^2+y^2\right) \ dy \ dx\right)$

The total moment of inertia of the space between these two segments evaluates to:

$I_L = I_{\mathrm{Lo}}-I_{\mathrm{Li}}$

Now, the same process has to be repeated for the portion of the washer above the given line. It is said that the washer is cut along the line and displaced. The cut washer is bounded by the line and two circular arcs of radius 1 and 2, centered at $(-1,-1)$ .

Consider the portion of the mass above the line $y = x +0.5$ . Consider the circle:

$(x+1)^2 + (y+1)^2 = 1$

To compute the moment of inertia of this circular segment about the Z-axis, it is necessary to find the points of intersection between the circle and the line $y = x +0.5$ . Solving for the points of intersection gives a quadratic equation. The solution of this quadratic equation gives the x-coordinates of the points of intersection between the circle and the line. They are:

$x_{\mathrm{UiL}} = \frac{-5-\sqrt{7}}{4}$ $x_{\mathrm{UiH}} = \frac{-5+\sqrt{7}}{4}$

Knowing the coordinates, it is now possible to compute the moment of inertia of the upper-inner area by inspecting the diagram. The expression is as such:

$I_{\mathrm{Ui}} = \left(\int_{-2}^{x_{\mathrm{UiL}}} \int_{-1-\sqrt{1-x^2}}^{-1+\sqrt{1-x^2}} \left(x^2+y^2\right) \ dy \ dx \right)+\left( \int_{x_{\mathrm{UiL}}}^{x_{\mathrm{UiH}}} \int_{x+0.5}^{-1+\sqrt{1-x^2}} \left(x^2+y^2\right) \ dy \ dx\right)$

Consider the outer circle above the line:

$(x+1)^2 + (y+1)^2 = 4$

To compute the moment of inertia of this circular segment about the Z-axis, it is necessary to find the points of intersection between the circle and the line $y = x +0.5$ . Solving for the points of intersection gives a quadratic equation. The solution of this quadratic equation gives the x-coordinates of the points of intersection between the circle and the line. They are:

$x_{\mathrm{UoL}} = \frac{-5-\sqrt{31}}{4}$ $x_{\mathrm{UoH}} = \frac{-5+\sqrt{31}}{4}$

Knowing the coordinates, it is now possible to compute the moment of inertia of the upper-outer area by inspecting the diagram. The expression is as such:

$I_{\mathrm{Uo}} = \left(\int_{-3}^{x_{\mathrm{UoL}}} \int_{-1-\sqrt{4-x^2}}^{-1+\sqrt{4-x^2}} \left(x^2+y^2\right) \ dy \ dx \right)+\left( \int_{x_{\mathrm{UoL}}}^{x_{\mathrm{UoH}}} \int_{x+0.5}^{-1+\sqrt{4-x^2}} \left(x^2+y^2\right) \ dy \ dx\right)$

The total moment of inertia of the space between these two segments evaluates to:

$I_U = I_{\mathrm{Uo}}-I_{\mathrm{Ui}}$

Finally, the total moment of inertia of the cut washer about the Z-axis is:

$\boxed{I = I_L+I_U \approx 31.557235}$

All integrals were solved using Wolfram-Alpha.

I'm very impressed by how @Karan Chatrath was able to keep track of all the boundaries of integration, etc and get a good result. My approach was to apply an "if-else" statement to determine the cut washer coordinates $(x',y')$ based on the un-cut washer coordinates $(x,y)$ . The moment of inertia is computed based on $(x',y')$ . The concern is not for the lines bounding the integration regions, but rather for the locations of the infinitesimal masses themselves.

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import math

N = 1000

sigma = 1.0

dr = 1.0/N
dtheta = 2.0*math.pi/N

######################################

r = 1.0
A = 0.0
I = 0.0

while r <= 2.0:

    theta = 0.0

    while theta <= 2.0*math.pi:

        x = r*math.cos(theta)
        y = r*math.sin(theta)

        dA = r*dr*dtheta
        A = A + dA
        dm = sigma*dA

        if y >= x + 0.5:
            xp = x - 1.0
            yp = y - 1.0
        else:
            xp = x
            yp = y

        rpsq = xp**2.0 + yp**2.0

        dI = dm*rpsq

        I = I + dI

        #print xp,yp

        theta = theta + dtheta

    r = r + dr

######################################

print N
print A
print I


#>>> 
#1000
#9.44363694147
#31.6172055294
#>>> ================================ RESTART ================================
#>>> 
#2000
#9.43420509493
#31.5872693768
#>>> ================================ RESTART ================================
#>>> 
#5000
#9.42666291635
#31.5645258293
#>>> ================================ RESTART ================================
#>>> 
#10000
#9.42572043859
#31.5608931922
#>>>