Moment of inertia of a rotated line

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SUMMARY

The moment of inertia (MOI) for a composite drum consisting of a cylinder and a rotated line is calculated using the formula I = 1/2M(a² + b²) for the cylinder, where 'a' and 'b' are the inner and outer radii. For the rotated line represented by the equation y = mx + c, the MOI is derived by integrating the elemental section, resulting in I = 1/2 ρ π ∫(y²)² dx. The final expression for the total MOI combines both components: I = 1/2 M(a² + b²) + 1/2 ρ π ∫(y₂⁴ - y₁⁴) dx, where y₂ and y₁ correspond to the upper and lower bounds of the line's height.

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rock.freak667
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Homework Statement


I have a drum that comprises of basically a cylinder and rotated line in the form y=mx+c

http://img22.imageshack.us/img22/6108/druma.jpg


Homework Equations



[tex]I=\int r^2 dm[/tex]

The Attempt at a Solution



I found the MOI for the cylinder as 1/2M(a2+b2) where a and b are the inner and outer radii respectively.

I am getting trouble forming the MOI for the line. I am supposed to consider an elemental section, but I am not sure if I should consider a disc or some other shape.

Disc: (Width dx and radius 'y')

dI=1/2(dm)y2

dm=ρ dV = ρ(πy2dx)

dI=1/2ρπ y4 dx

[tex]I_{line}= \frac{1}{2} \rho \pi \int_{x_1} ^{x_2} y^4 dx[/tex]


Since I have two lines, the inertia would be


[tex]I = \frac{1}{2} \rho \pi \int_{x_1} ^{x_2} (y_2^4 -y_1^4) dx[/tex]

I can appropriately select x2 and x1 based on height of the cylinder and the height of the total drum. I can find the ρ for the material. But I'd need to calculate the equations for y2 and y1.

So my MOI would be

[tex]I = \frac{1}{2} M(a^2+b^2)+ \frac{1}{2} \rho \pi \int_{x_1} ^{x_2} (y_2^4 -y_1^4) dx[/tex]

Is this correct or did I make a mistake in my element or in the algebra?
 
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