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further maths mechanics - moments of inertia
hey... i haven't got the question with me, but can someone PLEASSEEE explain to me how you find the moment of inertia of an isosceles (sp?)triangle with its line of symmetry as the axis? The question says to use integration.. i think the mass was m, the base 2b and height h... pleaseee help explain it properly cos i'm really thick when it comes to imagining images!!!! 
Reply 1
15
the triangle is attached.
the equations of the triangles sides are x = +- (b(y-h))/h
σ = m/bh (the mass-per-unit-area of the triangle)
moment of inertia is: I = ∫σ x² dA
dA = dx dy
=> I = ∫[0 -> h] ∫[(b(y-h))/h -> -(b(y-h))/h] m/bh x² dx dy
=> I = ∫[0 -> h] m/3bh . -2b³/h³ (y-h)³ dy
=> I = m/12bh . 2b³/h³ . (-h)^4 = mb²/6
the equations of the triangles sides are x = +- (b(y-h))/h
σ = m/bh (the mass-per-unit-area of the triangle)
moment of inertia is: I = ∫σ x² dA
dA = dx dy
=> I = ∫[0 -> h] ∫[(b(y-h))/h -> -(b(y-h))/h] m/bh x² dx dy
=> I = ∫[0 -> h] m/3bh . -2b³/h³ (y-h)³ dy
=> I = m/12bh . 2b³/h³ . (-h)^4 = mb²/6
Reply 2
elpaw
the equations of the triangles sides are x = +- (b(y-h))/2h
the equations of the triangles sides are x = +- (b(y-h))/2h
i'm sorry.. could u please explain this part?!??
Reply 3
15
Sorry, the 2 was a mistake (I had originally thought the base of the triangle was just b, not 2b, so the x axis originally went from -b/2 to b/2 and that is where the 2 cropped in), i have corrected my original solution.
I've drawn on the diagram that y = hx/b + h and y = -hx/b + h are the equations of ther lines of th sides of the triangle (using just y=mx+c). rearranging these to make them in terms of x you get:
y-h = +- hx/b
b(y-h) = +- hx
x = +- (b(y-h))/h
the + solution is the left side, the - solution is the right side.
I've drawn on the diagram that y = hx/b + h and y = -hx/b + h are the equations of ther lines of th sides of the triangle (using just y=mx+c). rearranging these to make them in terms of x you get:
y-h = +- hx/b
b(y-h) = +- hx
x = +- (b(y-h))/h
the + solution is the left side, the - solution is the right side.
Reply 4
thanks! 
one more thing tho,
elpaw
moment of inertia is: I = ∫σ x² dA
moment of inertia is: I = ∫σ x² dA
is this a standard formula??? the textbook i have uses a slightly diff method.. urs seems so much easier, so if this is a standard formula, i'd rather use it! again, i apologise for being such a thickhead!!
Reply 5
oh u know what.. i stared at it for a few minutes and tried working it out myself.. and i think i understand now..
thank u sooo much! i'm sorry i'm such a bother! thank u thank u thank u!!!
thank u sooo much! i'm sorry i'm such a bother! thank u thank u thank u!!!
Reply 6
15
its alright. i never did further mechanics, so i dont know what they teach you, but that is the physics-degree-level definition of the moment of inertia (symplified slightly - i.e. in one dimension/no tensors etc....)
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