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Unformatted text preview: Solutions to Homework Set 2 1) Bent bars : First some elementary calculus: the curvature κ , and the radius R of the osculating circle, at a point ( x, y ) on a curve y ( x ) are given by κ = 1 R = dψ ds = 1 √ 1 + y prime 2 d tan 1 y prime dx = y primeprime (1 + y prime 2 ) 3 / 2 . Here the arc length s and the angle ψ between the curve and the x axis constitute the intrinisic coordinates of the curve. ψ y x R Osculating circle and intrinsic coordinates. For all parts of the present problem y prime is assumed small enough that we can get away with the approximation R 1 = y primeprime . a) Euler’s problem : We observe that if a rod of fixed length L is bent into a curve y ( z ), we have ds 2 = dz 2 + dy 2 , ⇒ dz = radicalBig 1 ˙ y 2 ds = parenleftbigg 1 1 2 ˙ y 2 + ··· parenrightbigg ds. Here the overdot indicates differentiation with respect to the arclength s . For curves with small y prime , we may approximate y prime = ˙ y and ds ≈ dz in the last expression. The length of the projection of the rod on the z axis is therefore L integraldisplay L 1 2 y prime 2 dz. Combining the resulting expresion for the potential energy of the load with the bending energy gives the functional U [ y ]. Inserting the mode expansion for y ( z ) and doing the integral gives U [ y ] = L 2 ∞ summationdisplay n =1 a 2 n braceleftBigg Y I 2 parenleftBigg n 4 π 4 L 4 parenrightBigg Mg 2 parenleftBigg n 2 π 2 L 2 parenrightBiggbracerightBigg ....
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This homework help was uploaded on 01/29/2008 for the course PHYS 598 taught by Professor Stone during the Fall '07 term at University of Illinois at Urbana–Champaign.
 Fall '07
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