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solutionhw2 - Solutions to Homework Set 2 1) Bent bars :...

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Unformatted text preview: Solutions to Homework Set 2 1) Bent bars : First some elementary calculus: the curvature , and the radius R of the osculating circle, at a point ( x, y ) on a curve y ( x ) are given by = 1 R = d 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 between the curve and the x axis constitute the intrinisic coordinates of the curve. 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) Eulers 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 arc-length 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 4 L 4 parenrightBigg- Mg 2 parenleftBigg n 2 2 L 2 parenrightBiggbracerightBigg ....
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solutionhw2 - Solutions to Homework Set 2 1) Bent bars :...

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