HW10_prob2_GPS15_G4

HW10_prob2_GPS15_G4 - Problem Goldstein 5-4 (3rd ed. 5.15)...

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Problem Goldstein 5-4 (3 rd ed. 5.15) You can save a lot of e f ort by orienting the triangle in a way that ef- fectively exploits its symmetry. My preferred choice is shown in the f gure. 0 1 Y -1 1 X In this f gure, the x and y variables are normalized by the factor h = a/ 2 , so that X = x/h and Y = y/h ,whe re h is the height of the triangle and a is thelengthofas ide . Inth isor ientat ion, there are only three non-zero elements of the inertia tensor, the diagonal elements. All of the o f -diagonal elements are zero; the xy and yx components vanish by symmetry; the xz , zx , yz ,and zy components vanish because the plate is in the z =0 plane. Of course, the same results would also be true if you oriented the triangle on its base. To see explicitly that the o f -diagonal elements vanish, start with the general expression for the αβ component of the inertia tensor I αβ = Z ρ ( r )( r 2 δ αβ x α x β ) d 3 r. (1) For the body in question, we can write the density as ρ ( r )=( M/h 2 ) δ ( z )= σδ ( z ) , (2) where M is the mass of the plate and a 2 / 2= h 2 is its area, so σ is just the mass per unit area of the plate. The delta function δ ( z ) localizes all of the mass to the x - y plane. The key to evaluating Eq.(1) is to get the correct integration
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HW10_prob2_GPS15_G4 - Problem Goldstein 5-4 (3rd ed. 5.15)...

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