By Casey J.
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Extra info for A treatise on the analytical geometry
109) in the (realistic) stretched string case; ka = YA, and a 1 = , m µ v2 = give YA . 114) Here Y is the “effective Young’s modulus” in the particular stretched condition. 8 Field Theoretical Treatment and Lagrangian Density It was somewhat artificial to treat a continuous string as a limiting case of a beaded string. The fact is that the string configuration can be better described by a continuous function η ( x, t) rather than by a finite number of discrete generalized coordinates ηi (t). It is then natural to express the kinetic and potential energies by the integrals T= µ 2 ∂η ∂t L 0 2 dx, V= τ 2 L 0 ∂η ∂x 2 dx.
The formulation in the previous section only makes sense for motions per bead small compared to the extension per bead ∆L/N. Stated differently, the instantaneous tension τ must remain small compared to the standing tension τ0 . A solid, on the other hand, resists both stretching and compression and, if there is no standing tension, the previously assumed approximations are invalid. To repair the analysis one has to bring in Young’s modulus Y, in terms of which the length change ∆L of a rod of length L0 and cross sectional area A, subject to tension τ, is given by ∆L = L0 τ/A .
1. The position vectors of three point masses, m1 , m2 , and m3 , are r1 , r2 , and r3 . Express these vectors in terms of the alternative configuration vectors sC , s3 , and s12 shown in the figure. Define “reduced masses” by m12 = m1 + m2 , M = m1 + m2 + m3 , µ12 = m1 m2 , m12 µ3 = m3 m12 . 79) M Calculate the total kinetic energy in terms of s˙ , s˙3 , and s˙ 12 and interpret the result. Defining corresponding partial angular momenta l, l3 , and l12 , show that the total angular momentum of the system is the sum of three analogous terms.