Physics 325 Lecture 6 - Physics 325 Lecture 6 Finally the...

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Physics 325 Lecture 6 Finally, the most general case is: f) ( ) , , F F r v t = G G G G In general, we can’t solve this analytically. But there are some special cases which can be. If G can be factorized such that the variables separate then things simplify considerably. We work in one dimension for simplicity. Suppose F ( ) ( ) ( , F v t f v g t = ) . Then ( ) ( ) ( ) ( ) dv dv f v g t m g t dt m dt f v = = And with a little luck, these integrals can be done. Similarly, suppose ( ) ( ) ( ) , F x v h x f v = . Then ( ) ( ) dv h x f v m dt = Invoking again the chain rule dv dx dv dv v dt dt dx dx = = Gives ( ) ( ) ( ) ( ) dv vdv h x f v mv h x dx m dx f v = = The integration yields ( ) v x , and x(t) is then found from dx dt = ( ) dx dt v x = Finally, the last variation on this, ( ) ( ) ( ) , F x t h x g t = , is not solvable analytically unless either h or g is a constant. Instead, numerical techniques must be applied. Next I want to illustrate another technique for solving equation of motion problems other than the integration or energy techniques we’ve just used a few times. This problem will also allow us to exercise our tools in a non-cartesian coordinate system. Bead on a Rod Example (courtesy Jim Wiss) Consider a bead on a thin rod that is rotating around a frictionless pivot such that it has a constant angular velocity: t φ ω = r
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