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Unformatted text preview: 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 noncartesian 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 t φ ω = r This problem begs for cylindrical coordinates, so that’s what we’ll use. The bead can move along the rod (through a hole in the bead) without friction. Our job is to solve for r(t) for the bead. We will ignore gravity. We can’t use conservation of energy techniques because energy is being supplied to the system from the outside to keep the rod rotating at constant angular velocity. In fact, it isn’t obvious what the force on the bead is. However, we can make some progress by writing down the general equation for acceleration in cylindrical coordinates. From Example 1.8 in the text: ( ) ( ) 2 ˆ ˆ ˆ 2 a r r r r r z φ φ φ φ = − + + + G ¡ ¡ ¡ ¡ ¡¡ ¡ ¡¡ z The motion here is in the xy plane (if you’ll allow me to mix coordinate system terminology) so there’s nothing happening in z and we’ll drop the last term (which means we are really in polar coordinates. Even without knowing too much about the forces, this equation gets us going because we do know that there are no forces along...
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This note was uploaded on 10/06/2011 for the course PHYS 325 taught by Professor Staff during the Fall '08 term at University of Illinois, Urbana Champaign.
 Fall '08
 Staff
 Physics, mechanics

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