Physics 325 Lecture 12 - Physics 325 Lecture 23 Calculus of...

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Physics 325 Lecture 23 Calculus of Variations So far in Physics 325 we have followed the procedures that you learned in Physics 111/211 of finding all the forces acting on a system and then solving for the equation of motion using Newton’s 2 nd law. We learned some new things in Special Relativity, but the basic principle for solving for the motion of a particle are the same. Now we are going to learn to address a different set of problems, such as “What is the shortest path between two points?”, or “What path results in the shortest time to go from one point to another?”. Ultimately, through Lagrangian dynamics and Hamilton’s principle, this will lead us back to solving for the equation of motion of a particle, but with new and powerful tools. The basic problem we want to solve is to determine a function y(x) that minimizes (typically) or maximizes the integral () 2 1 ,, x x Jf y y x d = x where yd y d ′ ≡ x . Let’s start with a simple example. Suppose we want to find α that minimizes ( ) 2, 2 2 0,0 ' J x ππ = over paths of the form sin yx x x =+ 0 1 2 3 4 5 6 7 01234567 (2 ,2 ) x y This is a straightforward problem because we can calculate y’ , evaluate the integral and minimize it. Here goes: 2 2 0 2 22 0 ( ) sin( ) ' 1 cos( ) 1c o s ( ) 1 2 cos( ) cos ( ) 2 20 0 2 y xx xy x Jd x x dx x x dJ d π απ πα = + = ++ = + == = So the line y=x is the path that minimizes the integral of the slope squared out of all the choices given by . But is it the minimal path of all possible paths? sin x x
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Example: A particle is moving from point A to point B under the influence of gravity. Its motion is constrained by a rigid wire. What path represents the fastest way from A to B? A B g a b c Path a is the shortest path, c is the longest and b is in between a and c in length. However, the average velocity along c is the largest. So which is the fastest? Let’s first calculate the time it takes for the particle to go from A to B along a and c . If a is tilted at an angle θ with respect to the horizontal, (we’re going to take the horizontal direction as y , and the vertical as x – it’s arbitrary but if we do it the opposite way we will end up with all our x and y s swapped from what the book has) then the force on the particle along the path is sin Fm g = The distance traveled is 22 1 sin 2 sg t xy 2 = =Δ+ Δ Solving for the time, we find () 12 2 sin tx g + Δ y Let’s plug in some numbers so we can easily compare a to c . Let Δ y=12m and
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Physics 325 Lecture 12 - Physics 325 Lecture 23 Calculus of...

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