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Erich Poppitz

Lecture 22: Energy Conservation in 3D, Cen- tral Forces ▯ relevant text sections: Morin 5.3, 7.1-7.2 Conservative 3D Forces ▯ What 3D forces are conservative? MUST only be functions of position: F(~r) but must also meet the requirement that: r ▯ F = 0 (1) i.e. that the CURL of the function is 0. In order to prove this, you need to use Stoke’s Theorem (which you will learn in multivariable calculus near the end of the course). I will assume you haven’t seen this yet and therefore you are not responsible for proving the above. But you are responsible for being able to use Equation (1) to test whether a force is conservative. You can read about the proof in the text if you like. ▯ Since some of you haven’t seen the \Curl" before, its just a cross prod- uct of the gradient operator and a function. I’ve provided the formula in the supplementary notes and Deepak will cover some examples in next Monday’s lecture. Also check out Appendix B.6. ▯ It turns out that one class of 3D forces: \Central Forces" are ALWAYS conservative. A central force is de▯ned to be a force that points radially and whose magnitude depends only on the distance from the origin (i.e. on the magnitude jrj = r). This means you can write a central force in polar coordinates as: F = F(r)r^ (2) Notice that there is no ▯ component to the force and the force only depends on the scalar r. You can prove that all central forces are ~ conservative by showing that r ▯ F = 0 for all central forces. This is done on pg 151 of Morin, but I’ve done a more detailed version of this in the supplementary notes so please look at that. 1 ▯ The fact that central forces are all conservative is a very important result because it says we can de▯ne potential energy functions for any central force. Some important ones you might remember: GRAVITY FORCE, ELECTRIC FORCE. Basically, many important forces have this form and we can de▯ne potentials for all of them and therefore study their energy equation fairly simply. ▯ We will not be covering sections 5.4-5.8 on gravity and momentum conservation in this course so you are not responsible for the types of questions in these sections for the exam (e.g. ▯nding the gravitational force for di▯erent mass distributions, using momentum conservation to solve problems, collision problems). Angular Momentum Conservation with Central Forces ▯ Since central forces are conservative, we can write them in terms of a potential V (r) with: F = F(r)r ^ = ▯rV (r) = ▯ dV ^ dr dV (r) so taking Rhr r 0omp0nent gives: F(r) = ▯ dr and integrating: V (r) = ▯ r0dr F(r )+V . 0ere the gradient has a component only in the radial direction. ▯ Central forces have special properties. The ▯rst is that the angular momentum L of a particle in a central force ▯eld is conserved. If we de▯ne the linear momentum p ~ = m~ v, then the angular momentum is ~ L = ~r ▯ ~: 2 z L out of page p r y x ▯ The vector L is perpendicular to the instantaneous plane of motion of the ▯eld, which contains r and ~. In the ▯gure,r and ~ are in the y ▯z plane and L points in the x direction, using the right hand rule. ▯ In a central ▯eld, the angular momentum of a particle is conserved. We can show this by taking the time derivative of angular momentum and showing that it is 0: _ _ _ ~ L = ~r ▯ ~ + r ▯ ~ = v ▯ (m~v) + r ▯ F = 0: The ▯rst term ~v ▯ (m~v) is zero by de▯nition of the cross product, and the second term ~r ▯ F is zero because F points in the ~ r direction. This tells us that the magnitude and the orientation of L are constant following the particle in a central ▯eld. ▯ Conservation of angular momentum provides a big constraint on the motion. One thing it means is that a particle in a central ▯eld remains in the plane of motion, because the DIRECTION of the angular mo- mentum vector cannot change. Since the angular momentum vector points perpendicular to the plane of motion, this means the plane of motion can’t change. Energy Equation and E▯ective Potential ▯ Since we have a central force, it turns out to be easiest to work in polar coordinates to derive an energy equation. It also turns out we can derive an energy equation that only involves r and r _ (so no ▯ stu▯). This means it e▯ectively becomes a 1D equation. Here is how: 3 1. Start with our de▯nition for mechanical energy: 1 E = K + V = mv + V (r) 2 2. Express v in terms of the derivative of the position coordinate. Recall that position in polar
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