Probs5 - Physics 316, Problem Set 5, due Tuesday November 1...

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Unformatted text preview: Physics 316, Problem Set 5, due Tuesday November 1 in class 1. Noethers theorem for a snakey rope A flexible rope of fixed length L has a mass-per-unit-length l , and negligible bending energy. It is free to move in space with no gravity. It is charged along its length with uniform linear density so that it repels itelf. Its configuration in space is labeled by the function r ( u ), where u is the position along the rope. a) What is the Lagrangian of this system? b) Now suppose that the rope has the form of a uniform loop of an arbitrary shape. Use Noethers theorem to find a functional of r ( u ) and r ( u ) that is a constant of the motion. Solution: a)The Lagrangian is (in SI units): L = Z L 1 2 l [ ~ r ( u )] 2 du + 1 2 Z du 1 Z du 2 2 (4 )- 1 | r ( u 1 )- r ( u 2 ) |- 1 . b) We can easily get the conservation of energy, angular momentum, etc. But we are most interested in the conserved quantity that cant be easily guessed but can be obtained by Noethers theorem. Since the rope forms a loop, the Lagrangian is unchanged by the mapping: h s ( ~ r ( u )) = ~ r ( u + s ). By Noethers theorem, we have a corresponding conserved quantity. But we have to be careful here because the form of Noethers theorem in the textbook is for finite number of local coordinates q and what we have now is an infinite number of local coordinates ~ r ( u ) on the loop. A way to do it is to divide the loop into a finite number of parts and use the Noethers theorem on them and then take the limit. So L N = N n =0 1 2 l [ ~ r ( u n )] 2 du and we get the conserved quantity I N : I N = L ~ r dh s ( ~ r ) ds s =0 = N X n =0 L ~ r ( u n ) dh s ( ~ r ( u n )) ds s =0 = N X n =0 [ l ~ r ( u n ) du ] d~ r ( u ) du u = u n . Taking the limit, we get the conserved quantity I as: I = lim N I N = Z L l ~ r ( u ) d~ r ( u ) du du. 2. Circular meniscus A vertical glass rod of radius b is extends upward from a large dish of water. The angle between the glass surface and the water surface at the circle of contact is a material constant called ....
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Probs5 - Physics 316, Problem Set 5, due Tuesday November 1...

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