410ahw1soln - PHYS 410 Classical Mechanics, Fall 2011,...

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Unformatted text preview: PHYS 410 Classical Mechanics, Fall 2011, University of Maryland, College Park Instructor: Prof. T Jacobson TA: Yuchen Peng 1.33 The total force on loop 2 from loop 1 is We have: if are vectors, then . Let be and , separately, then the integral above becomes: The first integral is just , and the term vanishes when we integrate around loop 1. So Exchange the subscript index 1 with 2 to get the expression of , then nothing changes except that , and you get 4.4 The angular momentum of the ball is conserved in this problem. (a) , so (b) The force to pull the string equals to the centripetal force, i.e. , in which . The total work (c) The change in KE We have 4.24 (a) By analyzing symmetry we know that the force only has x-component. , and the direction is toward the rob. (d) The potential . In which is the zero potential point 4.36 (a) (finally we drop the constant Mgl, which is uninteresting) (b) Let . There will be an equilibrium point at as long as ....
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This note was uploaded on 12/29/2011 for the course PHYSICS 410 taught by Professor Jacobson during the Fall '11 term at Maryland.

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410ahw1soln - PHYS 410 Classical Mechanics, Fall 2011,...

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