This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Phys 263/Amath 261 Assignment 6 SOLUTIONS 1. (a) To set up this problem, use the figure, where the mass m is a distance r from the center of the ring. Then the potential is given by d Φ = G dM b = Gaρ b dφ (1) where b is the distance between dM (on the ring) and m (the dotted green line in the figure), and ρ = M/ 2 πa . Let r and r be the position vectors to dM and m , as illustrated: b =  r r  =  a cos φ ˆ x + a sin φ ˆ y r ˆ x  (2) = ( a cos φ r )ˆ x + a sin φ ˆ y  (3) = [( a cos φ r ) 2 + a 2 sin 2 φ ] 1 / 2 (4) = ( a 2 + r 2 2 ar cos φ ) 1 / 2 = a " 1 + r a 2 + 2 r a cos φ # 1 / 2 (5) Integrating gives Φ( r ) = ρaG Z dφ b (6) = ρG Z 2 π dφ q 1 + ( r a ) 2 2 r a cos φ (7) as required. (b) We are asked to expand the denominator: " 1 + r a 2 + 2 r a cos φ # 1 / 2 = 1 1 2 " r a 2 + 2 r a cos φ # (8) + 3 8 " r a 2 + 2 r a cos φ # + · · · (9) = 1 + r a cos φ + 1 2 r a 2 (3 cos 2 φ 1) + · · · (c) Now we can integrate the expression: Φ(...
View
Full
Document
This note was uploaded on 09/20/2010 for the course AMATH 261 taught by Professor Rogermelko during the Spring '10 term at Waterloo.
 Spring '10
 RogerMelko

Click to edit the document details