hw2.solns - Physics 318: Solutions to Homework #2 February...

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Physics 318: Solutions to Homework #2 February 11, 2008 1 M = N s n =1 m n R ( t ) = 1 M N s n =1 m n r n ( t ) r n ( t ) = r n ( t ) R ( t ) (1) a.) Simplifying: T = 1 2 M ˙ R 2 + 1 2 N s i =1 m i ˙r 2 i (2) = 1 2 M ˙ R 2 + 1 2 N s i =1 m i ( ˙r 2 i 2 ˙r i · ˙ R + ˙ R 2 ) (3) = 1 2 M ˙ R 2 + N s i =1 1 2 m i ˙r 2 i 1 M N s i,j =1 m i m j ˙r i · ˙r j + 1 2 M ˙ R 2 (4) = 1 2 M ˙ R 2 + N s i =1 1 2 m i ˙r 2 i M ˙ R 2 + 1 2 M ˙ R 2 (5) = N s i =1 1 2 m i ˙r 2 i (6) b.) Simplifying: L = M R × ˙ R + N s i =1 m i r i × ˙r i (7) = M R × ˙ R + N s i =1 m i ( r i × ˙r i R × ˙r i r i × ˙ R + R × ˙ R ) (8) = M R × ˙ R + N s i =1 m i r i × ˙r i 1 M N s i,j =1 ( m i m j r j × ˙r i + m i m j r i × ˙r j ) + M R × ˙ R (9) = M R × ˙ R + N s i =1 m i r i × ˙r i M R × ˙ R M R × ˙ R + M R × ˙ R (10) = N s i =1 m i r i × ˙r i (11) 1
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c.) The strong form of Newton’s third law can be written F nm = F mn = f nm ( | r n r m | )( r n r m ) (12) Let’s see if the Frst term of Eqn.(2) is conserved (time independent). d dt ( 1 2 M ˙ R 2 ) = M ˙ R · ¨ R (13) With ¨ R = 1 M N s i =1 m i ¨ r i (14) = 1 M N s i =1 N s j =1 F ij = 1 M N s i =1 N s j =1 F ji = 1 M N s j =1 N s i =1 F ji (15) ¨ R = 0 (16) d dt ( 1 2 M ˙ R 2 ) = 0 (17) We have used the fact that a (symmetric) sum over antisymmetric indices must be zero. Now let’s check the Frst term in Eqn.(7). d dt ( M R × ˙ R ) = M ˙ R × ˙ R + M R × ¨ R (18) = M R × ( 1 M N s n =1 m n ¨ r n ) (19) = R × N s n =1 N s m =1 F nm (20) = 0 (21) (Again by symmetry). We will now show that the second term is conserved by showing that L is conserved and using the previous result. d dt L = d dt ( N s n =1 m n r n × ˙r n ) (22) = N s n =1 m n ˙r n × ˙r n + N s n =1 m n r n × ¨ r n (23) = N s n =1 r n × N s m =1 F nm (24) = N s n =1 r n × N s m =1 f nm ( r n r m ) (25) = N s n =1 N s
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hw2.solns - Physics 318: Solutions to Homework #2 February...

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