HW1_prob3_GPS03_G06

HW1_prob3_GPS03_G06 - dt = r 1 p 1 + r 2 p 2 = r 1 F ( e )...

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Problem 1.6 (3 rd ed. 1.3) For a system of two particles, we are given that the overall equations of motion (EOM), M d 2 R dt 2 = F ( e ) = F ( e ) 1 + F ( e ) 2 , (1) and d L dt = N ( e ) = r 1 × F ( e ) 1 + r 2 × F ( e ) 2 , (2) are valid. The individual EOM are m 1 d 2 r 1 dt 2 = F ( e ) 1 + F 21 , (3) and m 2 d 2 r 2 dt 2 = F ( e ) 2 + F 12 . (4) Now, simply add together Eqs.(3) and (4) to obtain m 1 d 2 r 1 dt 2 + m 2 d 2 r 2 dt 2 = F ( e ) 1 + F ( e ) 2 + F 12 + F 21 . (5) It follows from the de f nition of the center of mass and from Eqs.(1) and (5) that F 12 + F 21 = 0 , (6) which is the weak law of action and reaction for a pair of particles. From the de f nition of the total angular momentum L = r 1 × p 1 + r 2 × p 2 , it follows that d L
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Unformatted text preview: dt = r 1 p 1 + r 2 p 2 = r 1 F ( e ) 1 + r 2 F ( e ) 2 + r 1 F 21 + r 2 F 12 . (7) Since Eq.(2) is satis f ed, it follows that r 1 F 21 + r 2 F 12 = , (8) which can be rewritten as ( r 1 r 2 ) F 21 = , (9) with the help of Eq.(6). Equation (9) implies that F 21 lies strictly in the direc-tion of the vector ( r 1 r 2 ) , which is the strong law of action and reaction. 1...
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