Analytical Mech Homework Solutions 111

# Analytical Mech Homework Solutions 111 - 9 1 2 2 2 2 2 1...

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9 () 1 2 2 22 21 11 1 53 2 1 Ma Mga αβ α β π ααβ    −+ + −      =  −−  D T (a) 2 2 1 1 1 12 1 T T =≈ D to 1 st order in (b) 10 mg = 1 M kg = 1.27 am = 5 bc m = 0.01 = 0.0394 = 1 1 0.9992 12 T ≈− = D T (actually 0.9994 using complete expression) 8.10 The period of the “seconds” pendulum is 2 I Mgl == Ts The period of the modified pendulum is 20 In Mgl n ′′ T where I , M , l refer to parameters of pendulum with m attached and is the number of seconds in a day. n ( 24 60 60 =×× ) 2 m I Im l =+ m M lm l l M + ′ = where l is the distance of the attached mass from the pivot point. m m So 2 2 20 1 m m 2 I ml I nM g lM l m l ππ + −== + g
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## This note was uploaded on 01/06/2012 for the course PHYSICS 4360C taught by Professor Johnanderson during the Fall '11 term at UNF.

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