Analytical Mech Homework Solutions 156

Analytical Mech Homework Solutions 156 - ( ) m ( r + 2r r...

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() 22 2 sin r mr r r F φθ θ −− ±± = 2 2s i n c o s mr r r F θθ φ +− = ± ± ± sin 2 sin 2 cos r r F ++ ± ± ± ± = The equations agree. 10.22 For a central field, ( ) r = VV , so 0 θφ = = ∂∂ and V F r = − . For spherical coordinates, using equation 1.12.12: 222 2 sin vrr r 2 =+ + ± 2 2 sin 2 m LTV r r r Vr =−= + + ± L mr r = ± ± , dL mr dt r  =   ± , 2 sin LV mr mr rr 2 sin r mr mr mr F = 2 L mr = ± ± , 2 2 mr mrr dt ± ± ± , sin cos L mr = ± 2 i n c o mr mrr mr ± ± ± s 0 = sin L mr = ± ± , 2 2 sin 2 sin 2 sin cos mr mrr mr dt =+ + ± ± ± 0 L = 2 2 sin 2 sin 2 sin cos 0 mr mrr mr + += ± ± 10.23 Since α == constant, there are two degrees of freedom, and r . , vr r = ± sin = ± 2 2 11 sin Tm v m r r == + ± ± cos Vm g r = sin cos 2 m m g r αα + ± ± L mr r = ± ± , mr dt r = ± , sin cos L mr mg r =− ± sin cos mr mr mg ± sin L mr = ± ± , 0 L = 18
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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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