Analytical Mech Homework Solutions 152

Analytical Mech Homework Solutions 152 - f 2m 2l 2 sinh t...

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22 2s i n h c o s h sinh f ml t t F rl t λ ωω θω == ω 2 2c o s h t = 10.19 (See Example 4.6.1) The equation of constraint is ( ) ,0 fr r a θ = −= ( 2 2 m Tr r ) =+ ± ± cos Vm g r = r a θ () 2 cos 2 m Lr r m g r =+− ± ± 0 LdL f rd tr r ∂∂ += ∂∂∂ ± 0 f dt θθ −+ ± = 1 f r = 0 f = Thus 2 cos 0 mr mg mr −− + ± ±± = = 2 sin 2 0 mgr mr mrr −− = ± ± Now , rr so ra = 0 ± 2 cos 0 ma mg ± 2 sin 0 mga ma sin g a = and d d = ± ± so sin g dd a = ∫∫ or 2 cos 2 gg aa = ± hence, 3cos 2 mg λθ =− and when 0 particle falls off hemisphere at 1 2 cos 3  =   D 10.20 Let 2 x mark the location of the center of curvature of the movable surface relative to a fixed origin. This point defines the position of m 2 . 1 r marks the position of the particle of mass m 1 . r is the position of the particle relative to the movable center of curvature of m 2 . Kinetic energy 1 1 11
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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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