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Analytical Mech Homework Solutions 152

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

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2 2 2 sinh cosh sinh f m l t t F r l t λ ω ω θ ω = = ω 2 2 cosh m l t ω ω = 10.19 (See Example 4.6.1) The equation of constraint is ( ) , 0 f r r a θ = = ( 2 2 2 2 m T r r ) θ = + ± ± cos V mgr θ = r a θ ( ) 2 2 2 cos 2 m L r r mgr θ θ = + ± ± 0 L d L f r dt r r λ + = ± 0 L d L f dt λ θ θ θ + ± = 1 f r = 0 f θ = Thus 2 cos 0 mr mg mr θ θ + ± ±± λ = λ = 2 sin 2 0 mgr mr mrr θ θ θ = ±± ± ± Now , r r so r a = 0 = = ± ±± 2 cos 0 ma mg θ θ + ± 2 sin 0 mga ma θ θ = ±± sin g a θ θ = ±± and d d θ θ θ θ = ± ±± ± so sin g d d a θ θ θ = ± ± θ or 2 cos 2 g g a a θ θ = − + ± hence, ( ) 3cos 2 mg λ θ = and when 0 λ particle falls off hemisphere at 1 2 cos 3 θ = 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
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