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Analytical Mech Homework Solutions 146

Analytical Mech Homework Solutions 146 - F = ma 2m v m r jy...

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( ) 2 F ma m v m r ω ω ω = + × + × × G G G G G G G G ( ) ( ) ( ) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 2 F m ix jy kz m k ix jy kz m k k ix jy kz ω ω ω ˆ = + + + × + + + × × + + ±± ±± ± ± ±± ± ( ) 2 2 x F m x y x ω ω = ±± ± ( ) 2 2 y F m y x y ω ω = ±± ± z F m = ±± z 10.10 ( ) 2 2 2 1 2 T m r r θ = + ± ± θ r ( ) 2 1 cos 2 V k r l mgr θ = D ( ) ( ) 2 2 2 2 cos 2 2 m k L T V r r r l mgr θ θ = = + + D ± L mr r = ± ± , d L mr dt r = ±± ± ( ) 2 cos L mr k r l mg r θ θ = + D ± ( ) 2 cos mr mr k r l mg θ = + D ±± ± 2 L mr θ θ = ± ± sin L mgr θ θ = − ( ) 2 sin d mr mgr dt θ θ = − ± 10.11 (See Example 4.6.2) ( ) 2 sin 2 4 a x θ θ = + ( ) 1 cos2 4 a y θ = at 0 θ = & 0 x y = ( ) 2 2 1 2 T m x y = + ± ± V mgy = ( ) ( ) cos2 s2 x 1 co 2 2 a a θ θ θ θ = + ± ± ± θ = + ± sin 2 2 a y θ θ = ± ± ( ) ( ) 2 2 2 2 1 cos2 sin 2 1 cos2 8 4 ma mga
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