Analytical Mech Homework Solutions 58

# Analytical Mech Homework Solutions 58 - da = r r r 2 r 2 r...

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22 rot da rrr r dt ωω ω  ′′′ ′ =+×+×+ ×+ ×   G G G G G GG ±±± ±± ± ± ± ± r G () ( ) ( ) rrr ′′′ × +× × +× × G GG G ± ( ) 2 ar r r r ′′ ×=×+× × + × × +× × × GG GG G GG G G GG ± ± G G Now is to r × and r . Let this define a direction : ˆ n GG GG ˆ rr ×= × G G n Since ˆ n , is in the plane defined by ( r ×× G ) G and r G and G G ˆ rn r ×× =× ×= × G G G G GGG G . r Since r ⊥× × 2 ××× = ×   G And is in the direction ( r ) ˆ n G Thus ( ) 2 ××× = × G G ( ) ( ) 2 2 r r ×=×+× × + × × − × ± ± GG G GG r ( ) 33 r r r r ωωω =+×+ ×+ ×+× × ± ± ± ( ) 2 23 +××+××− × r 5.16 With xy , and 0 zxy ′′′′′ ===== DDDDD zv = D ± D Equations 5.4.15a – 5.4.15c become: 32 1 cos cos 3 xt g t tv λω λ =− D 0 yt = 2 1 2 zt g t v t + D When the bullet hits the ground ( ) 0 = , so 2 v t g = D 84 1 cos cos 3 vv xg v gg DD D 3 2 4 cos 3 v x g ′=− D x is negative and therefore is the distance the bullet lands to the west . 5.17 With
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