Analytical Mech Homework Solutions 146

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
( ) 2 Fm a m vm r ωω ω ′′ =+ × +×× G GG G G G G G () ( ) ( ) ˆˆ ˆ ˆ ˆ 2 F m ix jy kz m k ix jy kz m k k ix jy kz ˆ =+ + + × + + + + ±± ± ± ± 2 2 x x y x =−− ± 2 2 y y x y ± z = z 10.10 22 2 1 2 Tm r r θ ± ± θ r 2 1 cos 2 Vk r lm g r =− D 2 2 cos mk LTV r r rl m g r =−= + + D ± L mr r = ± ± , dL mr dt r  =   ± 2 cos L mr k r l mg r + D ± 2 cos mr mr k r l mg + D ± 2 L mr = ± ± sin L mgr 2 sin d mr mgr dt ± 10.11 (See Example 4.6.2) 2s i n 2 4 a x 1c o s 2 4 a y at 0 = & 0 xy = 1 2 x y Vm g y = cos2 s 2 x 1 co aa θθ ± = + ± sin 2 2 a y = ± ± 2 2 o s 2 s i n2 o s 2 84 ma mga  + +  ± [] 1 2cos2 1 1 ma mga + ± cos sin ma mga   ± where we used the trigonometric identies … 2 2cos 1 and 2 2sin 1 8
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online