Analytical Mech Homework Solutions 38

Analytical Mech Homework Solutions 38 - sin = gb v2 cos 2 =...

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2 sin gb v θ =− D 42 22 4 cos 1 sin vg b v θθ = D D 2 242 2 2 2 max 2 gb v g b gb v h vv =+ DD D 2 g Measured from the ground, max 2 gb v hb + D D The mud leaves the wheel at 1 2 sin gb v    D 4.8 cos xR φ = and ( ) cos x x vt v t α == so co R t s cos v = D sin yR = and () 11 sin y v tg tv = y t α φ 2 cos 1 cos sin sin cos 2 cos RR Rv g φα D 2 cos sin tan cos 2c o s gR v D 2 2 cos 2 cos tan cos sin sin cos cos sin cos cos R gg αα φφ From Appendix B, sin sin cos cos sin φθ += + 2 2 o s sin cos v R g D R is a maximum for () ( 2 2 2 0s i n s i n c o s c o s cos dR v dg ) αφ − + D Implies that ( ) cos sin sin 0 −− −= cos From appendix B, cos cos cos sin sin so cos 2 0 2 2 π = + 2 max 2 2 cos sin cos 4 2 4 2 v R g φπ D
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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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