Analytical Mech Homework Solutions 40

Analytical Mech Homework Solutions 40 - 2sin 2 = 2 gh = 1 2...

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2 22 2 00 2 0 1 2 2sin 1 1 csc 1 gh gh gh vv v α  == − + =   + Finally … 2 2 0 2 1 gh v =+ csc (b) Solving for R max max sin 1 1 2 csc hh h R ϕ αα = −− Substituting for 2 csc and solving … 2 0 max 2 0 1 v gh R gv 4.10 We can again use the results of problem 4.8. The maximum slope range from problem 4.8 is given by … () 2 0 max 1s i n s i n v h R g + Solving for sin sin 1 gh gh =− Thus … max max cos cos sin xR h We can calculate cos from the above relation for sin 1 1 2 2 2 cos 1 sin 1 2 1 gh gh ϕϕ = − Inserting the results for sin and cos into the above … 1 2 2 0 max 2 0 cos 12 sin v gh xh
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