s11f1032

s11f1032 - 2 A B 1.5 1 v[m/s 0.5 0-0.5-1-1.5-2-2-1.5-1-0.5...

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Unformatted text preview: 2 A B 1.5 1 v [m/s] 0.5 0 -0.5 -1 -1.5 -2 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 θ [radiant] PHY 4936 – Solution 32 The energy is E= 1 ˙ 3 − 8 cos θ + M gR 1 − 4 cos θ . M R2 θ 2 2 2 9π 9π ˙ Solving for v 2 = R2 θ2 gives ˙ v 2 = R2 θ 2 = 2 [E − M gR (1 − 4 cos θ/(9π )] M [3/2 − 8 cos θ/(9π )] A. E = M gR: v2 = 8gR cos θ/(9π ) , 3/2 − 8 cos θ/(9π ) √ v = ± v2 . (1) √ B. E = M gR [1 − 4/(9π 2)]: v 2 √ 8gR [cos θ/(9π ) − 1/(9π 2)] = , 3/2 − 8 cos θ/(9π ) √ v = ± v2 . (2) With g = 9.81 [m/s] and R = 1 [m] plots of v (θ) for (1) and (2) are given in the figure. ...
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This note was uploaded on 12/11/2011 for the course PHY 4936 taught by Professor Berg during the Fall '11 term at FSU.

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