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s11f1022

# s11f1022 - x 1 2 = 1 ± √ 1 1 = 1 ± √ 2(7 of which...

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Solution for assignment 22: Turning points of the spherical pendulum for a special case. The Energy is E = m 2 R 2 ˙ θ 2 + m 2 R 2 sin 2 ( θ ) ˙ φ 2 + U 0 cos( θ ) (1) with U 0 given by U 0 = mg R = E/ 2. The angular momentum M z = mR 2 sin 2 ( θ ) ˙ φ (2) is conserved. Substituting ˙ φ 2 = M 2 z / ( m 2 R 4 sin 4 θ ) the energy becomes E = m 2 R 2 ˙ θ 2 + M 2 z 2 mR 2 sin 2 θ + U 0 cos( θ ) (3) with M 2 z / (2 mR 2 ) = E . Turning points are then given by the solutions of 0 = m 2 R 2 ˙ θ 2 = E - E sin 2 θ - E 2 cos θ (4) 0 = sin 2 ( θ ) - 1 - 1 2 cos θ sin 2 θ = cos 2 ( θ ) - 1 2 cos θ + 1 2 cos 3 θ . (5) With x = cos θ 0 = + x ± x 2 - 2 x - 1 ² (6) with the solutions x 0 = 0 θ 0 = π/ 2 and
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Unformatted text preview: x 1 , 2 = 1 ± √ 1 + 1 = 1 ± √ 2 (7) of which only x 2 = 1-√ 2 is in the physical range of cos θ . So, we ﬁnd the turning points θ min = cos-1 (0) = π 2 = 1 . 5707963 ... , (8) θ max = cos-1 (1-√ 2) = 1 . 9978749 ... . (9) As an addtional task a plot of f ( θ ) = E sin 2 θ + E 2 cos θ-E (10) for, e.g., E = 1 is instructive. 1...
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