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Analytical Mech Homework Solutions 157

# Analytical Mech Homework Solutions 157 - d mr 2 sin = 0 dt...

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( ) 2 sin 0 d mr dt φ α = ± Say constant 2 sin mr φ α = = ± A 2 3 cos mr mg mr α = A ±± 2 1 2 d dr d r r r dt dr dr = = = ± ±± ± ± ± r ( ) 2 2 3 cos 2 m dr d r mg dr m r α = A ± 2 2 2 cos 2 2 mr mgr C mr α = − + ± A The constant of integration C is the total energy of the particle: kinetic energy due to the component of motion in the radial direction, kinetic energy due to the component of motion in the angular direction, and the potential energy. ( ) 2 2 cos 2 U r mgr mr α = + A For , A , and turning points occur at 0 φ ± 0 0 r = ± Then 2 2 0 co 2 mgr C mr α = − + A s ( ) 2 3 2 cos 0 2 mg r Cr m α + = A The above equation is quadratic in r ( 2 r 4 A ) and has two roots. 10.24 Note that the relation obtained in Problem 10.23, ( ) 2 3 2 cos 0 2 l mg r Cr m α + = , defines the turning points. For the particle to remain on a single horizontal circle, there
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