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 φα = ± Say constant 2 sin mr == ± A 2 3 cos mr mg mr α =− A ±± 2 1 2 dd r d rrr dt dr dr === ± ± ± ± r 2 2 3 cos 2 md r dr m g d r mr A ± 22 2 cos mr mgr C mr + ± 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 Ur m g r mr =+ A For , A , and turning points occur at 0 φ ± 0 0 r = ± Then 2 2 0c o 2 mgr C mr + A s 2 32 cos 0 2 mg r Cr m −+= 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 cos 0 2 l mg r Cr m , defines the turning points. For the particle to remain on a single horizontal circle, there
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