Parabola and Springs - Unequal Masses
Two unequal massive beads are positioned on a smooth wire in the shape of the curve . The more massive bead (with mass ) is to the left of the axis, and the less massive bead (with mass ) is to the right of the axis.
A spring with force constant and natural length has one end attached to each bead. There is an ambient gravitational acceleration in the negative direction.
When the system is in stasis, how far from the axis is the more massive bead?
Details and Assumptions:
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All quantities in standard SI units
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Give your answer as a positive number
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Drawing not necessarily to-scale
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The answer is 0.16786.
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1 solution
If the lighter bead has coordinates ( x , x 2 ) , while the heavier bead has coordinates ( − u , u 2 ) , then the length of the spring is ( x + u ) 1 + ( x − u ) 2 . Thus the potential energy of the system is V = m g x 2 + M g u 2 + 2 1 k [ ( x + u ) 1 + ( x − u ) 2 − L 0 ] 2 = 1 0 x 2 + 2 0 u 2 + 1 5 [ ( x + u ) 1 + ( x − u ) 2 − 1 ] 2 The system will be in equilibrium when the spring is in compression and V is at a global minimum. Solving the equations ∇ V = 0 numerically, we see there is a unique solution x = 0 . 5 2 1 1 1 7 and u = 0 . 1 6 7 8 5 6 .