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M4 - Equilibrium Stability

Four uniform rods, each of mass m and length 2a are smoothly joined together to form a rhombus ABCD. a light elastic string of modulus 2mg and natural length a connect A and C. the vertex A is smoothly pinned to a fixed support and the system hangs at rest. show that there is a position of stable equilbrium in which the angle BCD is 120 degrees.

thanks
Reply 1
Avatar for Zygroth
Zygroth
14
If V = total potential energy of the system, and GPE = 0 at A, and x = angle CAB (i.e. half angle BCD), then

V=2mg2a(4acos⁡x−a)2−2mg(acos⁡x)−2mg(3acos⁡x)V = \dfrac{2mg}{2a}(4a\cos x - a)^2 - 2mg(a\cos x) -2mg(3a\cos x)

The first bit is the elastic energy in the string, second bit is the GPE of the top 2 rods, and the last bit is the GPE of the bottom 2 rods.

In equilibrium, dV/dx = 0
Reply 2
Avatar for Jsbartlett
Jsbartlett
OP
kk thanks il try it in a bit, i think i was getting confused about the relative rotation of the rhombus, forgot about it rotating to line up with the center of mass and the support

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