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mwrigh58
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Homework Statement
This is for a third year Nonlinear ODE and Chaos course, but the question at hand is a first year physics question. If this post belongs somewhere else, please let me know.
"Consider the bead on a tilted wire, described below. Find the (equilibrium) equation from which fixed points can be found. Show that the equilibrium equation can be written in dimensionless form as
1 - [tex]\frac{h}{u}[/tex] = [tex]\frac{R}{\sqrt{1 + u^2}}[/tex]
for appropriate choice of R, h, and u."
The scenario (see attached image, if possible):
- a box with two sides and a bottom.
- a (stiff) wire is attached to either side of the box, attached higher on the left side than the right, at an angle [tex]\theta[/tex] to the horizontal.
- a bead of mass m is on the wire.
- a massless spring (with spring constant k) is attached to the bead and the bottom of the box (at a location to the left of the bead). [Note: I have called the length of this distance c to simplify equations/explanations, but c is not a given variable.]
- an imaginary line from a point on the wire (to the left of the bead) to where the spring is fixed to the bottom of the box has length a.
- the vector from where a and the wire intersect to the bead is x. x is positive to the right (oriented along the wire).
- the acceleration of the bead due to gravity is g.
Other variables I have defined:
- [tex]\gamma[/tex] is the angle between x and c (inside the triangle formed by x, c, and a).
Homework Equations
Final form:
1 - [tex]\frac{h}{u}[/tex] = [tex]\frac{R}{\sqrt{1 + u^2}}[/tex]
Force of gravity:
Fg = m g
Force due to stretched/compressed spring:
Fs = k [tex]\Delta[/tex]x
The Attempt at a Solution
Coordinate system:
- x-direction along the wire, positive to the right
- y-direction perpendicular to the wire, positive upwards
Assumptions:
- a is the length of the spring when it is neither stretched nor compressed.
- triangle a c x is a right angle triangle (the angle between a and x is a right angle). If I do not assume this, the equations get really messy and do not look promising.
Constraints:
- the wire is stiff, therefore the net force in the y-direction must be zero.
Equations for net force:
Fnet, x = Fg sin[tex]\theta[/tex] - Fs cos[tex]\gamma[/tex]
Fnet, y = Fnormal - Fg cos[tex]\theta[/tex] -Fs sin[tex]\gamma[/tex]
Looking for equilibrium positions, therefore, looking for the places where Fnet, x = 0. i.e.:
0 = Fg sin[tex]\theta[/tex] - Fs cos[tex]\gamma[/tex]
Substitute in the relevant equations:
0 = m g sin[tex]\theta[/tex] - k (c - a) cos[tex]\gamma[/tex]
From assumption triangle a c x is a right triangle (c = [tex]\sqrt{a^2 + x^2}[/tex] and cos[tex]\gamma[/tex] = [tex]\frac{x}{c}[/tex]):
0 = m g sin[tex]\theta[/tex] - k ([tex]\sqrt{a^2 + x^2}[/tex] - a) [tex]\frac{x}{c}[/tex]
0 = m g sin[tex]\theta[/tex] - k x ([tex]\sqrt{1 + (\frac{x}{a})^2}[/tex] - 1) [tex]\frac{1}{\sqrt{1 + (\frac{x}{a})^{2}}}[/tex]
0 = [tex]\frac{m g sin\theta}{k x}[/tex] - 1 + [tex]\frac{1}{\sqrt{1 + (\frac{x}{a})^{2}}}[/tex]
1 - [tex]\frac{m g sin\theta}{k x}[/tex] = [tex]\frac{1}{\sqrt{1 + (\frac{x}{a})^{2}}}[/tex]
If you let u = [tex]\frac{x}{a}[/tex] and h = [tex]\frac{m g sin\theta}{k a}[/tex], then the equation of equilibrium is:
1 - [tex]\frac{h}{u}[/tex] = [tex]\frac{1}{\sqrt{1 + u^2}}[/tex]
As you can see, this is the form I am supposed to have, where R = 1, but for later parts of the problem I need to vary R. Can anyone see where this R comes in or what it is (physically)?
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