The figure is in the top right corner Problem 1 A bead of mass m is constrained

Question

The figure is in the top right corner

Problem 1 A bead of mass m is constrained to slide (subject to a drag force) along a straight wire inclined at an angle with respect to the horizontal. The mass is attached to a spring of stiffness k and relaxed length Lo and also acted on by gravity As shown in the figure, we choose to be the position of the mass along the wire, sothat the. 0 occurs at the point cloost to thesupport point of the spring; let a be the distance between this support point and the wire. Note that the nature and number of equilibrium points for the mass depends on whether a is greater or smaller than Lo. In other words, the equilibria undergo a bifurcation as a is varied. wire , (a) Show that the equilibrium positions of the bead satisfy (b) Show that this equilibrium equation can be written in dinmensionless form as 12 for appropriate choices of R, , and w (c) Graphically analyze the dimensionless equation for the cases R 1. how many equilibria can exist in each case? for small r, h, and u. tions) for this system are given by From this, you can find an expression for he(R). Reexpress this in terms of the original dimensional variables and solve for sin o. What sort of "catastrophe does this equation describe the onset of?

Explanation / Answer

given

angle of wire with horizontal = theta

mass of bead = m

spring stiffness = k

spring relaxed length = Lo

distance of the closest point of the spring support from the wire = a

a. lets say the bead is at equilibrium at some value of x

then

from force balance

mgsin(theta) = k(sqroot(x^2 + a^2) - Lo)x/sqroot(x^2 + a^2)

mgsin(theta) = kx(1 - Lo/sqroot(x^2 + a^2))

b. the above equation can be written as

mgsin(theta) = kx(1 - Lo/sqroot(x^2 + a^2))

kx - mgsin(theta) = kx(Lo/sqroot(x^2 + a^2))

now let u = a/x

kx - mgsin(theta) = k(Lo/sqroot(1 + u^2))

h = mgsin(theta)a/kx^2

Lo/x = R

then

mgsin(theta) = h*kx^2/a

so,

kx - hkx^2/a = k*Rx/sqroot(1 + u^2)

u - h = Ru/sqroot(1 + u^2)

c. for R < 1, there is one equilibrium for the beat at u = 0

i.e. x = 0

for R > 1, there are three equilibrium positions including u = 0, i.e. x = 0

d. let r = R - 1 = Lo/x - 1

R = r + 1

u - h = (r + 1)u/sqroot(1 + u^2)

squaring both sides

u^2 + h^2 - 2uh = (r^2 + 1 + 2r)u^2/(1 + u^2)

u^2 + h^2 - 2uh + u^4 + h^2*u^2 - 2u^3h = r^2u^2 + u^2 + 2ru^2

h^2 - 2uh + u^4 + h^2*u^2 - 2u^3h = r^2u^2 + 2ru^2

h^2 is approximately 0, u^3h is 0 , r^2 is approximately 0,

- 2h + u^3 = 2ru

h + ru - u^3/2 = 0

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