Problem 1 A bead of mass m is constrained to slide (subject to a drag force) alo

Question

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 r to be the position of the mass along the wire, so that the r -0 occurs at the point closest to the support 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 i:s 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 0 2a2 (b) Show that this equilibrium equation can be written in dimensionless form as Lu for appropriate choices of R, h, and «u

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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