Bead on a titled wire, connected to a spring

In summary, the problem at hand is to find the equilibrium equation for a bead on a tilted wire, which can be written in dimensionless form as 1 - \frac{h}{u} = \frac{R}{\sqrt{1 + u^2}} for appropriate choice of R, h, and u. This is achieved by considering the forces acting on the bead, assuming a right angle triangle between the wire, spring, and imaginary line, and substituting variables to simplify the equation. The value of R is still unknown and needs to be determined for later parts of the problem.
  • #1
mwrigh58
1
0

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)?
 

Attachments

  • Bead_on_tilted_wire.png
    Bead_on_tilted_wire.png
    6 KB · Views: 565
Last edited:
Physics news on Phys.org
  • #2
I thought it might be related to the angle \theta but substituting in m g sin\theta for R does not work (the equations no longer satisfy Fnet, x = 0). If you could provide an explanation or a hint towards an explanation, I would be very grateful.Thanks for your time!
 

1. What is a "bead on a titled wire, connected to a spring"?

A "bead on a titled wire, connected to a spring" is a physical system used in experiments to demonstrate the effects of gravity, tension, and spring force on an object.

2. How does the system work?

The system consists of a bead placed on a tilted wire, which is then connected to a spring. When the system is released, the bead will move down the wire due to the force of gravity, while the spring will stretch and exert a force in the opposite direction. This creates an oscillating motion as the forces balance each other out.

3. What is the purpose of using this system?

This system is commonly used in physics experiments to study the principles of oscillation, such as amplitude, frequency, and period. It can also be used to demonstrate the relationship between potential and kinetic energy.

4. What factors affect the behavior of the system?

The behavior of the system is affected by several factors, including the angle of the wire, the mass of the bead, the stiffness of the spring, and the surface friction between the bead and the wire. These factors can be manipulated to observe the effects on the motion of the system.

5. Can this system be used to study real-life phenomena?

Yes, this system can be used to model and study real-life phenomena, such as the motion of a pendulum or the behavior of a spring-mass system. It can also be used to understand the movement of objects in gravitational fields or systems with varying degrees of tension and force.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
184
  • Introductory Physics Homework Help
Replies
16
Views
335
  • Introductory Physics Homework Help
Replies
3
Views
309
  • Introductory Physics Homework Help
Replies
5
Views
226
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
858
  • Introductory Physics Homework Help
Replies
17
Views
280
  • Introductory Physics Homework Help
Replies
2
Views
549
  • Introductory Physics Homework Help
Replies
2
Views
417
  • Introductory Physics Homework Help
Replies
20
Views
975
Back
Top