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Homework Help: Stable or not stable equilibrium

  1. Sep 14, 2010 #1
    Clearly, I need to look at the potential function and differentiate it to determine extreme and then the stability of the extrema. However, I'm not sure how to do this mathematically at the moment.

    http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-14203321.jpg?t=1284516380 [Broken]
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 14, 2010 #2
    What do you mean? You don't know how to take two derivatives, or you don't know how to set the problem up?
  4. Sep 14, 2010 #3
    Sorry. How to setup the problem. The potential in this case would be gravitational and the height is 2R.
  5. Sep 14, 2010 #4
    Not quite, you want the height to the center of mass of the square (the middle of the square). Do you see why?
  6. Sep 14, 2010 #5
    Well, for stable equilibrium, you don't want there to be a torque on the cylinder from the effective gravitational force on the block. If the gravitational force is directed downwards, there will be a normal force from the cylinder. I assume this is Newton's Third Law.
  7. Sep 14, 2010 #6
    Well, assume that the cylinder is fixed in place it's attached by a screw to a metal table, and that the cylinder is extremely, extremely rough, such that it will not slide off the cylinder. All you are looking at now is at what angle from the center of the cylinder will cause the square's center of mass to fall.
  8. Sep 14, 2010 #7
    The force would be mg sin(theta). The negative of the force equals the gradient of the potential. In this case it would be U(theta). I'll integrate the effective gravitational force.
  9. Sep 14, 2010 #8
    You'll still want to make a potential function, but you might want to consider making a potential like this


    You'll want to figure how the height of the center of mass varies with theta.
  10. Sep 14, 2010 #9
    Dammit. What I had is not right. I'm tired. That's not helping.

    R cos(theta) + (b/2) cos(theta)
  11. Sep 14, 2010 #10
    Yeah, that's pretty close, but you need one last little bit. Make a really huge picture and you'll see.

    [tex]h(\theta)=(R+b/2)cos(\theta)+R\theta sin(\theta)[/tex]

    Voila, you're on your way.
  12. Sep 14, 2010 #11
    My big picture helped me see the first term R cos(theta) + (b/2) cos(theta). lol. I don't see the second term R*theta*sin(theta).
  13. Sep 14, 2010 #12
    I almost missed it too. :)
  14. Sep 15, 2010 #13
    Well, I missed it! lol.

    I'm pretty sure the homework is due today, too. There's one problem where I need to do a numerical calculation, but I don't know if I have access to one. The only program I used very briefly was MathLab in my ODE class last fall.
  15. Sep 15, 2010 #14
    Differentiate the potential energy seems to be quite long in this problem. Why don't you try dynamics analysis? Let the cube move away from the equilibrium position by a very small angle. If the torque of gravitational force about the point of contact tends to make the cube move back to the equilibrium position, it's stable equilibrium; otherwise, it's unstable.
    Consider the case that the cube is at the critical angle (i.e. the angle right between stable and unstable equilibrium). In this case, gravitational force lies in the line joining the cube's center of mass and the point of contact. Notice that this is a very small angle. From here, you can calculate the condition of that critical case, and thus, deduce the conditions for stable and unstable equilibrium.
  16. Sep 15, 2010 #15
    The professor worked it out today the same way we did. He immediately caught that little third height.
  17. Sep 15, 2010 #16
    Well, that's just my suggestion for a more elegant, natural and easy-to-visualize way :smile:
  18. Sep 15, 2010 #17
    Hmm. I wasn't quite following you, honestly. Wouldn't the torque from the gravitational force always move it away from equilibrium? For your critical case, how would the gravitational force lie in the line connecting the center of mass with the point of contact?
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