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Normal Modes; Rod suspended from strings

  1. Jan 22, 2013 #1
    1. The problem statement, all variables and given/known data

    A thin uniform rod of length 2b is suspended by two vertical light strings, both of fixed length l, fasted to the ceiling. Assuming only small displacements from equilibrium, find the Lagrangian of the system and the normal frequencies. Find and describe the normal modes. [Hint: A possible choice of generalized coordinates would be x, the longitudinal displacement of the rod, and y1 and y2, the sideways displacement's of the rod's two ends. You'll need to find how high the two ends are above their equilibrium height and what angle the rod has turned through.]


    2. Relevant equations

    L = T - U


    3. The attempt at a solution

    I’m not sure how to find the kinetic energy and potential energy in terms of the suggested generalized coordinates.

    My idea for using the suggested generalized coordinates was to write the longitudinal displacement as:
    x = l sin[[itex]\phi[/itex]]

    The sideways displacements of the two ends are
    y1 = l sin[[itex]\vartheta_{1}[/itex]]
    y2 = l sin[[itex]\vartheta_{2}[/itex]]

    where [itex]\vartheta_{1}[/itex] and [itex]\vartheta_{2}[/itex] are the ‘sideways’ angle of the string for each of the rod’s ends. I can differentiate these to get the velocities, but the kinetic energy of the rod should be the sum of the kinetic energy of the center of mass and the kinetic energy of rotation about the center of mass and I don’t understand how to phrase this in terms of these coordinates.

    Also, the potential energy is simply due to the height of the center of mass which is equal to the average height of the two ends of the rod. I think that the height of each end is
    z1 = l cos[[itex]\phi[/itex]+[itex]\vartheta_{1}[/itex]]
    z2 = l cos[[itex]\phi[/itex]+[itex]\vartheta_{1}[/itex]]

    Does that seem right?

    Any help is appreciated. I haven't had any problem solving normal mode problems that involve only point masses.
     
  2. jcsd
  3. Jan 22, 2013 #2

    haruspex

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    How exactly are you defining ϕ?
    Are you assuming the strings are vertical when unperturbed? I guess you have to.
    You seem to be trying to switch to the angles as being your coordinate system, which is not in the spirit of the hint. If you stick to the x and ys, you can write down the KE reasonably easily. The challenge then is the PE.
    Consider one end of the rod, displaced x one way and y at right angles. What is its net displacement? What height change will that result in?
     
  4. Jan 23, 2013 #3
    Thanks for the reply.

    The angle [itex]\phi[/itex] is the angle responsible for the longitudinal displacement (which must be the same for each string). Sorry for not being clear about that.

    Yes. The unperturbed strings are vertical.

    To answer your question about one end of the rod being displaced by an amount x and y: Taking the origin to be the ceiling where the string is attached, the initial height of the rod is z = -l. As one end of the rod is displaced by x longitudinally and y “sideways” (to use the language of the problem) then the new height of the rod will be z=- l + [itex]\sqrt{l^{2}-x^{2}-y^{2}}[/itex].

    Because there are similar expressions for both ends of the rod I suppose that the potential energy is mg[itex]\frac{z1+z2}{2}[/itex] (where [itex]\frac{z1+z2}{2}[/itex] is the average height of each end of the rod above its equilibrium height.

    The kinetic energy still has me puzzled. Perhaps I’m over thinking it, but it definitely seems like the rotational inertia of the rod must be included, but it is not clear to me how to do this in terms of the suggested coordinates. For a moment let’s just consider that the rod is confined to move in the longitudinal direction so that there is one degree of freedom. The situation is very similar to a simple pendulum. Would the kinetic energy simply be 1/2m(xdot[itex]^{2}[/itex])?

    For the sideways displacements there are two options: the rods ends can swing in phase or out of phase (which must be two of the normal modes). If they are in phase then the situation is much like the simple pendulum again, but if they are out of phase the rod is essentially rotating about its center of mass.
     
  5. Jan 23, 2013 #4

    haruspex

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    It can't be that, if you're measuring height from the ceiling. That's more like a displacement from neutral position (which is actually what you want), but I still think it's wrong. I get (x2+y2)/2l.
    Yes.
    Yes.
    The context perhaps makes it appear more complex than it is. You can think of it as a rod moving and rotating in a plane. You have the linear KE based on the speed of the mass centre, and the rotational KE based on rotation about the mass centre.
     
  6. Jan 26, 2013 #5
    You are absolutely right. That’s what I get for the displacement of one end from its equilibrium position.

    The net displacement in the plane is [itex]\sqrt{x^2+y^2}[/itex]. I figured I could find the height above equilibrium by using [itex]l =\sqrt {x^2+y^2+z^2}[/itex]. This is how I found the height above equilibrium
    [itex]z = -l +\sqrt{l^2-x^2-y^2}[/itex].

    I don’t understand how you found this.


    Ok. The kinetic energy is then [itex] \frac{1}{2} m \dot{x^2}+\frac{1}{2}m (\frac{(\dot{y_1}+\dot{y_2})}{2})^2 +\frac{1}{2} I \dot{\theta^2}[/itex]

    For the second term I have used the fact that the total sideways displacement of the center of mass is the average of the sideways displacements of each end. In the third term [itex]\dot{\theta}[/itex] is the angle that the rod has turned through in the plane.

    Now I just have to relate the angle that the rod has turned in the horizontal plane to the suggested coordinates. It looks like [itex]\theta=Arcsin[\frac{y_1-y_2}{2b}] [/itex] will do the trick.

    I'll give it a shot hopefully before the weekend is over (with your height above equilibrium as well, though I would love an explanation of how you found this) and see if I get anything sensible.
     
  7. Jan 26, 2013 #6

    haruspex

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    Right, but you want an approximation for small x and y. Use the binomial expansion.
     
  8. Jan 26, 2013 #7
    Of course.

    Just one last thing. We have to expand [tex](l^2-x^2-y^2)^\frac{1}{2}[/tex] so if I let [tex]a^2=x^2+y^2[/tex] this becomes [tex](l^2-a^2)^\frac{1}{2}[/tex]. But the series expansion for this to first order is [tex]l-\frac{a^2}{2l}[/tex]. Or [tex]-\frac{x^2+y^2}{2l}[/tex] We differ in sign, but yours seems correct.
     
  9. Jan 26, 2013 #8

    haruspex

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    If you look back at how you defined z, you were measuring it downwards. But you want the gain in height.
     
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