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Homework Help: Moment of intertia, lagrangian, etc

  1. Nov 15, 2009 #1
    A uniform ladder of mass M and length 2L is leaning aainst a frictionless vertical wall with its feet on a frictionless horizontal floor. Initially the stationary ladder is released at an angle [tex]\theta_{0}[/tex] = 60o to the floor. Assume that the gravitation field g acts vertically downward.

    1) Show that the moment of intertia of the ladder is about its midpoint is I = (1/3)ML2

    2) Derive the Lagrangian of the system.

    3) Derive the equations of motions using the Lagrangian.

    4) Find the angle [tex]\theta[/tex] at which the ladder loses contact with the vertical wall.

    Ive been reading my textbooks but there arent any similar examples for me to relate to.
  2. jcsd
  3. Nov 15, 2009 #2
    For (1), you need to calculate the moment of inertia or a rod: [itex]I=cmr^2[/itex] where c is some number. Methods of calculating the moment of inertia of any object should be found in your 1st year physics textbooks.
    For (2), you have a constant-length rod that is falling from some initial angle. Sounds like it would require rotational kinetic energy, [itex]T=\frac{1}{2}I\omega^2[/itex], added to the potential energy for your Lagrangian: [itex]L=\frac{1}{2}I\omega^2+V[/itex].
    For (3) and (4), once you have (2), you should be able to do these two.
  4. Nov 15, 2009 #3
    I was able to obtain part 1.
    and concerning part 2, the potential would just be MgL or in our case 2MgL?
  5. Nov 15, 2009 #4
    While the latter has the appropriate form, you are not taking into account the angle for which this ladder is leaning. Multiply the above by the appropriate trigonometric function, and there's your potential.
  6. Nov 15, 2009 #5
    I dont understand. Do u mean that the potential V = 2MgLSin60 ?
  7. Nov 15, 2009 #6
    I wouldn't necessarily put [itex]\theta_0[/itex] in there, I would use just plain [itex]\theta[/itex]. As for the actual trigonometric function, it completely depends on where the angle is being measured, but it would be either cosine or sine.
  8. Nov 15, 2009 #7
    and would it be negative potential since the gravity is acting vertically downward?
  9. Nov 15, 2009 #8
    and what is meant by the equations of motion?
  10. Nov 15, 2009 #9
    How far into Lagrangian mechanics have you studied in class? The equations of motion are the reason for utilizing Lagrangian mechanics. Have you seen this equation:

    \frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}}\right)-\frac{\partial L}{\partial q}=0
  11. Nov 15, 2009 #10
    yes I've seen that before.
  12. Nov 15, 2009 #11
    Well if you let [itex]\dot{q}=\dot[\theta][/itex] and [itex]q=\theta[/itex], you take the respective derivatives of your Lagrangian and find that you will have something of the order


    where [itex]f(q,\dot{q},t)[/itex] is something that will depend on coordinates, velocities, and possibly time. This is your equation of motion.
  13. Nov 15, 2009 #12
    so its only in terms of [tex]\dot{\theta}[/tex] ? we dont do it also in terms of the length of the rod/ladder?
  14. Nov 15, 2009 #13
    More than likely those constants will be factors in that right hand side, however the more important factor in determining the acceleration would be the position and velocity of the rod, that is [itex]\theta[/itex] and [tex]\dot{\theta}[/tex].

    Also, I don't think I mentioned this part, but in the kinetic energy term, [tex]\omega=\dot{\theta}[/tex]. If you don't set these equal to each other, you won't get the answer necessary.
  15. Nov 15, 2009 #14
    ahhhh ok got it!! How do I determine if we will be using Cos or Sin? Mechanics and Classical Physics arent my thing, im more interested in atomic and electronics
  16. Nov 15, 2009 #15
    You'll determine which trig function to use by using trigonometry; it entirely depends on where your angle is being measured from.
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