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Tricky problem: Falling plank originally leaning against a wall

  1. Aug 1, 2011 #1
    1. The problem statement, all variables and given/known data

    A plank of lengt 2l leans against a wall. It starts to slip downward without friction. Show that the top of the plank loses contact with the wall when it is at two-thirds of its initial height.
    Hint: Only a single variable is needed to describe the system. Note the motion of the center of mass.

    2. Relevant equations

    [tex]\tau=I\alpha[/tex]
    Energy conservation

    3. The attempt at a solution

    I dont get very far at all.

    First of all I evaluate the torque of the system: [tex]\tau=lMg\sin \theta=I\ddot{\theta}=\frac{Ml^{2}}{3}\ddot{\theta},[/tex] which leads to [tex]\ddot{\theta}=\frac{3g}{l}\sin \theta.[/tex] This is not possible to solve exactly and therefore is of no help.

    Next, I try energy conservation to evaluate the angular velocity at a height of two-thirds of the original: [tex]Mgl\cos\theta=\frac{1}{2}\frac{Ml^{2}}{3}\omega^{2}+\frac{2}{3}Mgl\cos\theta,[/tex] which leads to the result [tex]\omega^{2}=\frac{2g}{l}\cos\theta.[/tex] Now this is of no use either since it cannot be combined with the first equation. It is a relation relative to the original angle, whereas the first equation above is a general differential equation.

    Now if neither torque evalutations nor energy conservation leads anywhere, I dont know how to go on. Does anyone have a hint?
     
  2. jcsd
  3. Aug 1, 2011 #2

    SammyS

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    Draw a free body diagram.

    You are finding the torque about which point?
     
  4. Aug 1, 2011 #3

    Doc Al

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    It's a tricky problem. Use the hint: What path does the CM follow as the plank falls?
     
  5. Aug 2, 2011 #4
    Here is a FBD as I think it should be. In the diagram I also think that the normal forces are [tex]N_{1}=Mg\sin\theta[/tex] and [tex]N_{2}=Mg\cos\theta[/tex] where the latter might be most interesting. See my post below for further thoughts.
     

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  6. Aug 2, 2011 #5

    Doc Al

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    How did you determine this?
     
  7. Aug 2, 2011 #6
    Sure, I can try that. First of all I can evaluate the center of mass in the x direction. [tex]x_{cm}=l\cos\theta[/tex] and differentiating this gives [tex]\frac{dx_{cm}}{dt}=-l\sin\theta\omega[/tex] Now since theta is evaluated in the positive direction, whereas I am interested in the negative direction, I revise this to [tex]\frac{dx_{cm}}{dt}=l\sin\theta\omega[/tex] Differntiating once again yields [tex]\frac{dx^{2}_{cm}}{dt^{2}}=l\cos\theta\omega^{2}+l\sin\theta\alpha[/tex] Now putting in relations for alpha and omega in my original post, slightly changed, I get [tex]\frac{dx^{2}_{cm}}{dt^{2}}=6g\cos\theta(\cos\theta-cos\phi)-3g\sin^{2}\theta,[/tex] where phi is the starting angle.

    At first I thought Í could find where the acceleration is zero to find where the plank loses contact, but I find that the acceleration is always positive. Instead I might use the results from my free body diagram in post above to solve the equation [tex]M\frac{dx^{2}_{cm}}{dt^{2}}=N_{1}=Mg\cos\theta[/tex] which leads to the equation [tex]6\cos\theta(\cos\theta-\cos\phi)-3\sin^{2}\theta=\cos\theta[/tex] to get a relation between theta and phi. Unfortunately it seems like a very difficult equation to solve.

    I can do the same thing with the center of mass in the y direction. Now the result is [tex]y_{cm}=l\sin\theta[/tex] [tex]\frac{dy_{cm}}{dt}=l\cos\theta\omega[/tex] [tex]\frac{dy^{2}_{cm}}{dt^{2}}=l\sin\theta\omega^{2}-l\cos\theta\alpha[/tex] But unless my equation I found above is correct, I am stuck once again.
     
  8. Aug 2, 2011 #7

    Doc Al

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    Describe in words the path of the CM. It's surprisingly simple. (The path, not the problem.)
     
  9. Aug 2, 2011 #8
    On closer thought I need to revise this to [tex]N_{2}=\frac{\sin\theta\cos\theta}{2}[/tex] where the trignonometric part is due to the force being applied perpendicular to the gravity force through the plank. The factor 2 is due to the normal force being at twice the length. I think this should be correct.
     
  10. Aug 2, 2011 #9
    I don't know how this helps. Will think about it. But as for know I can readily see that it is a circle.
     
  11. Aug 2, 2011 #10

    Doc Al

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    Excellent! Now apply conservation of energy to find the velocity of the CM as a function of angle.
     
  12. Aug 2, 2011 #11
    I still have no clue why this is good. I try to think it through and although it seems like something good I cant see the end of the string.

    I can express it as a function of a fraction of the original height, or, as I guess you wanted me too, express it as a difference of the original and the final angle: [tex]v^{2}_{CM}=2lg(\sin\phi-\sin\theta)[/tex] or if a is a fraction of the original height [tex]v^{2}_{CM}=2lg\sin\phi(1-a)[/tex]
     
  13. Aug 2, 2011 #12

    Doc Al

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    There's only one angle here: The angle the plank makes with the wall. (Assume it starts out perfectly flat against the wall, and is then given a nudge.)

    To apply conservation of energy, you must include both rotational and translational KE terms.
     
  14. Aug 2, 2011 #13
    Ooops, sorry about that. In that case i san write [tex]v_{cm}^{2}=2gl(1-\sin\theta)-\frac{1}{3}l^{2}\omega^{2}[/tex] or [tex](\omega l)^{2}=\frac{3}{2}gl(1-\sin\theta)[/tex]

    By the way, i'm off tomorrow, so maybe I need to wait a week to solve this problem if I don't today.
     
  15. Aug 2, 2011 #14

    Doc Al

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    Good. Hint: Consider the horizontal component of this velocity.
     
  16. Aug 9, 2011 #15
    Ok, now I'm back from my little vacation. I still don't know how velocities can help, though. I want to deal with accelerations. So I need to differentiate the equation above.

    Firstly I separate the horizontal and vertical velocities.

    [tex]\omega^{2}(\sin^{2}\theta+\cos^{2}\theta)=\frac{3}{2}gl(1-\cos\theta)[/tex] I can then differntiate this to [tex]-2\omega\sin\theta a_{x_{CM}}l+2\omega\cos\theta a_{y_{CM}}l=-\frac{3}{2}gl\cos\theta\omega[/tex] Continuing this way, first dividing by omega, then substituting for axcm=N2 (where it loses contact), omega2 (angular velocity) and alpha (angular acceleration), I get the equation [tex]\sin^{2}\theta+2\sin\theta\cos\theta-\frac{3}{2}=0[/tex] This leads to the result that the plank loses contact at about 5% of its initial height (analyzing the equation numerically).

    Am I on the right track or am I supposed to do something completely different?
     
  17. Aug 9, 2011 #16

    Doc Al

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    Yes, you'll need to differentiate.

    Start by giving an expression for the horizontal component of the velocity. Then differentiate that.
     
  18. Aug 9, 2011 #17
    Yes, I needed the free body diagram to reevaluate the angular acceleration, and using that the horizontal acceleration should be zero I arrived at the equation [tex]-2\sin\theta+2\sin^{2}\theta-\cos^{2}\theta=-1[/tex] with the non-trivial solution [tex]\sin\theta=\frac{2}{3}[/tex] Thanks for help Doc!
     
  19. Aug 9, 2011 #18

    Doc Al

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    Excellent! :approve: (A tricky problem indeed.)
     
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