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Rod about fixed pivot

  1. Oct 25, 2005 #1
    I have a little conceptual question about a rod-pivot system.

    We have this rod (mass M, length l) pivoting about a fixed point. It is released at 30 degrees above the horizontal, and we're asked for the magnitude and direction force on the pivot when the rod reaches the horizontal.

    So i figured one of the forces on the pivot would be the centripital force keeping the rod in a circular motion. This I can "easily" calculate, and it's always pointing in the -r direction (opposite to the rod).

    But I can't figure out what and if there are other forces. In fact, I "believe" there are no others. But the question is asked in such a way as to hint there might be something else.

    If you have any insights, please share tehm with me.
     
  2. jcsd
  3. Oct 25, 2005 #2
    further complications

    It actually turns out even calculating the centripetal force is not as trivial as I had expected.

    I'm trying to calculate the acceleration knowing that [itex] I \alpha=F r (center of mass) [/itex]. I know [itex] I = \frac {1}{2} m l^2 [/itex], [itex]r=\frac {l}{2}[/itex] and that [itex] F=-m g sin(\theta) [/itex], where the angle is the angle between the rod and the horizontal.

    But this means that
    [tex] \frac {d^2\theta}{dt^2}=\frac{-3 g sin (\theta)}{2l} [/tex]
    and I don't know how to solve this ODE.

    Usually, in class, the teacher approximates [itex] sin(\theta)=\theta[/itex], but in my case the angle is quite big for this.

    What can I do about this?
     
    Last edited: Oct 25, 2005
  4. Oct 26, 2005 #3
    I don't think you should set it up as a second order differential equation. What class is it?
     
  5. Oct 29, 2005 #4
    differential equation

    This is an intro to classical mechanics problem.

    I agree that I shouldn't run into this sort of differential equation. However, this is what comes out of the motion equations once I set them up. If you see any mistake in the setup, let me know.

    On the other hand, this wouldn't be the first time the teacher expects us to know the solution of a crazy diffrential equation, even though we've never seen them before.

    I still haven't figured out what other forces are applied on the pivot. Can it be that some of the weight of the rod also acts on it?
     
    Last edited: Oct 29, 2005
  6. Oct 29, 2005 #5

    Tide

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    If it's a classical mechanics class then you have learned about energy conservation. Use it! :)
     
  7. Oct 29, 2005 #6
    tentative solution

    ok, so from the conservation of energy i know that [itex]K(\theta)+U(\theta)=E_T[/itex].

    Since
    [tex] U(\theta)=\int_{0}^{L}\!\frac{M}{L}g l sin(\theta){dl}=\frac{1}{2}MgLsin(\theta) [/tex]

    and
    [tex] K(\theta)=\frac{1}{2}I\omega^2=\frac{1}{6}ML^2\left(\frac{d\theta}{dt}\right)^2 [/tex]

    and initially all energy was potential, therefore
    [tex] E_T=U(30\degree)=\frac{1}{4}MgL [/tex]

    I have that
    [tex] \left(\frac{d\theta}{dt}\right)^2=\frac{3g}{L}\left(\frac{1}{2}-sin(\theta)\right)[/tex]

    Now, to come back to the total force exerted on my pivot, it should have two components: the gravitational component [itex] Mgsin(\theta) [/itex], which becomes 0 for [itex]\theta=0[/itex], and a force opposite to centripital force (which acts on the rod). This force should be
    [tex] F(\theta)=-F_c=-Ma_r=-M\left(-r\left(\frac{d\theta}{dt}^2\right)\right) [/tex]
    where the effective radius [itex]r=\frac{L}{2}[/itex]

    Substituting for r and the derivative of theta found above, I get
    [tex]F(\theta)=\frac{3Mg}{2}\left(\frac{1}{2}-sin(\theta)\right)[/tex]
    [tex]F(0)=\frac{3Mg}{4} [/tex]
    Does this make any sense?
     
    Last edited: Oct 29, 2005
  8. Oct 29, 2005 #7

    lightgrav

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    (most people refer to the "time derivitive of theta"
    as the angular velocity, omega . I'm using "w" for it.)

    in a simple pendulum at the bottom, T = Mg + Ma = Mg + Mrw^2 .

    I think that your ½Iw^2 = ½MgL sin(30) + ½MgL ,
    since the PE at bottom should equal -½MgL (pivot is h=0)
     
    Last edited: Oct 29, 2005
  9. Oct 30, 2005 #8
    I'm not sure what you mean. The angle i'm considering is the angle with the horizontal. My starting point is 30 degrees above the horizontal and my endpoint is the horizontal. Therefore, I believe
    [tex]\frac{1}{2}I\omega^2=U(0)-U(30)=\frac{1}{2}MgLsin(0)-\frac{1}{2}MgLsin(30)=0-\frac{1}{4}MgL[/tex]

    which I realize now comes to the same as what I was doing before, it's just simpler.
     
  10. Oct 30, 2005 #9

    lightgrav

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    Whoa, sorry ... I read that one wrong!:eek:
    I was going from your 30 to your -90
    I'd better go to bed ...:zzz:
     
  11. Oct 30, 2005 #10
    Don't worry man. No problem. Actually forced me to review my calculations, which is never a bad thing to do :tongue2:
     
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