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Vertical stick falls, rotates about CM, derive v(y,theta)

  1. Nov 29, 2009 #1
    1. The problem statement, all variables and given/known data
    A uniform stick of mass m and length L, initially
    upright on a frictionless horizontal surface, starts falling. The circle at the center of the
    stick marks the center of mass. Derive an expression for the speed of the center of mass
    as a function of y and θ if the stick falls as shown (with the center of mass moving
    straight downward).


    2. Relevant equations
    [tex]v=\frac{dy}{dt}[/tex]; [tex]\omega=\frac{d\theta}{dt}[/tex]

    [tex]v_{cm}[/tex][tex]=r\omega[/tex]; [tex]I=\frac{1}{12}[/tex][tex]mL^{2}[/tex]

    [tex]K_{rot}[/tex]=[tex]\frac{1}{2}[/tex][tex]I\omega^2[/tex]

    [tex]K=\frac{1}{2}mv^2[/tex]

    3. The attempt at a solution
    There are no non-conservative forces so [tex]E_{mech}[/tex] is conserved.

    Therefore I figure: [tex]U_{i}+K_{i}=U_{f}+K_{f}[/tex]

    So: [tex]mg\frac{1}{2}L=\frac{1}{2}mv^{2}_{cm}+\frac{1}{2}I\omega^2[/tex]

    Then subed in [tex]v=\frac{dy}{dt}[/tex] and [tex]\omega=\frac{d\theta}{dt}[/tex] and

    [tex]I=\frac{1}{12}[/tex]mL[tex]^{2}[/tex], canceled out the (1/2) and m and attempted to integrate the equation.

    mg and L are all constants right? So I got [tex]0=2m\frac{dy}{dt}y+\frac{1}{12}2mL^2\frac{d\theta}{dt}\theta[/tex] which I don't think is right..
     
  2. jcsd
  3. Nov 29, 2009 #2

    Doc Al

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    Staff: Mentor

    The basic idea of using conservation of mechanical energy is correct. Hint: Find the relationship between ω and vcm.
     
  4. Nov 29, 2009 #3
    Like [tex]v=r\omega[/tex] or [tex]v=\frac{1}{2}L\omega[/tex]

    So [tex]\frac{dy}{dt}=\frac{1}{2}L\frac{d\theta}{dt}[/tex] ?

    Or is there another relationship I'm supposed to be looking for?
     
  5. Nov 29, 2009 #4

    Doc Al

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    Staff: Mentor

    That's not quite right. Vcm is not the tangential speed of something rotating in a circle. (The center of mass falls straight down.)

    Another hint: Find an expression for the height of the center of mass in terms of θ.
     
  6. Nov 29, 2009 #5
    Ahhh yes thats right.

    Another hint: Find an expression for the height of the center of mass in terms of θ.[/QUOTE]

    so like [tex]y=\frac{1}{2}L-cos(\theta)[/tex]
     
  7. Nov 29, 2009 #6

    Doc Al

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    You're getting warmer, but that expression is not quite right.
     
  8. Nov 29, 2009 #7
    hm.. Yeah, if I say [tex]\theta[/tex] is 0, then [tex]y=\frac{1}{2}L-1[/tex] which I don't want..

    So is it [tex]y=\frac{1}{2}L-sin(\theta)[/tex] ?
     
  9. Nov 29, 2009 #8

    Doc Al

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    Nope, not yet. (You can't have a sinθ term, which has no units, added to a length term. That's a tip off that something is wrong.)

    Do this. Draw a diagram of the stick when it makes an angle θ with the ground. What's the height of the center? (Find the right triangle.)
     
  10. Nov 29, 2009 #9
    phys.jpg

    This is what is supplied with the problem. Well, its my version of it :redface: but it's pretty accurate. Um, neglecting that part of the angled stick is somehow below the surface of the table :uhh:

    oh heh heh.
    Ok so its [tex]y=\frac{1}{2}L-\frac{1}{2}Lcos(\theta)[/tex]..
     
  11. Nov 29, 2009 #10

    Doc Al

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    That will work. (My diagram would just measure the position of the center from the ground, thus y = L/2 cosθ.) Now take the derivative of both sides.
     
  12. Nov 29, 2009 #11
    So
    [tex]\frac{dy}{dt}=\frac{1}{2}Lsin(\theta)[/tex] .... in terms of my diagram? It would be negative for yours because as time increases, y is decreasing so the change is negative where as mine is positive because the change in y is increasing?
     
  13. Nov 29, 2009 #12

    Doc Al

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    You left off dθ/dt.
     
  14. Nov 29, 2009 #13
    oops.. ya so

    [tex]\frac{dy}{dt}=\frac{1}{2}Lsin(\theta)\frac{d\theta}{dt}[/tex]

    So do I solve for omega and sub that into my energy equation [tex]mg\frac{1}{2}L=\frac{1}{2}mv^{2}_{cm}+\frac{1}{2}I\omega^2[/tex] ?
     
  15. Nov 30, 2009 #14

    Doc Al

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    Staff: Mentor

    Yes. Express ω in terms of vcm.
     
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