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Dipole oscillation

  1. Sep 12, 2007 #1
    1. The problem statement, all variables and given/known data

    #3 on this PDF

    2. Relevant equations

    [tex] \tau = Frsin(\theta) = I \alpha [/tex]
    [tex] I=mr^2 [/tex]

    3. The attempt at a solution

    Here's what I've done:
    [tex] \tau = Frsin(\theta) = I \alpha [/tex]
    [tex]F=QE[/tex]

    [tex] 2QEsin(\theta) \frac{A}{2} = 2M\left(\frac{A}{2}\right)^2 \alpha [/tex]
    simplify:
    [tex] QEsin(\theta)A = \frac{1}{2}MA^2 \frac{d^2\theta}{dt^2} [/tex]
    [tex]\frac{d^2\theta}{dt^2} - \left(\frac{2QE}{MA}\right) \theta = 0[/tex]

    I'm not sure how to get the period.

    I know that [tex]\omega = 2 \pi \frac{1}{T} [/tex]

    but I'm stuck here. Any help would be much appreciated.
     
    Last edited: Sep 13, 2007
  2. jcsd
  3. Sep 12, 2007 #2

    learningphysics

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    You're almost there... [tex]\alpha[/tex] here equals [tex]-\frac{d^2\theta}{dt^2}[/tex]

    once you fix that your equation is of the form:

    [tex]\frac{d^2\theta}{dt^2} + {\omega}_0^2\theta = 0[/tex]

    so you can get [tex]{\omega}_0[/tex]
     
  4. Sep 13, 2007 #3
    why is alpha negative?
    and is the form [tex]\frac{d^2\theta}{dt^2} + {\omega}_0^2\theta = 0[/tex] standard?

    where did this equation come from?

    also, how would i show the system to have small oscillations?
     
  5. Sep 13, 2007 #4

    learningphysics

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    The way you calculate torque... it is acting clockwise... therefore [tex]\alpha[/tex] should be the angular acceleration in the clockwise direction... however as the object turns clockwise, your theta angle decreases.... examine what happens to theta as the object rotates in the direction of the torque...

    Yes, it is the standard equation for a harmonic oscillator.

    The question asks you to assume small oscillations. ie approximate sin(theta) as theta... as you did.
     
  6. Sep 13, 2007 #5
    I'm sorry, i did not mean to ask about small oscillations, I was wondering more about how to show the system exhibits simple harmonic motion, before ever going to get the period of oscillation.

    and the torque will sometimes be negative, and sometimes be positive. How do you know alpha is negative?
     
    Last edited: Sep 13, 2007
  7. Sep 13, 2007 #6

    learningphysics

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    The fact that the equation is in this form for small theta:
    [tex]\frac{d^2\theta}{dt^2} + {\omega}_0^2\theta = 0[/tex]

    shows that there's simple harmonic motion...

    As for the torque... let me ask this... what angle is [tex]\theta[/tex]... and how does it change with positive torque... how does it change with negative torque...

    when the torque is positive [tex]\alpha[/tex] should be positive... when the torque is negative [tex]\alpha[/tex] should be negative.
     
  8. Sep 13, 2007 #7
    right, so by saying [tex]-\frac{d^2\theta}{dt^2}[/tex]

    this means the torque is always negative, but it isnt is it?
     
  9. Sep 13, 2007 #8

    learningphysics

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    No, not when [tex]\frac{d^2\theta}{dt^2}[/tex] is negative...
     
  10. Sep 13, 2007 #9
    i'm just really dense right now. I still dont know why you would need a negative sign.
    could you explain it differently? i'm not seeing the whole picture
     
  11. Sep 13, 2007 #10

    learningphysics

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    let's put this problem aside and look at a different one... suppose I have a wrench on the ground connected to some pivot... the wrench forms an angle of let's 45 degrees with the east axis... so theta is 45. Let's say the wrench is 0.5m. I exert a force of 10N at the end of the wrench... eastward...

    What is the torque about the pivot?
     
  12. Sep 13, 2007 #11
    T=.5*10*sin45
     
  13. Sep 13, 2007 #12

    learningphysics

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    cool. now is [tex]\frac{d^2\theta}{dt^2}[/tex] positive or negative here? taking theta as the angle made with the east axis...
     
  14. Sep 13, 2007 #13
    angle is decreasing so it's negative?
     
  15. Sep 13, 2007 #14

    learningphysics

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    Yes... the angular acceleration is clockwise... in other words the angle the wrench makes with the vertical... call it [tex]\alpha[/tex]... [tex]\frac{d^2\alpha}{dt^2}[/tex] is positive... theta = 90 - alpha.... [tex]\frac{d^2\theta}{dt^2} = -\frac{d^2\alpha}{dt^2}[/tex]

    You found the torque clockwise... but [tex]\frac{d^2\theta}{dt^2}[/tex] is the counterclockwise angular acceleration...

    you could have done the opposite and found the torque counterclockwise... and said T=-.5*10*sin45 (that's the counterclockwise torque)... now in this case you can say [tex]T = I\frac{d^2\theta}{dt^2}[/tex]... but since you said T = +0.5*10*sin45 which is the clockwise torque... you must say [tex]T = -I\frac{d^2\theta}{dt^2}[/tex]
     
  16. Sep 13, 2007 #15
    so say we assumed the counter clockwise notation for the torque for the question for the dipole-dumbbells. so [tex]T = I\frac{d^2\theta}{dt^2}[/tex]

    then I would still get:
    [tex]\frac{d^2\theta}{dt^2} - {\omega}^2\theta = 0[/tex]
     
  17. Sep 13, 2007 #16

    learningphysics

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    no, because then

    [tex]T = - 2QEsin(\theta) \frac{A}{2} [/tex] in order for torque to be counterclocwise.
     
  18. Sep 13, 2007 #17
    ok sure, the wrench is fixed and applying a torque will always rotate it clockwise, however, the dipole-dumbbell rotates both clockwise and counterclockwise. This means that that torque can be negative and positive even if we took a clockwise/counterclockwise notation. Since it's not just going in a circle, why would alpha be negative inthe clockwise notation then?
     
  19. Sep 14, 2007 #18

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    It's not about whether the torque turns out clockwise or counterclockwise... torque is actually a vector... it has a magnitude and a direction... one way we deal with the direction is by using clockwise... counterclockwise...

    It is just like with vectors... suppose I have an object with just one force acting on it is labelled F with an arrow to the right... but F could be a negative number... in which case the force is actually acting towards the left... F is the force acting to the right... but it could be negative...

    Now suppose I label the acceleration to the left a... I have a with an arrow pointed to the left... again a itself can be positive or negative... if a is negative then the acceleration is actually to the right... a is the acceleration to the left... but a could be negative...

    What is the relationship between F and a here given a mass m?
     
  20. Sep 14, 2007 #19
  21. Sep 14, 2007 #20

    learningphysics

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    Seems like you're mixing up clockwise and counterclockwise...

    Do you agree that [tex]\frac{d^2\theta}{dt^2}[/tex] is the counterclockwise angular acceleration? because counterclockwise is the direction of increasing theta...
     
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