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Know acceleration as a function of position, can I find velocity?

  1. May 1, 2014 #1
    Hello I know acceleration of a pendulum as a function of angle. Can I find velocity by integration with respect to theta? My intuition says yes, but dimensional analysis tells me otherwise. What is right and wrong?

    And what about work? Can I integrate tau = Ia with respect to theta and find work?

    Thanks!

    Kind regards,
    Marius
     
  2. jcsd
  3. May 1, 2014 #2

    Nugatory

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    Staff: Mentor

    Yes, to within a constant of integration.

    Show us your dimensional analysis and we'll be more able to help.

    yes (assuming that ##\tau## is a torque and ##a## is the angular acceleration).
     
    Last edited: May 1, 2014
  4. May 1, 2014 #3
    I have acceleration, ##\vec{\alpha}(\theta)## for the pendulum given by:

    $$
    \vec{\alpha}(\theta) = -\hat{k}\left( \frac{L\,m\,g}{I} \right) \sin\theta
    $$


    If I could use integration to find ##\omega##:

    $$
    \vec{\omega}(\theta) = \hat{k}\left( \frac{L\,m\,g}{I} \right) \cos\theta + \vec{C}
    $$

    But If could integrate again:

    $$
    \vec{\theta}(\theta) = -\vec{\alpha} + \vec{C}\theta + \vec{D}
    $$

    Question 1: Surely that cannot make sense? ## \vec{\theta}## as a function of itself?

    Question 2: And looking at units, ##\alpha## is angular acceleration, so it must have units ##\rm s^{-2}##, if I integrate twice, i basically multiply by ##\theta## twice, which is dimensionless so I get ##\vec{\theta}(\theta) : \rm s^{-2}##. #\vec{\theta}## should have no dimensions. So think I have a contradiction. What is correct?

    What have I misunderstood?

    Thank you for your time.

    Kind regards,
    Marius
     
    Last edited: May 1, 2014
  5. May 1, 2014 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Your first integration is wrong. [itex]\alpha[/itex] is [itex]d\omega/dt[/itex], not [itex]d\omega/d\theta[/itex] so you cannot integrate the right side with respect to [itex]\theta[/itex] as it stands. You can use the chain rule to write [itex]d\omega/dt= (d\omega/d\theta)(d\theta/dt)= \omega d\omega/d\theta[/itex]. So you can write your equation as
    [tex]\omega d\omega/d\theta= -k\left(\frac{Lmg}{I}\right)sin(\theta)[/tex]
    which, when you separate variables, becomes
    [tex]\omega d\omega= -k\left(\frac{Lmg}{I}\right)sin(\theta)d\theta[/tex]

    and integrate both sides.
     
    Last edited: May 1, 2014
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