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Finding the equation of motion of an oscillator

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  1. Jan 29, 2016 #1
    1. The problem statement, all variables and given/known data 2l8as8k.png
    A simple pendulum consists of a mass m suspended by a ball to a yarn (massless) of length l. We neglect friction forces.

    Give the list of every forces applied to this system and then the motion of equation.

    Why is the following equations necessary to find the motion equation true?


    I don't know where does the ##\alpha^2## comes from...
    2. Relevant equations

    \begin{align*}
    \vec P + \vec T &=m\vec \alpha\\
    \Leftrightarrow\begin{pmatrix}
    {\vec ||P|| \cos \alpha}\\{- ||\vec P|| \sin \alpha}
    \end{pmatrix}
    +
    \begin{pmatrix}
    {-||\vec T||}\\{0}
    \end{pmatrix}
    &=
    m\begin{pmatrix}
    {l\alpha ^2}\\{l\alpha}
    \end{pmatrix}
    \end{align*}

    1. The problem statement, all variables and given/known data 2l8as8k.png
    A simple pendulum consists of a mass m suspended by a ball to a yarn (massless) of length l. We neglect friction forces.

    Give the list of every forces applied to this system and then the motion of equation.

    Why is the following equations necessary to find the motion equation true?


    I don't know where does the ##\alpha^2## comes from...

    Furthermore, I don't understand why the motion equation is found from the ##(2)## equation: ##\alpha + \frac{g\sin \alpha}{l}##.
    2. Relevant equations

    \begin{align*}
    \vec P + \vec T &=m\vec \alpha\\
    \Leftrightarrow\begin{pmatrix}
    {\vec ||P|| \cos \alpha}\\{- ||\vec P|| \sin \alpha}
    \end{pmatrix}
    +
    \begin{pmatrix}
    {-||\vec T||}\\{0}
    \end{pmatrix}
    &=
    m\begin{pmatrix}
    {l\alpha ^2}\\{l\alpha}
    \end{pmatrix}
    \end{align*}

    \begin{cases}
    {m\vec g\cos \alpha - ||\vec T||= -ml\alpha^2}(1)\\
    {- mg sin \alpha + 0 = ml\alpha} (2)
    \end{cases}
    3. The attempt at a solution

    I tought That we would find the motion equation out from:

    \begin{cases} mg + l \cos \alpha = 0\\ l \sin \alpha = 0 \end{cases}

    from the fact that:

    \begin{cases}
    \vec T_x=(l\cos \alpha + mg) \vec i=\vec 0\\
    \vec T_y=(l \sin \alpha) \vec j=\vec 0
    \end{cases}
     
    Last edited: Jan 29, 2016
  2. jcsd
  3. Jan 29, 2016 #2

    haruspex

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    There seems to be some confusion regarding alpha. In the first Relevant Equations you quote, alpha is being used for linear acceleration, which does not match the usage in the diagram. In the second quoted equation, the alpha on the left is as in the diagram. On the right, the alpha at the top should be angular velocity, while that at the bottom should be angular acceleration.
    Also, there is a wrong sign in the second equation.
    To sort it all out, I'll use theta as the angle in the diagram and a for linear acceleration.
    ##\vec P+\vec T=m\vec a##
    ##||\vec a||=l\ddot {\theta}##
    ##T-P\cos(\theta)=ml\dot{\theta}^2##
    ##-P\sin(\theta)=ml\ddot{\theta}##

    I cannot make any sense of your own equations. You seem to be treating l as a force instead of a length.
     
  4. Jan 30, 2016 #3
    First sorry, I misused the copy and paste tool so it gave this weird template which I can edit no more. The right one is the second one.
    In my attempt I did a list of the forces applied to the system:

    ##\vec P =m\vec g## toward the botom and which displays the weight.
    ##\vec T =
    \begin{cases}
    \vec T_x =l\cos \alpha \vec i\\
    \vec T_y =l\sin \alpha \vec j
    \end{cases}##

    At the equilibrium, all forces compense themselves which leads to

    ##
    \begin{cases}
    mg + l\cos \alpha =0\\
    l \sin \alpha = 0
    \end{cases}
    ##
    And I was stuck from there.

    And that was not the way the correction goes.
    From what you said, I presume that

    \begin{align*}
    \begin{pmatrix}
    ||\vec P||\cos \alpha \\
    -||\vec P||\sin\alpha
    \end{pmatrix}+
    \begin{pmatrix}
    {-||\vec T||}\\{0}
    \end{pmatrix}
    =
    m\begin{pmatrix}
    {l\alpha ^2}\\{l\alpha}
    \end{pmatrix}
    \end{align*}


    here we have on the top ##\alpha^2## the angular velocity and ##\alpha## the angular acceleration.

    That's exactly the point I don't understand. Where do these comes from?

    I understand the firs two lines of the equation you gived.

    Then is there a mistake in the third one? From the equation above, I thought it would have been ##ml\ddot\theta^2##
     
    Last edited: Jan 30, 2016
  5. Jan 30, 2016 #4

    haruspex

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    We're not making progress here. All the comments I first made still apply.
    What is this "##l##" in your equations? The way you are using it it appears to be a force, but what force?

    The target equation, the one in matrix form, still has the variable alpha used in three different ways. Is that the way it was given to you? It's crazy to work with that.
     
  6. Jan 30, 2016 #5
    Oh, yes, your right: I didn't specified that ##l## is the length of the yarn, not a force. And yes it is the way it was given to me to find the motion equation which is:

    ##\alpha + \frac{g\sin\alpha}{l}=0## and which I don't understand...

    Thank you for your patience!
     
  7. Jan 30, 2016 #6

    haruspex

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    Ok, but I'm going to ask you to stop using alpha for multiple meanings in this thread. Use theta for the angle of the string. For the angular velocity you can use either ##\omega## or ##\dot \theta##, and for angular acceleration use ##\alpha## or ##\dot \omega##, or ##\ddot \theta##. These are all standard notations. I believe the form of equation you were told to find is a misprint, and it should read
    ##\alpha + \frac{g\sin\theta}{l}=0##.

    Turning now to your attempt, your very first equations make no sense. You write ##\vec T_x=l\cos(\theta)\vec i##. Tx is a force (horizontal component of tension, I assume), but the right hand side is a distance. A force cannot equal a distance.
    The equation should take the form net force = mass * acceleration, so you need an expression for the horizontal acceleration in terms of theta and its derivatives.
    The other equation, Ty, has the same problem, and also omits gravity.
     
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