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Classical mechanics - small movements around equilibrium

  1. Jun 9, 2014 #1
    1. The problem statement, all variables and given/known data
    Two cylinders, that can rotate around their vertical axis, are connected with a spring as shown in the picture. Moment of inertia ##J## is the same for both and they also have the same radius ##R##. Distance between the axes is ##L##. Spring with constant ##k## is ##d## long (not deformed). Also note that ##d<L-2R##. Describe small movements of the system.

    Capturedd.JPG
    2. Relevant equations



    3. The attempt at a solution

    Let's say that we will measure the rotation of left cylinder with ##\varphi ## and rotation of the right one with ##\vartheta ##.

    The kinetic energy is than ##T=\frac{1}{2}J(\dot{\varphi }^2+\dot{\vartheta }^2)##

    Now lets use notation ##\eta =\begin{bmatrix}
    \varphi \\
    \vartheta
    \end{bmatrix}##, therefore kinetic energy is ##T=\frac{1}{2}\dot{\eta }^T \tilde{T}\eta ## where ##\tilde{T}=\begin{bmatrix}
    1 & 0\\
    0& 1
    \end{bmatrix}##.

    Now for kinetic energy we have to find out how much the spring deforms. Let's take a closer rook at the left cylinder:

    The deformation here is ##x_1=R(1-cos\varphi )##. Using taylor expansion around equilibrium state which is ##\varphi =0## tells me that ##x_1=R(1-(1-\frac{1}{2}\varphi ^2))=\frac{R}{2}\varphi ^2##

    The same goes for the other side, therfore ##x_2=\frac{R}{2}\vartheta ^2##

    So potential energy of the system is ##V=\frac{1}{2}k(x_1+x_2)^2=\frac{kR^2}{8}(\varphi ^2+\vartheta ^2)^2##.

    Now what worries me is that there is no way (or at least I don't see it) to write that last expression for ##V## using vector ##\eta ##.

    Why do I want to do that? Because using ##det(V-\omega ^2T)=0## will tell me everything I need to know about this system.

    Where did I get it wrong?
     
  2. jcsd
  3. Jun 9, 2014 #2

    AlephZero

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    You have left out the key things that cause the motion of the system.

    The displacement of the end of the spring is not just ##x_1##, but also ##y_1 = R\sin\varphi##.

    Also the spring has an initial potential energy, because it is stretched between the cylinders. The initial stretch is ##L - 2R - d##. When the system moves the length of the spring changes from ##L - 2R## to ##\sqrt{(L -2R + x_2 - x_1)^2 + (y_2-y_1)^2}##.

    Since the question says the displacements are small, you can approximate everything to first order in ##\varphi## and ##\vartheta##.



    Since the question sa
     
  4. Jun 9, 2014 #3
    hmmm... Ok, if I approximate everything to first order, than

    ##x_1=R(1-cos\varphi )\doteq 0## and also ##x_2=R(1-cos\vartheta )\doteq 0## are both ##0##?

    ##y_1=Rsin\varphi \doteq R\varphi ## and ##y_2=Rsin\vartheta \doteq R\vartheta ##.

    Than the deformation of the spring is ##\Delta L= \sqrt{(L-2R+(x_1+x_2))^2+(y_2-y_1)^2}-(L-2R)##. (I think that's what you meant and that ##-x_1## is just a typo. right?
     
  5. Jun 9, 2014 #4
    If yes, than ##\Delta L= \sqrt{(L-2R+(x_1+x_2))^2+(y_2-y_1)^2}-(L-2R) \doteq (L-2R)(1+\frac{1}{2}\frac{R^2}{(L-2R)^2}(\varphi -\vartheta )^2)-(L-2R)=\frac{R^2}{2(L-2R)}(\varphi -\vartheta )^2##

    Where I can't see how this helps me since ##V=\frac{1}{2}k(\Delta L)^2##
     
  6. Jun 9, 2014 #5

    AlephZero

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    I was taking ##x_1## and ##x_2## positive to the right, and ##y_1## and ##y_2## positive up.

    The sign convention doesn't matter so long as you are consistent, of course. Be careful about the signs of the ##y##'s and the angles!

    No, the spring is stretched initially. If the length increases by ##\Delta L##, the strain energy changes from ##\frac 1 2 k (L - 2R - d)^2## to ##\frac 1 2 k (L - 2R - d + \Delta L)^2##.
     
  7. Jun 9, 2014 #6
    You've lost me. I have come so far:

    Initial stretch: ##s_0=(L-2R)-d##

    For rotation ##\varphi ## and ##\vartheta ## the length of the spring changes for ##\Delta L=\sqrt{(L-2R)^2+R^2(\varphi - \vartheta )^2}-(L-2R)##

    Than the total stretch is
    ##s=s_0+\Delta L=(L-2R)-d+\sqrt{(L-2R)^2+R^2(\varphi - \vartheta )^2}-(L-2R)=\sqrt{(L-2R)^2+R^2(\varphi - \vartheta )^2}-d##.

    If I put this in ##V=\frac{1}{2}ks^2=\frac{1}{2}k(\sqrt{(L-2R)^2+R^2(\varphi - \vartheta )^2}-d)^2## I get all sorts of nasty things where the most horrible part is that I don't get rid of that sqrt....
     
  8. Jun 9, 2014 #7

    AlephZero

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    Let ##A = L - 2R## to simplify the algebra a bit.

    The length of the spring at an arbitrary position (assuming the angles are small) is ##\sqrt{A^2 + R^2(\varphi - \vartheta)^2}##.

    The unstretched length was ##d##, so the strain energy is ##\frac k 2 \left[ \sqrt{A^2 + R^2(\varphi - \vartheta)^2} - d\right]^2##.

    ##V = \frac k 2\left [A^2 + R^2(\varphi - \vartheta)^2 - 2d\sqrt{A^2 + R^2(\varphi - \vartheta)^2} + d^2\right]##.
    The constant terms in ##V## don't matter.
    ##\frac k 2 R^2(\varphi - \vartheta)^2## is a nice quadratic form already.
    The remaining term is ##-kd\sqrt{A^2 + R^2(\varphi - \vartheta)^2}
    = -kdA\sqrt{1 + (\frac R A)^2(\varphi - \vartheta)^2}##.
    Expand the square root to first order using the Binomial theorem (since the angles are small) and you get another constant, and another nice quadratic form.
     
    Last edited: Jun 9, 2014
  9. Jun 9, 2014 #8
    Jup, that makes much more sense than my notes. I will try tu use the same notation as you did.

    ##V = \frac k 2\left [A^2 + R^2(\varphi - \vartheta )^2 - 2d\sqrt{A^2 + R^2(\varphi - \vartheta)^2} + d^2\right]##
    if we forget about the constant terms
    ##V = \frac k 2 R^2(\varphi - \vartheta )^2 - kd\sqrt{A^2 + R^2(\varphi - \vartheta)^2}=##
    ##= \frac k 2 R^2(\varphi - \vartheta )^2 - kdA \sqrt{1 + (\frac R A)^2(\varphi - \vartheta )^2}=##
    ##= \frac k 2 R^2(\varphi - \vartheta )^2 - kdA(1 + \frac 1 2 (\frac R A)^2(\varphi - \vartheta )^2)##
    again forgetting about constants leaves me with

    ##V=\frac k 2 R^2(\varphi - \vartheta)^2 -\frac{kdR^2}{2A}(\varphi - \vartheta)^2=##
    ##=(\varphi - \vartheta)^2(\frac k 2 R^2-\frac{kdR^2}{2A})## for less writing lets say that ##C=\frac k 2 R^2-\frac{kdR^2}{2A}##

    so finally

    ##V=C(\varphi ^2-2\varphi \vartheta + \vartheta ^2)##

    Does this sound ok?
     
  10. Jun 9, 2014 #9

    AlephZero

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    That looks good to me.

    If you write the constant as ##\frac k 2 R^2(1 - \frac d A )##, you can see why the question said that ##d < A##. :smile:
     
  11. Jun 9, 2014 #10

    AlephZero

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    Actually, I think there are some terms missing from V. If you keep the ##x## displacements, the length of the spring is ##\sqrt{(A + \frac R 2 (\varphi^2 + \vartheta^2))^2 + R^2(\varphi - \vartheta)^2}##.

    If you go through the same process as before, that will give some more second order terms in ##\varphi## and ##\vartheta##, for example ##AR(\varphi^2 + \vartheta^2)##.

    The terms in ##\varphi## and ##\vartheta## higher than second order can be ignored.
     
  12. Jun 10, 2014 #11
    Huh, I agree.

    I did some calculations, and hopefully you won't find any mistakes:

    ##x_1=\frac 1 2 R\varphi ^2## and ##x_2=\frac 1 2 R\vartheta ^2## and
    ##y_1=R\varphi ## and ##y_2=R\vartheta ##

    The total stretch of the spring is than ##s=s_0+\Delta L##
    ##s=(A-d)+(\sqrt{[A+\frac R 2 (\varphi ^2+ \vartheta ^2)]^2+R^2[\varphi -\vartheta ]^2}-A)=##
    ##=\sqrt{[A+\frac R 2 (\varphi ^2+ \vartheta ^2)]^2+R^2[\varphi -\vartheta ]^2}-d##.

    Now ##s^2=[A+\frac R 2 (\varphi ^2+ \vartheta ^2)]^2+R^2[\varphi -\vartheta ]^2+d^2-2d\sqrt{[A+\frac R 2 (\varphi ^2+ \vartheta ^2)]^2+R^2[\varphi -\vartheta ]^2}=## (In every step possible I will simply forget about all the constants)
    ##=AR (\varphi ^2+ \vartheta ^2)+R^2[\varphi -\vartheta ]^2-2d\sqrt{A^2[1+\frac{R}{2A}(\varphi ^2+ \vartheta ^2)^2+\frac{R^2}{A^2} (\varphi -\vartheta )^2]} =##
    ##=AR (\varphi ^2+ \vartheta ^2)+R^2[\varphi -\vartheta ]^2-2dA(1+\frac 1 2[\frac{R}{2A}(\varphi ^2+ \vartheta ^2)^2+\frac{R^2}{A^2}(\varphi -\vartheta )^2])=##
    ##=AR (\varphi ^2+ \vartheta ^2)+R^2[\varphi -\vartheta ]^2-\frac{dR}{2}(\varphi ^2+ \vartheta ^2)-\frac{dR^2}{A}[\varphi -\vartheta ]^2##

    Using ##C=AR-\frac{dR}{2}## and ##D=R^2-\frac d A R^2##

    ##s^2=C(\varphi ^2+ \vartheta ^2)+D[\varphi -\vartheta ]^2##

    So finally ##V=\frac 1 2 k[\varphi ^2(C+D)-2\varphi \vartheta D +\vartheta ^2(C+D)]## which is the same as

    ##V=\frac 1 2 k \eta ^T\begin{bmatrix}
    C+D & -D\\
    -D&C+D
    \end{bmatrix}\eta##


    Finally I can start calculating ##det(V-\omega ^2 T)=det(V-\frac{J\omega ^2 }{k}T)=0## where ##\omega _k^2=\frac k J##.

    If everything is ok, I get two eigenvalues of the system: ##\frac{\omega ^2}{\omega _k^2}=\lambda=C+D\pm D## and accordingly two eigen vectors that describe possible movements of my system

    ## \eta_1=(1,1)## and ##\eta _2=(1,-1)## as expected.
     
    Last edited: Jun 10, 2014
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