Classical mechanics - small movements around equilibrium

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Homework Statement


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

Homework Equations





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?
 

Answers and Replies

  • #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
 
  • #3
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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?
 
  • #4
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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##
 
  • #5
AlephZero
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I think that's what you meant and that ##-x_1## is just a typo. right?
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!

Where I can't see how this helps me since ##V=\frac{1}{2}k(\Delta L)^2##
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##.
 
  • #6
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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....
 
  • #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:
  • #8
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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?
 
  • #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:
 
  • #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.
 
  • #11
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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.
 
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