Classical mechanics - small movements around equilibrium

[skrat]

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.

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 let's 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, therefore $x_2=\frac{R}{2}\vartheta ^2$

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?

 

[AlephZero]

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

 

[skrat]

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?

 

[skrat]

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 $V=\frac{1}{2}k(\Delta L)^2$

 

[AlephZero]

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 $\frac 1 2 k (L - 2R - d + \Delta L)^2$.

 

[skrat]

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...

 

[AlephZero]

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.

 

[skrat]

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 let's 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?

 

[AlephZero]

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 $d < A$.

 

[AlephZero]

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.

 

[skrat]

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.