Question about using the Lagrangian for a spinning rubber band problem

In summary: L = T-V = 1/2 M v^2 - 1/2 k x^2 = 1/2 {\cal L}^2/r^3 + 1/2 M \dot r^2 - 1/2 k ( 2 \pi r - 2\pi R)^2$$
  • #1
guv
123
22
Hello, I am having trouble applying Lagrangian to this problem:

A uniform thin circular rubber band of mass ##M ## and spring constant k has an original radius ##R##. Now it is tossed into the air. Assume it remains circular when stabilized in air and rotates at angular speed ##\omega## about its center uniformly. Which of the following gives the new radius of the rubber band?

The answer for the equilibrium solution is $$(4\pi^2kR)/(4\pi^2k − M\omega^2 ) $$

This can be found relatively easily through force analysis and Newton's 2nd law.

When trying to find the same solution through Lagrangian, I found the solution comes out to be ##(4\pi^2kR)/(4\pi^2k + M\omega^2 ) ##

Here is how I am using Lagrangian, suppose the radius of the ring is ##r## as it's spinning and it's radial velocity is ##\dot r##
$$L = T-V = 1/2 M v^2 - 1/2 k x^2 = 1/2 M ((r \omega)^2 + \dot r^2) - 1/2 k ( 2 \pi r - 2\pi R)^2$$
Applying Eular-Lagrange equation:
$$\frac{d}{d t} \frac{\partial L}{\partial \dot r} = \frac {\partial L}{\partial r}$$

Because ##\omega## is a function of ##r## and the angular momentum of the system cannot change, this allows the Lagrangian to be written in terms of angular momentum ##\cal L##.
$$L = T-V = 1/2 M v^2 - 1/2 k x^2 = 1/2 {\cal L}^2/r^3 + 1/2 M \dot r^2 - 1/2 k ( 2 \pi r - 2\pi R)^2$$

The algebra quickly leads to
$$M \ddot r = - {\cal L}^2/M r^3 - 4 \pi^2 k (r - R) = - M r \omega^2 - 4 \pi^2 k (r - R)$$

This leads to the equilibrium solution when ##\ddot r = 0##
$$r = (4\pi^2kR)/(4\pi^2k + M\omega^2 ) $$

What am I doing wrong?

Thanks,
 
Last edited:
Physics news on Phys.org
  • #2
guv said:
$$L = T-V = 1/2 M v^2 - 1/2 k x^2 = 1/2 M ((r \omega)^2 + \dot r^2) - 1/2 k ( 2 \pi r - 2\pi R)^2$$
You could just write$$\mathcal{L} = \frac{1}{2}Mr^2 \omega^2 - 2\pi^2 k(r-R)^2$$then euler lagrange gives straight away$$M\omega^2 r - 4\pi^2 k (r-R) = 0$$
guv said:
$$L = T-V = 1/2 M v^2 - 1/2 k x^2 = 1/2 {\cal L}^2/r^3 + 1/2 M \dot r^2 - 1/2 k ( 2 \pi r - 2\pi R)^2$$

Here the first term should be ##\frac{1}{2}Mr^2 \omega^2 = \frac{1}{2} \frac{L^2}{Mr^2}##, and not ##\frac{1}{2}\frac{L^2}{r^3}##. Also note that ##\dot{r}(t) = 0##
 
  • Like
Likes vanhees71 and berkeman
  • #3
@etotheipi Thanks!

I had initially thought about writing Lagrangian this way which indeed yields the correct solution. However, a thought experiment convinced me, that there may be more to this problem.

Imagine a spinning cylinder with radius ##r_{eq} = (4 \pi^2 kR)/(4 \pi^2 k − M \omega^2 ) ## where ##R## is the radius of a relaxed rubber band. It would suggest if the spin angular velocity of the cylinder is ##\omega##, the rubber band will not press onto the cylinder, normal force ##F_N = 0 ##, this I think we can all agree. Let's ignore gravitational force for now (there are many ways to counter gravitational force)

Next imagine if the cylinder's radius is not ##r_{eq}## but ##r_{eq}+\delta r## and spins at angular velocity ##\omega_1 = r_{eq}^2 \omega / (r_{eq} + \delta r)^2##. By the above analysis, we can agree now the normal force is no longer zero, the rubber band can press onto the cylinder if the following SHM condition is satisfied ##4 k \pi^2 - 3 M \omega^2 > 0## Careful analysis will show with such a set up the tension in the rubber band exceeds what would be needed for centripetal acceleration so it would press onto the cylinder.

Next imagine the cylinder suddenly disappears, this will cause the rubber band to contract and expand in a SHM with equilibrium , in fact we could find this SHM's angular frequency to be $$\omega_{osc} = \sqrt{\frac{4 k \pi^2 - 3 M \omega^2}{M}}$$

This is where the Lagrangian in my first post comes from. Right now I am not convinced that the equilibrium radius of the 2nd dynamics setup should be significantly different (##\pm##) from the 1st static setup. I am wary that the solution from the Lagrangian seems to miss a Coriolis term when the rubber band moves through the equilibrium position (defined as in the rubber band's non-inertial frame, acceleration of an observer is 0) as the rubber band is rotating. If we were to add an angular coordinate, it seems to contradict with the symmetric set up of the problem...

Hope this clarifies things.
 
Last edited:
  • #4
There are two degrees of freedom: the radius ##r## and the angular orientation ##\theta##. The mistake in the original post was to use angular momentum conservation to eliminate ##\dot \theta## (or ##\omega##) in the Lagrangian. This makes it look like the system has only one degree of freedom, ##r##. You do not get the correct equation of motion for ##r## this way.

Instead, express the Lagrangian in terms of both degrees of freedom. Derive equations of motion for both ##\theta(t)## and ##r(t)##. The equation of motion for ##\theta(t)## will yield the conservation of angular momentum. The differential equation for ##r(t)## will have the correct equilibrium solution ## r_{eq} =(4\pi^2kR)/(4\pi^2k − M\omega^2 ) ##.

When I use the equations of motion to derive ##\omega_{osc}## for small oscillations about the equilibrium radius ##r_e##, I get ##\large \omega_{osc} = \sqrt{\frac{4 k \pi^2 + 3 M \omega^2}{M}}## instead of ##\large \omega_{osc} = \sqrt{\frac{4 k \pi^2 - 3 M \omega^2}{M}}##.
 
  • Like
Likes guv and JD_PM
  • #5
guv said:
Here is how I am using Lagrangian, suppose the radius of the ring is ##r## as it's spinning and it's radial velocity is ##\dot r##
$$L = T-V = 1/2 M v^2 - 1/2 k x^2 = 1/2 M ((r \omega)^2 + \dot r^2) - 1/2 k ( 2 \pi r - 2\pi R)^2$$

Taking ##\theta## and ##r## as generalized coordinates (As TSny stated) completely describes the configuration of the system. This choice leads to the following Lagrangian

$$\mathcal{L} = \frac 1 2 M r^2 \dot \theta^2 - 2 k \pi^2 (r-R)^2$$
 
  • #6
JD_PM said:
$$\mathcal{L} = \frac 1 2 M r^2 \dot \theta^2 - 2 k \pi^2 (r-R)^2$$
There will also be the term ##\frac 1 2 M \dot r^2##
 
  • Like
Likes JD_PM
  • #7
Thanks Tsny, this is helpful, I have wondered about introducing angular position in ##L## as I mentioned in the previous post. Recall angular momentum is a constant, as a constraint this can reduce the degree of freedom of the problem. Since the setup is symmetric about angular position, I am not convinced why adding a ##\theta## dependence would be necessary yet. Another thing that I still do not see is Coriolis term, I believe as the rubber band moving radially passing equilibrium position, there should be a Coriolis term.

About the oscillation frequency, it will come out '-' if you use the constant angular momentum constraint.

TSny said:
There are two degrees of freedom: the radius ##r## and the angular orientation ##\theta##. The mistake in the original post was to use angular momentum conservation to eliminate ##\dot \theta## (or ##\omega##) in the Lagrangian. This makes it look like the system has only one degree of freedom, ##r##. You do not get the correct equation of motion for ##r## this way.

Instead, express the Lagrangian in terms of both degrees of freedom. Derive equations of motion for both ##\theta(t)## and ##r(t)##. The equation of motion for ##\theta(t)## will yield the conservation of angular momentum. The differential equation for ##r(t)## will have the correct equilibrium solution ## r_{eq} =(4\pi^2kR)/(4\pi^2k − M\omega^2 ) ##.

When I use the equations of motion to derive ##\omega_{osc}## for small oscillations about the equilibrium radius ##r_e##, I get ##\large \omega_{osc} = \sqrt{\frac{4 k \pi^2 + 3 M \omega^2}{M}}## instead of ##\large \omega_{osc} = \sqrt{\frac{4 k \pi^2 - 3 M \omega^2}{M}}##.
 
  • #8
guv said:
Thanks Tsny, this is helpful, I have wondered about introducing angular position in ##L## as I mentioned in the previous post. Recall angular momentum is a constant, as a constraint this can reduce the degree of freedom of the problem.
But this changes the functional dependence of the Lagrangian ##L## on ##r##, so the equation of motion for ##r## will change (to an incorrect form).

Take a very trivial example of a particle moving freely along the x-axis. So, ##L = \frac 1 2 m \dot x^2##. Linear momentum will be conserved, so ##m \dot x = p = ##constant. What happens if you replace ##\dot x## by ##p/m## in ##L##? Can you then get the correct equation of motion for ##x(t)##?

Another thing that I still do not see is Coriolis term, I believe as the rubber band moving radially passing equilibrium position, there should be a Coriolis term.

There will not be any Coriolis force in the lab frame. Coriolis forces arise in rotating reference frames.

Suppose that in the lab frame the rubber band is spinning at a rate ##\omega_e## such that ##r## is constant at ##r_e##. Now produce a small perturbation such that the band executes radial SHM about ##r_e##. In the lab frame, ##\omega## varies as ##r## changes in a way that keeps the angular momentum constant.

But, suppose we switch to a rotating frame of reference that has a constant angular speed of ##\omega_e## relative to the lab. So, before the perturbation, the rubber band is at rest in this frame. After the perturbation, the rubber band will execute radial SHM and ##\omega## will vary as ##r## varies. But in this frame, the angular momentum will not be constant. At instants of time when ##r = r_e##, the rubber band will have zero angular momentum in this rotating frame. But, for other values of ##r## the angular momentum will be nonzero. The varying angular momentum in this frame can be accounted for by the Coriolis force which acts tangentially to the rubber band and provides the necessary torque.
 
  • Like
Likes JD_PM and etotheipi
  • #9
Hi TSny thanks for the correction, I indeed forgot about the radial velocity term.

TSny said:
When I use the equations of motion to derive ##\omega_{osc}## for small oscillations about the equilibrium radius ##r_e##, I get ##\large \omega_{osc} = \sqrt{\frac{4 k \pi^2 + 3 M \omega^2}{M}}## instead of ##\large \omega_{osc} = \sqrt{\frac{4 k \pi^2 - 3 M \omega^2}{M}}##.

I am afraid here comes a trivial question: how did you get ##\omega_{osc}##?

Once I got the equations of motion

$$\frac{d}{dt} \Big( M r^2 \dot \theta \Big) =0 \Rightarrow M r^2 \dot \theta = L \tag1$$

$$M(\ddot r -r \dot \theta^2) + 4k \pi^2(r - R) = 0 \tag2$$

I assumed the rubber did not rotate (##L=0##) and got the expected result ##\omega_{osc} = \sqrt{\frac{4 k \pi^2}{M}}##. OK.

To get ##\omega_{osc}## when the rubber rotates, l proceeded as follows

I solved for ##\dot \theta## in ##(1)## and plugged it into ##(2)## to get

$$M(\ddot r -r \omega^2) + 4k \pi^2r = 4k \pi^2R$$

$$\ddot r +\Big(\frac{-M \omega^2 + 4k \pi^2}{M}\Big)r = \frac{4k \pi^2R}{M} \tag3$$

The solution of this differential equation is

$$r(t) = \frac{ 4k \pi^2 R}{ 4k \pi^2-M\omega^2} + A\exp\Big(t\sqrt{\frac{ M\omega^2 - 4\pi^2k}{M}}\Big) + B\exp\Big(-t\sqrt{\frac{ M\omega^2 - 4\pi^2k}{M}}\Big) \tag4$$

Thus I get ##\omega_{osc} = \sqrt{\frac{-4 k \pi^2 + M \omega^2}{M}}##
 
  • #10
JD_PM said:
I am afraid here comes a trivial question: how did you get ##\omega_{osc}##?
This is not trivial. It requires some manipulations and making an assumption of small displacements from the equilibrium radius ##r_e##.

To get ##\omega_{osc}## when the rubber rotates, l proceeded as follows

I solved for ##\dot \theta## in ##(1)## and plugged it into ##(2)## to get

$$M(\ddot r -r \omega^2) + 4k \pi^2r = 4k \pi^2R$$
Ok, but note that ##\omega## is not a constant. It is a function of ##r##. From the conservation of angular momentum, you can see that ##r^2 \omega = r_e^2 \omega_e##, where ##\omega_e## is the rate of spin when the band is in equilbrium at radius ##r_e##. So, you can express ##\omega## in terms of ##\omega_e##, ##r_e##, and ##r##.

Substituting for ##\omega## in the differential equation for ##r##, you get ##\ddot r## expressed as a function of ##r##. To get a differential equation corresponding to SHM, you need to make a small-displacement approximation. Let ##s## be the displacement of ##r## from ##r_e##, so ##r = r_e+s##. The equation for ##\ddot r## can then be converted to an equation that expresses ##\ddot s## as a function of ##s##. This can then be approximated to first order in the small quantity ##s/r_e##.
 
  • Like
Likes JD_PM
  • #11
@TSny I'm struggling a little with this too. If we have$$\mathcal{L} = \frac{1}{2}Mr^2 {\dot{\theta}}^2 + \frac{1}{2}M\dot{r}^2 - 2\pi^2 k(r-R)^2$$Then for the coordinate ##r##,$$\frac{\partial \mathcal{L}}{\partial r} = Mr\dot{\theta}^2 - 4\pi^2 k (r-R)$$ $$\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{r}} = \frac{d}{dt} M\dot{r} = M\ddot{r}$$and for ##\theta##,$$\frac{\partial \mathcal{L}}{\partial \theta} = 0$$ $$\frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{\theta}} = \frac{d}{dt} Mr^2 \dot{\theta} = 2Mr \dot{r}\dot{\theta} + Mr^2 \ddot{\theta}$$So the equations of motion would be$$M\ddot{r} = Mr\dot{\theta}^2 - 4\pi^2 k(r-R)$$ $$2\dot{r}{\dot{\theta}} + r\ddot{\theta} = 0$$Now for the equilibrium radius ##r_e##, where ##\ddot{r}(r_e) = 0##, we have$$Mr_e [\dot{\theta}(r_e)]^2 = 4\pi^2 k (r_e - R)$$Angular momentum conservation also gives$$r_e^2 [\dot{\theta}(r_e)] = r^2\dot{\theta}$$Maybe then we say $$Mr\dot{\theta}^2 = Mr\left(\frac{r_e^2}{r^2}[\dot{\theta}(r_e)]\right)^2 = \frac{M r_e^4 [\dot{\theta}(r_e)]^2}{r^3}$$Which gives us, from the original equation of motion,$$M\ddot{r} = \frac{M r_e^4 [\dot{\theta}(r_e)]^2}{r^3} - 4\pi^2 k(r-R) = \frac{r_e^3}{r^3} \left[4\pi^2 k (r_e - R)\right] - 4\pi^2 k(r-R)$$ $$M\ddot{r} = 4\pi^2 k \left(r\left[\frac{r_e^4}{r^4} - 1\right] + R\left[1 - \frac{r_e^3}{r^3}\right] \right)$$But I don't know what to do next. I suppose at some stage we will require the change of coordinates ##r = r_e + \varepsilon##. Is this along the right lines so far? It seems to give the expected behaviour of ##\ddot{r}## in the cases ##r = r_e##, ##r > r_e## and ##r<r_e##.
 
Last edited by a moderator:
  • Like
Likes JD_PM
  • #12
etotheipi said:
$$M\ddot{r} = \frac{M r_e^4 [\dot{\theta}(r_e)]^2}{r^3} - 4\pi^2 k(r-R) = \frac{r_e^3}{r^3} \left[4\pi^2 k (r_e - R)\right] - 4\pi^2 k(r-R)$$
OK. Looks good. To find the equation for small oscillations about ##r = r_e##, let ##s## denote the deviation in ##r## from ##r_e##. So, ##r = r_e+s##. Hence, ##\ddot r = \ddot s##. Substitute ##r = r_e+s## in the right side of the equation of motion. Note $$\frac 1 {r^3} = \frac 1 {r_e^3(1+\frac s {r_e})^3} \approx \frac 1 {r_e^3}(1-3\frac s {r_e})$$ Things will simplify.
 
  • Like
Likes JD_PM and etotheipi
  • #13
Cool, that works, I get$$M\ddot{\varepsilon} = -4\pi^2 k \left(4- \frac{3R}{r_e} \right)\varepsilon$$And since$$4- \frac{3R}{r_e} = 4- \left(3 - \frac{3M\omega_0^2}{4\pi^2 k} \right) = 1 + \frac{3M\omega_0^2}{4\pi^2 k}$$we obtain$$\omega_{osc} = \sqrt{\frac{4\pi^2 k + 3M\omega_0^2}{M}}$$which was your result. Thank you!
 
  • Like
Likes JD_PM
  • #14
etotheipi said:
Cool, that works, I get$$M\ddot{\varepsilon} = -4\pi^2 k \left(4- \frac{3R}{r_e} \right)\varepsilon$$And since$$4- \frac{3R}{r_e} = 4- \left(3 - \frac{3M\omega_0^2}{4\pi^2 k} \right) = 1 + \frac{3M\omega_0^2}{4\pi^2 k}$$we obtain$$\omega_{osc} = \sqrt{\frac{4\pi^2 k + 3M\omega_0^2}{M}}$$which was your result. Thank you!
Nice!
 
  • Like
  • Love
Likes JD_PM and etotheipi
  • #15
I am uneasy with the expression for the equilibrium radius$$r_{eq}=\frac{(4\pi^2kR)}{(4\pi^2k − M\omega^2 )}.$$It relates ##r_{eq}## as a dependent variable to the spin angular speed ##\omega## as a dependent variable. When I see negative signs in denominators, I need to understand what happens when the denominator becomes zero or turns negative. Usually, there is some physical constraint that prevents this from happening. Here, I see no such constraint because it seems to me that the rubber band can be set spinning with an arbitrarily large ##\omega.## My first question is, "how does one interpret the denominator going to zero?"

A second question I have has to do with the potential energy term in the Lagrangian. It was written down as ##V=\dfrac{1}{2}k(2\pi r-2\pi R)^2##. This is the potential energy stored in a straight rubber band obeying Hooke's law when is stretched by a linear force from initial length ##2\pi R## to final length ##2\pi r## and can be obtained by doing a standard line integral. Here we have a radial symmetric force stretching the rubber ring. Isn't it more appropriate to find the work done by this radial force as the ring expands from ##R## to ##r## and use the result to find the potential energy?
 
  • Like
Likes JD_PM
  • #16
kuruman said:
A second question I have has to do with the potential energy term in the Lagrangian. It was written down as ##V=\dfrac{1}{2}k(2\pi r-2\pi R)^2##. This is the potential energy stored in a straight rubber band obeying Hooke's law when is stretched by a linear force from initial length ##2\pi R## to final length ##2\pi r## and can be obtained by doing a standard line integral. Here we have a radial symmetric force stretching the rubber ring. Isn't it more appropriate to find the work done by this radial force as the ring expands from ##R## to ##r## and use the result to find the potential energy?

Wouldn't it give the same result? If a segment subtending ##d\theta## experiences a radial force of ##N = Td\theta##, then if ##T = k\delta = k(r-R)##$$dW = 2\pi k (r-R) d\theta dr \implies W = 2\pi^2 k (r-R)^2$$Is the issue that Hooke's law is no longer a good model for a circular spring?
 
  • #17
etotheipi said:
s the issue that Hooke's law is no longer a good model for a circular spring?
No, the issue is whether the circular geometry gives the same result as the linear geometry. Stated differently, if you stretch a linear band from ##2\pi R## to ##2\pi r## is it obvious that you do zero work to then bend it into a circle of radius ##r##?

What I am trying to ascertain is whether the first question in post #15 arises because the wrong potential energy is used in the Lagrangian resulting in an incorrect form for ##r_{eq}##.
 
  • Like
Likes etotheipi
  • #18
kuruman said:
I am uneasy with the expression for the equilibrium radius$$r_{eq}=\frac{(4\pi^2kR)}{(4\pi^2k − M\omega^2 )}.$$ When I see negative signs in denominators, I need to understand what happens when the denominator becomes zero or turns negative.
I understand the uneasy feeling with the denominator. However, I think it's correct. A similar expression arises in the simpler system shown
1595789555790.png


The mass slides without friction on the rod that maintains a constant angular speed ##\omega## in a horizontal plane. The mass is attached to a spring of natural length ##R##. If you derive an expression for the mass's equilibrium radius of rotation, ##r_{eq}##, you get a result that is very similar to the rubber band.

Added note: The equilibrium radius occurs when the "centrifugal force" ##m \omega^2 r## balances the spring force ##k(r-R)##. If the rate of spin ##\omega## is such that ##m \omega^2 > k##, then you can see that ##m \omega^2 r > k (r-R)## for all non-negative values of ##r##. So, in this case, there is no equilbrium position.
 
Last edited:
  • Like
Likes JD_PM and kuruman
  • #19
TSny said:
Added note: The equilibrium radius occurs when the "centrifugal force" balances the spring force . If the rate of spin is such that , then you can see that for all non-negative values of . So, in this case, there is no equilbrium position.
Ah! That's the missing piece. Thanks.
 
  • #20
TSny said:
I understand the uneasy feeling with the denominator. However, I think it's correct. A similar expression arises in the simpler system shown
View attachment 266890

The mass slides without friction on the rod that maintains a constant angular speed ##\omega## in a horizontal plane. The mass is attached to a spring of natural length ##R##. If you derive an expression for the mass's equilibrium radius of rotation, ##r_{eq}##, you get a result that is very similar to the rubber band.

Added note: The equilibrium radius occurs when the "centrifugal force" ##m \omega^2 r## balances the spring force ##k(r-R)##. If the rate of spin ##\omega## is such that ##m \omega^2 > k##, then you can see that ##m \omega^2 r > k (r-R)## for all non-negative values of ##r##. So, in this case, there is no equilbrium position.

Hi TSny, thanks for posting an extra problem. I wanted to explicitly get ##r_{eq}## and compare it to the one obtained in the previous problem. Here's what I've done.

Let's pick ##r## and ##\theta## as our generalized coordinates.

1595789555790.png


The kinetic energy of the system is

$$T = \frac 1 2 \Big( \frac 1 3 M L^2 \Big) \dot \theta^2 + \frac 1 2 m \dot r^2 + \frac 1 2 m r^2 \dot \theta^2$$

The potential energy of the system is

$$V= Mg \frac L 2 \sin \theta + mgr \sin \theta + \frac 1 2 k (r-R)^2$$

Thus the Lagrangian is

$$\mathcal{L} = \frac 1 6 M L^2 \dot \theta^2 + \frac 1 2 m \dot r^2 + \frac 1 2 m r^2 \dot \theta^2 -Mg \frac L 2 \sin \theta - mgr \sin \theta - \frac 1 2 k (r-R)^2$$

Where ##M## and ##L## are rod's mass and length respectively. We assume the spring to be massless.

The equation of motion for ##\theta## (applying the small angle approximation) is

$$(ML^2 + 3mr^2) \ddot \theta + Mg \frac {3L}{2} + 3mgr = 0$$

The equation of motion for ##r## (applying the small angle approximation) is

$$m \ddot r - mr \dot \theta^2 -mg\theta -k(r-R)=0$$

Thus ##r_{eq}## is

$$r_{eq}=\frac{kR-mg \theta}{k+m \omega^2}$$

Where ##\omega = \omega(r)##

We're only left to find ##\omega_{\text{osc}}##. Before doing so though, let's think of what I've obtained so far.

1) I did not explicitly get conservation of angular momentum. I indeed expected to get it as we're in the lab frame and there is no Coriolis Force providing an external torque. I think I am missing something but still did not find exactly what.

2) There's no -ive sign in the denominator of ##r_{eq}##. Honestly, I see no issue with that but I may be missing something and there may indeed exist one. Besides, I get ##r_{eq} (\theta)## which I think is a sign I did something wrong, as ##r_{eq}## should not depend on nothing but constants.
 
  • Like
Likes etotheipi
  • #21
JD_PM said:
The potential energy of the system is
$$V= Mg \frac L 2 \sin \theta + mgr \sin \theta + \frac 1 2 k (r-R)^2$$

I think the x-y plane is horizontal in the original statement so the first two of these terms should not be there

JD_PM said:
1) I did not explicitly get conservation of angular momentum. I indeed expected to get it as we're in the lab frame and there is no Coriolis Force providing an external torque. I think I am missing something but still did not find exactly what.

Also if you do let the ##y## axis be vertical so that we worry about gravity, angular momentum is now rightly not a conserved quantity since the two weights constitute external torques about the origin.
 
Last edited by a moderator:
  • Like
Likes JD_PM
  • #22
etotheipi said:
I think the x-y plane is horizontal in the original statement so the first two of these terms should not be there
Also if you do let the ##y## axis be vertical so that we worry about gravity, angular momentum is now rightly not a conserved quantity since the two weights constitute external torques about the origin.

Argh ~ 1:00 a.m. over here but I couldn’t resist to answer now!

So your point is that we should take the whole system to be in the x-y plane, so that no gravitational potential energy plays a role. OK. If we do so we indeed get angular momentum conservation and the ##\theta## term in ##r_{eq}## goes away.

I just happened to approach it as we usually do with pendulums: take your zero of gravitational potential energy passing (horizontally) through the axis of rotation. I suspect TSny was thinking as you suggest though.
 
  • Like
Likes etotheipi
  • #23
JD_PM said:
So your point is that we should take the whole system to be in the x-y plane, so that no gravitational potential energy plays a role. OK. If we do so we indeed get angular momentum conservation and the ##\theta## term in ##r_{eq}## goes away.

I just happened to approach it as we usually do with pendulums: take your zero of gravitational potential energy passing (horizontally) through the axis of rotation. I suspect TSny was thinking as you suggest though.
Yes, no gravity. I was also making it even simpler by having the rod rotate with a constant angular speed ##\omega## by some external mechanism. So, the Lagrangian has only one degree of freedom: ##r## for the particle. The expression for the equilibrium value ##r_e## will be very similar to the expression for the rubber band. It will have the minus sign in the denominator. The rotating rod example was meant to illustrate a very simple system that has this divergent behavior for ##r_e##. @kuruman and I were a little concerned with this at first.
 
  • Like
Likes JD_PM and etotheipi
  • #24
Reply to @JD_PM post #20.

When I did this problem, I assumed a massless rod and no gravity, i.e. motion in the ##xy##-plane. I got the expression $$r_{eq}=\frac{\omega_0^2}{\omega_0^2-\omega^2}R$$where ##\omega_0=\sqrt{k/m}## is the natural frequency of the spring-mass system.
 
  • Like
Likes JD_PM, etotheipi and TSny
  • #25
Nice so let our system move over the horizontal plane, have a massless rod and rotate (with constant angular speed ##\omega##) due to some external mechanism. Then the system can be described only by the generalized coordinate ##r##.

The kinetic energy is

$$T = \frac 1 2 m \dot r^2 + \frac 1 2 m r^2 \omega^2$$

The potential energy is

$$V = \frac 1 2 k (r-R)^2$$

The Lagrangian is

$$\mathcal{L} = \frac 1 2 m \dot r^2 + \frac 1 2 m r^2 \omega^2 - \frac 1 2 k (r-R)^2$$

The equation of motion is

$$m (\ddot r - r \omega^2) + k(r-R)=0$$

The equilibrium position (i.e. when ##\ddot r = 0##) is

$$r_e = \frac{k}{k-m\omega^2}R, \ \text{or defining} \ k:=\omega_0^2m \Rightarrow r_e = \frac{\omega_0^2}{\omega_0^2-\omega^2}R$$

As kuruman stated.

PS: when ##\omega := \omega(r)## (i.e. not constant) and still assuming no gravity and massless rod, ##\omega_{\text{osc}}## can be obtained in a really similar way as before. The answer I get is

$$\omega_{\text{osc}} = \sqrt{\frac{k \omega_0^2 + 3k\omega^2}{m\omega_0^2}} \ \text{or} \ \omega_{\text{osc}} = \sqrt{\frac{k^2 + 3mk\omega^2}{mk}}$$
 
  • #26
I agree with your results. Here is a plot of the orbit over one revolution. The input parameters I used were
##\omega_0=10~##i.s.u. (i.s.u. = "in some units")
##\omega=0.04~\omega_0##
##R=1~##i.s.u.
##\epsilon_0 = 0.025~r_{eq}## (amplitude of small oscillations.)

The orbit appears closed but it isn't.

Orbit.png
 

Related to Question about using the Lagrangian for a spinning rubber band problem

1. What is the Lagrangian method and how is it used in physics?

The Lagrangian method is a mathematical approach used in classical mechanics to describe the motion of a system. It involves using the Lagrangian function, which is a combination of the kinetic and potential energies of the system, to derive the equations of motion.

2. Can the Lagrangian method be applied to spinning objects, such as a rubber band?

Yes, the Lagrangian method can be applied to spinning objects as long as the system can be described by a set of generalized coordinates and has a well-defined kinetic and potential energy.

3. What is the advantage of using the Lagrangian method over other methods in physics?

The Lagrangian method is advantageous because it allows for a more elegant and concise formulation of the equations of motion compared to other methods, such as Newton's laws of motion. It also takes into account all the constraints and forces acting on the system, making it a more comprehensive approach.

4. How do you incorporate the spinning motion into the Lagrangian for a rubber band problem?

To incorporate the spinning motion into the Lagrangian, you would need to add a term for the rotational kinetic energy to the Lagrangian function. This term would depend on the moment of inertia of the rubber band and its angular velocity.

5. Are there any limitations to using the Lagrangian method for a spinning rubber band problem?

One limitation of using the Lagrangian method for a spinning rubber band problem is that it assumes the rubber band is a continuous and homogeneous object. This may not hold true in real-life scenarios, and other methods may need to be used to accurately describe the motion of the rubber band.

Similar threads

  • Advanced Physics Homework Help
Replies
6
Views
1K
  • Advanced Physics Homework Help
Replies
16
Views
988
Replies
6
Views
1K
  • Advanced Physics Homework Help
Replies
0
Views
767
  • Advanced Physics Homework Help
Replies
11
Views
1K
Replies
12
Views
637
  • Advanced Physics Homework Help
Replies
2
Views
276
  • Advanced Physics Homework Help
Replies
9
Views
1K
  • Advanced Physics Homework Help
Replies
9
Views
608
  • Advanced Physics Homework Help
Replies
1
Views
175
Back
Top