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Lagrangian of a Rotating Mass on a Spring

  1. Oct 30, 2009 #1
    1. The problem statement, all variables and given/known data
    A point mass m slides without friction on a horizontal table at one end of a massless spring of natural length a and spring constant k. The other end of the spring is attached to the table so that it can rotate freely without friction. The spring is driven by a motor beneath the table so that the spring and mass are constrained to move around the origin with angular frequency [tex]\omega[/tex] (ignore bending of the spring and assume it always remains radially outward from the origin).

    a) Using Cartesian coordinates, write down expressions for the kinetic energy of the system.
    b) Change to polar coordinates. Also give the expression for potential energy and the Lagrangian.
    c) How many degrees of freedom do you have for this system? Name them(it).
    d) Give the equation of motion for the mass



    2. Relevant equations

    Please note this is my first time with LaTeX, so it might be a little bumpy!

    L = T - U
    Spring Potential: [tex]U = \frac{1}{2} k \ast x^2[/tex]



    3. The attempt at a solution

    The kinetic energy is going to be given by the rotating mass plus the mass extending (or compressing) with the spring. Since a is the natural length of the spring, I let b be the distance it extends (or compresses).

    [tex] T = \frac{1}{2} m[\dot{x}^2 + \dot{y}^2 + (b')^2] [/tex]

    [tex] U = \frac{1}{2} k(r-a)^2 [/tex] <--Note the potential energy doesn't have to be in Cartesian.

    Now change T from Cartesian to polar:
    [tex]\dot{x} = -r \omega sin(\omega t)[/tex]
    [tex]\dot{y} = r \omega cos(\omega t)[/tex]


    [tex]T = \frac{1}{2} m (r^2 \omega^2 sin^2 (\omega t) + r^2 \omega^2 cos^2 (\omega t) + (r')^2)[/tex]

    [tex]T = \frac{1}{2} m (r^2 \omega^2 + (r')^2)[/tex]

    [tex]L = T - U = r^2 \omega^2 + (r')^2 - \frac{1}{2} k(r^2 -2ra + a^2)[/tex]

    [tex]\frac{dL}{dr} = 2 \omega^2 r - (rk - ra) = 2\omega^2 r -rk +ra[/tex]

    [tex]\frac{dL}{dr'} = 2r'[/tex]

    [tex]\frac{d}{dt} \frac{dL}{dr'} = 2r''[/tex]

    [tex]\frac{dL}{dr} - \frac{d}{dt} \frac{dL}{dr'} = 0[/tex]

    [tex]2 \omega^2 r - rk + ra - 2r'' = 0
    [/tex]



    So the last line gives the equation of motion for the mass. There is one degree of freedom in the radial distance r.

    I am unsure if I set the problem up correctly. Can I just say let b be the distance the spring stretches or compresses? And then eventually make that r' (since it is the rate of change of r)
     
  2. jcsd
  3. Oct 30, 2009 #2
    https://www.physicsforums.com/showthread.php?t=349763

    Mmm, are you sure? I think, rather than starting with the kinetic energy you should first look at the constraints of the problem. How many are there and what are they?
     
  4. Oct 30, 2009 #3
    The only constraint that I can think of is given in the problem, that the spring and the mass are constrained to move around the origin with angular frequency [tex]\omega[/tex].

    We know that
    [tex]\omega =\frac{v}{r}[/tex]

    So should I solve for v, which will be dependent on r, and then use that in the kinetic energy formula? I really don't know what else to look for.
     
  5. Oct 31, 2009 #4
    There's also one not given in the problem that you have implicitly assumed: [itex]\dot{z}=z=0[/itex]. So there are two constraints for a single particle, making the total degrees of freedom 1.


    I wouldn't use this. Stick with x-dot, y-dot, and r-dot's. Using these, your kinetic energy should be

    [tex]
    T=\frac{1}{2}m\left(\dot{x}^2+\dot{y}^2\right)
    [/tex]

    Potential Energy

    [tex]
    U=\frac{1}{2}k\left(x^2+y^2\right)
    [/tex]

    Lagrangian

    [tex]
    L=T-U=T=\frac{1}{2}m\left(\dot{x}^2+\dot{y}^2\right)-\frac{1}{2}k\left(x^2+y^2\right)
    [/tex]

    which I just noticed you seem to have dropped your [itex]m[/itex]'s from your kinetic energy in your Lagrangian.
     
  6. Oct 31, 2009 #5
    I should be a little more careful in saying this. If we consider polar/cylindrical coordinates, [itex](r,\phi,z)[/itex], then since [itex]\phi=\omega t[/itex] and [itex]z=0[/itex], the only available coordinate to work with is [itex]r[/itex]. If we considered the Cartesian coordinates, there are still the 3 natural degrees of freedom, but only [itex]z[/itex] is constrained making us have 2 total degrees of freedom: [itex]x,y[/itex]. I used the first of the above two because you are asked to convert this into polar coordinates.
     
  7. Oct 31, 2009 #6

    Where in here does it take into account the kinetic energy of the spring extending or compressing? Or I think that is contained in the xdot and ydot term?

    [tex]T=\frac{1}{2}m\left(\dot{x}^2+\dot{y}^2\right)[/tex]

    Also in the potential energy term, wouldn't a have to appear somewhere, since this is the natural length, and when
    [tex]x^2 + y^2 = a^2[/tex]
    then
    [tex]U = 0[/tex]

    So shouldn't the potential be written as
    [tex] U = \frac{1}{2} k(r-a)^2[/tex]
     
  8. Oct 31, 2009 #7
    You don't put a potential energy term into the kinetic energy. The kinetic energy is the measure of how much is 'happening' (how much it is moving around) while the potential energy is the measure of how much could happen, but isn't yet. So no, the kinetic energy is just proportional to the time derivatives of the position of the mass.

    Your variable [itex]r[/itex] takes into account the variation of [itex]a[/itex], so it is not necessary in the potential.

    I also just noticed that you inadvertently added the correct term in the polar Lagrangian, but you should find by the conversion of Cartesian to polar coordinates,

    [tex]
    \dot{x}^2+\dot{y}^2=\dot{r}^2+r^2\omega^2
    [/tex]

    (I'll leave this for you to prove...)
     
  9. Oct 31, 2009 #8
    I was referring to the actual movement of the mass when the spring extends, not the potential, but I see that this is in the xdot and ydot terms.

    The potential would be 0 when r = a. So the potential you have written:
    [tex] U = \frac{1}{2} k(x^2 + y^2) = \frac{1}{2} kr^2 [/tex]

    Would not be equal to zero when r = a.

    Hmmm, I do not see this.

    [tex]x = rcos(\omega t)[/tex]
    [tex]y = rsin(\omega t)[/tex]
    [tex]\dot{x} = -r\omega sin(\omega t)[/tex]
    [tex]\dot{y} = r\omega cos(\omega t)[/tex]
    [tex]\dot{x}^2 + \dot{y}^2 = r^2 \omega^2[/tex]
     
  10. Oct 31, 2009 #9
    Well you could do one of the following: add the [itex]a[/itex] into the potential, let [itex]r=a[/itex] and solve accordingly, or replace all the [itex]r[/itex] terms with [itex]a[/itex]. I would say that one of the latter two would be a better choice (I choose the middle option).


    You are neglecting the fact that [itex]r[/itex] changes with time:

    [tex]x=r\cos[\omega t]\rightarrow\dot{x}=\dot{r}\cos[\omega t]-r\sin[\omega t]\omega[/tex]

    [tex]y=r\sin[\omega t]\rightarrow\dot{y}=\dot{r}\sin[\omega t]+r\cos[\omega t]\omega[/tex]

    [tex]\dot{x}^2=\dot{r}^2\cos^2[\omega t]+r^2\omega^2\sin^2[\omega t]-r\dot{r}\omega\sin[\omega t]\cos[\omega t][/tex]

    [tex]\dot{y}^2=\dot{r}^2\sin^2[\omega t]+r^2\omega^2\cos^2[\omega t]+r\dot{r}\omega\sin[\omega t]\cos[\omega t][/tex]

    [tex]\dot{x}^2+\dot{y}^2=\dot{r}^2+r^2\omega^2[/tex]
     
  11. Oct 31, 2009 #10

    Ah! How did I not see that. That makes sense.

    About the potential, it makes sense to me to write it like:
    [tex] U = \frac{1}{2} k(r-a)^2[/tex]

    so now we have
    [tex] L = T-U=\frac{1}{2} m(\dot{r}^2 + r^2 \omega^2) - \frac{1}{2}k(r^2 - 2ra + a^2)[/tex]

    [tex]\frac{dL}{dr} = mr\omega^2 - kr + ka[/tex]


    [tex] \frac{dL}{d\dot{r}} = m\dot{r} \frac{d}{dt} \frac{dL}{d\dot{r}} = m\ddot{r}[/tex]


    [tex]\frac{dL}{dr} - \frac{d}{dt} \frac{dL}{d\dot{r}} = 0 = mr\omega^2 - kr +ka -m\ddot{r}[/tex]

    The equation of motion for r:
    [tex]m\ddot{r} + (k-m\omega^2)r = ka[/tex]

    The final part of the question asks if the radial motion is a simple harmonic oscillator. Just from intuition, I would say yes, since there would be an outward force from the rotation at [tex]\omega[/tex] which cause a displacement from equilibrium in the spring, causing it to be an SHO.

    Looking at the equation of motion, a normal SHO would look like [tex] \frac{d^2r}{dt} + B\frac{\partialr}{\partialt} + r = F(t)[/tex]


    So in this case we have B = 0 because there is no damping, and the driving force is equal to ka.
     
    Last edited: Oct 31, 2009
  12. Oct 31, 2009 #11
    Be careful now, the equation you have here is for a driven harmonic oscillator. The simple harmonic oscillator is given by

    [tex]
    \ddot{x}=\frac{k}{m}x
    [/tex]

    The equation of motion you have is

    [tex]
    \ddot{x}=\frac{k}{m}a-\left(\frac{k}{m}-\omega^2\right)r
    [/tex]

    which is similar to the SHO, but not exactly.
     
  13. Oct 31, 2009 #12
    Ok, so this is not a SHO then, but it is a driven harmonic oscillator.

    Thanks for all of you're help!!
     
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