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Homework Help: SOS Problem with lagrange derivation

  1. Dec 25, 2009 #1
    SOS .. Problem with lagrange derivation!!

    1. The problem statement, all variables and given/known data

    Im having a hard time with the problem illustrated in the following figure:


    it said that solve the equations of motion for a coupled oscillators system consists of a metallic smooth ring of mass M and radius R oscillates in its own plane with one point fixed, along with it a particle of mass M sildes without friction on it. This particle is attached to the point of support of the ring by a massless spring of stiffness constant k and unstreched length 2R, taking in consideration only small oscillations about the equilibrium configuration
    In this problem the given parameters are as follows:

    R (radius) = 4.0 cm ,
    M (mass) = 13.0 g,
    k (spring constant) = 8.0 N/m

    2. Relevant equations

    I have to get somehow to the following equations


    3. The attempt at a solution

    i was thinking that for the sliding mass, x=2Rcos(theta+phi) and y=2Rsin(theta+phi)
    and for the ring its x=2Rsin(theta) and y = 2Rcos(theta).. and then the potential energy of the system would be = -2Rmgcos(theta)-2Rcos(theta+phi)
    and the kinetic energy is = 0.5*m*(x-dot^2(for the sliding mass) + x-dot^2(for the ring)

    I know what im doing is wrong coz no matter how i look at the questions I still cant find out how to reach the equations they reached..

    I appreciate any help .. thanks in advance ..
  2. jcsd
  3. Dec 25, 2009 #2
    Re: SOS .. Problem with lagrange derivation!!

    Wow! This is hard! Let me try:

    1. Find the potential energy of an arbitrary small arc that of the circle that is displaced from the vertical by an angle [tex]\theta[/tex] with respect to the vertical. Then line-integrate it to find the total potential energy. Keep in mind that if the metallic ring has a uniform mass, then its mass density is M/2[tex]\pi[/tex]R (Why?)

    2. Find the kinetic energy of the ring. Remember that, for a rigid body(the ring), the rotational kinetic energy is [tex]\frac{1}{2}I\dot{R}[/tex] where "I" is the moment of inertial around its pivot.

    3. Find the Kinetic Energy of the point-mass.

    4. Find the potential Energy of the point-mass.

    I am not sure how the "small oscillations" thing fits into it.
  4. Dec 25, 2009 #3
    Re: SOS .. Problem with lagrange derivation!!

    I appreciate your help .. thanks alot

    i think the point from small oscillation is that we can make the following approximation:
    sin(theta) = theta

    i still dont get how analyze this system >_< i was trying the past two days, but still not able to figure it out ..
  5. Dec 26, 2009 #4


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    Re: SOS .. Problem with lagrange derivation!!

    The first thing that probably should occur to you is that you can at least try to use the angles, [itex]\theta[/itex] and [itex]\phi[/itex], as generalized coordinates. That saves you from figuring out what [itex]x[/itex] and [itex]y[/itex] are. (It's not as simple as what you came up with)

    As Pinu7 was getting at, there are basically four terms:

    1. Potential energy of the ring, which is due to its height. You could do this with a spatial integral, but it's simpler to just use the center of mass.
    [tex]U = Mgh[/tex]
    where [itex]h[/itex] is the height of the center of mass. (If you pick the pivot point to be at zero height, [itex]h[/itex] will be negative.)

    2. Kinetic energy of the ring. Since one point (the pivot) is fixed, this is purely rotational. The formula is actually
    [tex]K = \frac{1}{2}I \omega^2[/tex]
    In this case, [itex]\omega[/itex] is the time derivative of one of your angular variables. [itex]I[/itex] is the moment of inertia about the pivot point.

    3. Kinetic energy of the point mass. Again, you'll want to express this in terms of the angles, which you can do directly by a clever choice of coordinates. I'd use polar coordinates with the origin at the pivot point. The formula for kinetic energy of a point mass in polar coordinates is
    [tex]K = \frac{1}{2}m r^2 \omega^2 + \frac{1}{2}m\dot{r}^2[/tex]
    where [itex]r[/itex] is the distance from the pivot to the point mass, and [itex]\omega[/itex] is the time derivative of the angular position of the point mass. I'll give you a big hint: [itex]\omega = \frac{\mathrm{d}}{\mathrm{d}t}[\theta + \phi][/itex]. (for this part only, of course; [itex]\omega[/itex] here is not the same as [itex]\omega[/itex] in part 2!)

    4. Potential energy of the point mass. There's a contribution from the spring,
    [tex]U = \frac{1}{2}k(x - x_0)^2[/tex]
    which you should be able to figure out easily once you've determined [itex]r[/itex] from part 3, and also a gravitational contribution, which is just [itex]mgh[/itex] with [itex]h[/itex] being the height of the point mass. That height is going to be a little tricky to calculate, but since at this point you'll already have figured out the polar coordinates of the point mass in part 3, you should be able to get a height from those without too much trouble.

    Once you've got all the energy terms, just construct the Lagrangian and apply the Euler-Lagrange equations. At this point, you'd want to make the approximations [itex]\sin\theta \approx \theta[/itex] and [itex]\cos\theta \approx 1 - \theta^2/2[/itex]. If all went well, you should wind up with the equations in your second picture. (If not, just post your work here and we'll see what went wrong; it's a reasonably complex problem so you could be forgiven for making a few mistakes along the way ;-)
    Last edited: Dec 26, 2009
  6. Dec 26, 2009 #5
    Re: SOS .. Problem with lagrange derivation!!

    Thanks diazona for ur notations .. i will try now again and then post what i have done..
  7. Dec 26, 2009 #6
    Re: SOS .. Problem with lagrange derivation!!

    I appreciate ur help both of u diazona and Pinu7.. but this problem is just too tough for my little brian to handle .. i may try later coz i want to have some rest now :`( ..
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