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Equation of Motion for a Disk inside a Rotating Ring

  1. Mar 28, 2017 #1
    [Mentor's note : No template as this thread was moved from the technical forums]

    Hi all! I am working on finding the Lagrangian for the situation stated in the title. This is actually a Wolfram Mathematica demonstration as well in which they give you the Lagrangian. I am working on re-deriving it.

    I seem to be missing four terms in my Lagrangina that I believe stems from my equation for Kinetic energy of the disk and ring. The answer can be found in the link provided. http://demonstrations.wolfram.com/DiskRollingInsideARotatingRing/

    My attempt at the solution is below.

    Position of disk: $$ r(t) = (R-r)\sin\theta \hat{x} - (R-r)\cos\theta \hat{y} $$
    Velocity of Disk $$\dot{r} = (R-r)\dot{\theta}\cos\theta \hat{x} + (R-r)\dot{\theta}\sin\theta \hat{y}$$
    Velocity of Ring $$\dot{r} = -R\dot{\phi}\sin\phi \hat{x} + R\dot{\phi}\cos\phi \hat{y}$$
    Total Kinetic Energy for ring and disk: T = $$\frac{1}{2}I_{ring}\dot{\phi}^2 + \frac{1}{2}I_{disk}\dot{\alpha}^2 + \frac{1}{2}m_{disk}((R-r)^2\dot{\theta}^2) $$
    Where alpha is given in the Wolfram Demonstartion that basically just says that since friction is sufficient enough for no slipping the disk should rotate with the ring.
    Potential Energy: $$ U = -m_{disk}g(R-r)\cos\theta $$

    The four terms I am missing are $$ \frac{1}{2}m_2(2Rr\dot{\theta}\dot{\phi}\cos\theta - 2R^2\dot{\theta}\dot{\phi}\cos\theta + I_{ring}R^2\dot{\phi}^2 + m_{ring}r^2R^2\dot{\phi}^2) $$

    Or, $$m_{disk}R\dot{\phi}\dot{\theta}( (r-R)\cos\theta) + R^2\dot{\phi}^2(I_{ring} + m_{ring}r^2)$$
    Thank you!
     
    Last edited by a moderator: Mar 29, 2017
  2. jcsd
  3. Mar 28, 2017 #2
    As is often the case, the most difficult part of the dynamics problem is in the kinematics!!

    Your work above does not reflect the rolling without slipping constraint that is said to exist between the ring and the disk. Go back to the Wolfram page and look at how they have incorporated this. It will make a world of difference!
     
  4. Mar 28, 2017 #3
    I thought I did incorporate it in my equation for kinetic energy where my ##\dot{\alpha}^2## is ##\dot{\Psi}^2##, I just nevery expanded it out in my kinetic energy equation. I still end up with no Cosine terms in my Lagrangian except for the potential energy of the disk. I was able to get an extra term so now I am missing three terms in my Kinetic energy and they are. $$\frac{1}{2}\frac{I_{ring}R^2\dot{\phi}^2}{r^2} - m_{disk}R\dot{\phi}\dot{\theta}(R-r)\cos\theta$$

    Thank you for replying!
     
  5. Mar 28, 2017 #4
    Here is my full Kinetic Energy equation from the information I gave above...
    $$T = \frac{1}{2}I_{ring}\dot{\phi}^2 + \frac{1}{2r^2}I_{disk}(r^2\dot{\theta}^2 - 2rR\dot{\theta}^2 + R^2\dot{\theta}^2 + 2Rr\dot{\theta}\dot{\phi} - 2R^2\dot{\theta}\dot{\phi} + R^2\dot{\phi}^2) + \frac{1}{2}m_{disk}((R-r)^2\dot{\theta}^2)$$

    Thanks!
     
    Last edited by a moderator: Mar 29, 2017
  6. Mar 30, 2017 #5
    After working on this problem for a while, both on a fresh formulation and also on trying to work backwards from the Mathematica website information, I can say with confidence that the posted solution on the Mathematica site is wrong. There should not be any cosine functions in the kinetic energy expression.

    As formulated on the Mathematica site, there are two bodies, the disk and the ring.
    1) The ring only rotates; it does not translate. This contributes only one term to the kinetic energy and nothing at all to the potential energy.
    2) The disk both rotates and translates (the latter, with the CM moving on a circular path). This contributes a rotational term and a translational term to the kinetic energy, and it contributes one term (involving a cosine) to the potential energy.

    As posted at the Mathematica site, the negative of the potential energy is the first term in the in the Lagrangian. It is correct as written there. That means that the remaining terms are all thought to be kinetic energy terms, but they are simply in error. It is unfortunate that the Mathematica author did not post more details, but we can only go by what he did post, and it is not correct. Therefore, the computer animation that he shows is entirely suspect. There is no reason to expect the solution to the wrong differential equations to give a correct picture.

    The lesson: Don't believe everything your read on the Internet, not even when it is on the Mathematica website. Those folks are only human also.
     
  7. Mar 30, 2017 #6

    haruspex

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    I agree.
     
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