Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Damn analytical mechanics

  1. Mar 26, 2005 #1
    Sorry for that title, but what can I say, i'm gonna be mad to get the motion equation for some systems but always failing

    system 1:

    a rope rounded to a cylinder, the rope ends with a mass m and creating a pendulum with a maximum angle creating on a vibration [tex]\theta[/tex], the length of the rope while [tex]\theta = 0[/tex] is [tex]\ell[/tex], the wight of the rope is neglected

    find the equation of motion using lagranges equation..

    Here is the CORRET answer:

    [tex](\ell + r\theta)\ddot{\theta} + r\dot{\theta} + g sin\theta = 0[/tex]

    here are the steps i took, i didn't get the same result, i want some body to give me step by step,

    As we know the lagrangian is:

    [tex]L = T - u[/tex]

    and lagranges equation of motion is:

    [tex]\frac{d}{dt} \frac{\partial L}{\partial \dot{q}} - \frac{\partial L}{\partial q} = 0[/tex]

    T is kinetic energy of the system,
    u is potential energy of the system
    q is a generalised coordinate

    Now to find the equation of motion, we will use polar coordinates, we first find that r is constant, and the change is only in the angle so:

    [tex]T = \frac{1}{2} mr^2 \dot{\theta}^2[/tex]

    and the potential energy is the mgh while h is the lenght of the rope plus the lenght added with the angle [tex]\theta[/tex]

    [tex]u = -mg(\ell + r\theta)[/tex]

    and now differentiating:

    [tex]\frac{d}{dt} \frac{\partial L}{\partial \dot{\theta}} = \frac{d}{dt}mr^2 \dot{\theta} = mr^2 \ddot{\theta} [/tex]

    [tex]\frac{\partial L}{\partial \theta} = -mgr[/tex]

    finally, the equation of motion is:

    [tex]\frac{d}{dt} \frac{\partial L}{\partial \dot{q}} - \frac{\partial L}{\partial q} = 0[/tex]

    [tex] mr^2 \ddot{\theta} + mgr = 0[/tex]
    [tex] \ddot{\theta} + \frac{g}{r} = 0[/tex]

    And this answer is very different to the correct one, WHY??? Any one can explain? and give me the correct steps?

    I'm gonna give other systems later, Please anyone try GIVING ME the way to understand this mechanics .. and thanks
  2. jcsd
  3. Mar 26, 2005 #2
    Does the length of the rope not change as the pendulum swings? Perhaps I'm just not understanding the situation.

    edit: could you maybe include a diagram?
    Last edited: Mar 26, 2005
  4. Mar 27, 2005 #3
    No, the lenght doesn't change, but the hight is changing due to the angle

    Attached Files:

  5. Mar 27, 2005 #4

    Dr Transport

    User Avatar
    Science Advisor
    Gold Member

    The cylinder is stationary, the mass swinging is the pendulum. The potential is not just [tex] mgr [/tex] but [tex] mgr \cos(\theta) [/tex]. As the mass swings from side to side, teh rope will change length giving the other term for the kinetic energy.
  6. Mar 27, 2005 #5
    No man, I didn't say it's (mgr), potential energy depends on the hight, I wrote the lenght of the rope plus the rope rounded around the cylinder because of the angle theta:

    [tex]u = -mg(\ell + r\theta)[/tex]
  7. Mar 27, 2005 #6
    Will you exaplin more, explain everything please !!!
  8. Mar 27, 2005 #7


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The first step is to define your coordinate system. What does [itex] \theta [/itex] mean? Where is 0?
  9. Mar 27, 2005 #8
    0 is in the center of the cylinder
  10. Mar 27, 2005 #9
    Have you tried including the rotational kinetic energy of the cylinder into T?

  11. Mar 27, 2005 #10


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I though [itex] \theta [/itex] was an angle?

    Your drawing seems to indicate that the cylinder is rotating? I think the problem is for a pendulum which swings off of the cylinder. Please explain your coordinate system. What is [itex] \theta [/itex] ?
    Last edited: Mar 27, 2005
  12. Mar 27, 2005 #11
    I've found a solution which is very near, ideas please:

    if we made:

    [tex]T = \frac{1}{2} mR^2 \dot{\theta}^2 = \frac{1}{2} m(\ell+r\theta)^2 \dot{\theta}^2[/tex]
    [tex]u = mgr(1-cos(\theta))[/tex]

    [tex]L = T - u = \frac{1}{2} m(\ell+r\theta)^2 \dot{\theta}^2 - mgr(1-cos(\theta))[/tex]

    Then substitute:

    [tex]\frac{d}{dt} \frac{\partial L}{\partial \dot{\theta}} = \frac{d}{dt}m(\ell+r\theta)^2 \dot{\theta} = 2m(\ell +r\theta)r\dot{\theta}^2 + m(\ell + r\theta)^2 \ddot{\theta} [/tex]

    [tex]\frac{\partial L}{\partial \theta} = m(\ell + r\theta)r\dot{\theta}^2 - mgr sin(\theta)[/tex]

    then the equation of motion:

    [tex]\frac{d}{dt} \frac{\partial L}{\partial \dot{q}} - \frac{\partial L}{\partial q} = 0[/tex]

    [tex]m(\ell + r\theta)^2 \ddot{\theta} + 2m(\ell + r\theta)r\dot{\theta}^2 - m(\ell + r\theta)r \dot{\theta}^2 + mgr sin(\theta) = 0 [/tex]

    deviding by [tex]m[/tex] and [tex](\ell + r\theta)[/tex]

    [tex](\ell + r\theta) \ddot{\theta} + r\dot{\theta}^2 + \frac{gr}{(\ell+r\theta)} sin(\theta)=0[/tex]

    WELL, the difference now is only the [tex]\frac{r}{(\ell+r\theta)}[/tex]

    Can anyone explain?
    Last edited: Mar 27, 2005
  13. Mar 27, 2005 #12
    MY GOD, the cylinder is rotating i meant with 0 the center of the coordinates, [tex]\theta[/tex] is the angle creating by the cylinder rotation in the 2 sides
  14. Mar 27, 2005 #13


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This is not clear to me, where is [itex] \theta = 0 [/itex] and how does it realate to the rope?
  15. Mar 27, 2005 #14


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I do not think that you have a handle on the potential energy, mainly because you are misinterpreting the problem. Consider a pendulum which swings from the cylinder, [itex] \theta [/itex] is the angle from the horizontal to the point of tangency of the rope. This is consistent with the initial point given in your problem definition of [itex] \theta = 0 [/itex] means the length of the rope = L . To find your potential energy you need to find the distance the mass is lifted by moving the point of tangency through [itex] \theta [/itex] radians. This is a geometry problem.
  16. Mar 27, 2005 #15
    Well, At the hight point is when [tex]\theta = 0[/tex] then potential energy is mgr, and note here, i made the begining of the potential energy in the line containing the tangency point and the center of the cylinder, and when the pendulum moves and creates the angle [tex]\theta[/tex] we will just add or subtract the lenght done by the angle [tex]\theta[/tex] the it will be [tex]mgr\cos \theta[/tex] and the difference between them is:

    [tex]u = mgr - mgr\cos \theta = mgr (1-\cos \theta)[/tex]

    am I wrong?
  17. Mar 27, 2005 #16
    Who said [tex]\theta = 0[/tex] ???????????????????

    I said the start point of the coordinates is the center of the cylinder because we are using the polar coordinates !!!!!! isn't that clear man??????
  18. Mar 27, 2005 #17


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Clearly you do not have a grasp of the problem. The first step is to understand your coordinate system. Re read you initial post you say

    Clearly [itex] \theta = 0[/itex] at SOME point. Remember [itex] \theta [/itex] is an angle, how does it relate to the rope. Read my previous post again and think about it. If you do not understand the physical system there is no way to derive a meaningful expression for the potential and kinetic energies.

    If the rope were simply moving up and down with rotations of the cylinder, then the change in potential energy would simply be r[itex] \theta[/itex] since this is not the case, maybe you need to make an effort at figuring out how the system is moving.
    Last edited: Mar 27, 2005
  19. Mar 28, 2005 #18
    Calm down, my english is not good, the rope lenght is not changing, but changing only because the theta angle,

    anyone found help for me on that result? (note i didn't yet select my coordinates system, everything is going random, then i proof the correct of it)

    And sorry integral if i disturbed you,

    can any one work with this problem and give me the T and U?

  20. Mar 28, 2005 #19


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    http://home.comcast.net/~rossgr1/Math/potential.pdf [Broken] and how I see the potential energy term. I have not used it to compute the equation of motion, but perhaps this will get us on the same page as far as the coordinate system goes.
    Last edited by a moderator: May 2, 2017
  21. Mar 29, 2005 #20
    Thanks integral, i'll give a try with this hight, and i'll return to tell you the result,
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook