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Work done by rolling inverted pendulum system

  1. Aug 29, 2009 #1

    I am trying to derive the equations of motion for a system that is essentially an inverted pendulum on a wheel. The pendulum is connected to the wheel by a motor. I have a motor torque M and a disturbance torque D. The radius of the wheel is r.

    My generalized coordinates are the angle of the pendulum, [tex]\theta[/tex], relative to vertical and the horizontal position of the center of the wheel, x.

    I'm having trouble wrapping my head around the generalized forces (which here, I think, are only the motor torque and disturbance moment). Do I need to consider both the motor torque and the reaction force at the wheel center, or only the motor torque? In other words, is the motor torque doing work in x, or only in [tex]\theta[/tex]. Should my virtual work equation look like this:

    [tex]\delta W = \frac{M + D}{r} \delta x + \left( M + D \left) \delta \theta[/tex]

    or this:

    [tex]\delta W = \left( M + D \left) \delta \theta[/tex]

    For a while I was considering using Lagrange multipliers (after looking at the example of the cylinder rolling down an incline), but I've convinced myself that since x and [tex]\theta[/tex] really are independent in my problem, I don't need to use Lagrange multipliers. Or, I need to add a generalized coordinate for the angle of the wheel and use Lagrange multipliers.

    So, I'm pretty confused... what am I missing? What is it that I don't understand?


  2. jcsd
  3. Aug 29, 2009 #2


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    First, this should really be in the homework forums.

    Second, I'm not sure I understand your problem description correctly. Can you post a free-body diagram for the system (include [itex]M[/itex] and [itex]D[/itex]and all other forces cting on the system)?
  4. Aug 30, 2009 #3
    No, this really shouldn't be in the homework forums. I am not a student - I'm trying to work this out for a project on my own.

    Free body diagram is attached.

    Thanks for the help.


    EDIT: This isn't the first time that I've asked a question that was mistaken for homework... should I use the homework forums even if it's not actually a homework problem? I suppose there's no harm in doing that?

    Attached Files:

  5. Aug 30, 2009 #4
    No, it is better to ask such questions here. If you post there, students who do have a homework problem might be helped later because some "homework helper" could be helping you with your problem instead. Also, we will help you here differently than if this were a homework problem.
  6. Aug 31, 2009 #5
    OK, so this is the place to ask such questions.

    So can anyone point out the missing concept? The more I read, the more it seems like a Lagrange multiplier would be appropriate, but I'm still not able to fully understand the reasons.


  7. Sep 2, 2009 #6
    No replies... so here's an update:

    I tried adding a third generalized coordinate for the wheel rotation angle. I called this angle [tex]\phi[/tex] and it is related to x by [tex]\phi = x / r[/tex]. When I do this, I found some terms were repeated in the virtual work equation. I don't think this is right.

    Since work done is force times distance (or moment times angle), I think the correct equation is probably

    [tex]\delta W = \frac{M}{r} \delta x - M \delta \theta[/tex]

    This neglects the disturbance input (for simplicity) and flips the sign on the term with the pendulum rotation, which makes sense since the reaction moment on the pendulum will be opposite the drive moment (and opposite the positive direction of [tex]\theta[/tex]).

    I would really appreciate any feedback!


  8. Sep 2, 2009 #7


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    Hi Kerry,

    I'm not sure what you mean by disturbance moment so I'll ignore that for now.

    If you apply the work-energy theorem to this problem and note that you have a rigid body in plane motion you should be able to come up with the answer you seek.

    [tex]T_1 + U_{1\rightarrow 2} = T_2 [/tex]

    T is the kinetic energy and U is the work done from position 1 to 2. Note that I'm assuming there are not any non-conservative forces (like friction) involved.

    If the moment M is constant the work done by it will be:

    [tex]W_M = M(\theta_2 - \theta_1)[/tex]

    The work done by gravity will act at the center of mass and will be:

    [tex]W_g = mg(y_2 - y_1)[/tex]

    Where y is the height of the CoM above the datum.

    The kinetic energy of a rigid body in plane motion is given by:

    [tex]T = \frac{1}{2}m \bar v^2 + \frac{1}{2}\bar I \omega^2 [/tex]

    Where v_bar is the tangential velocity of the CoM and I_bar is the mass moment of inertia of the body about its centroid.

    Combining all of these you get:

    [tex]\frac{1}{2}m \bar v_1^2 + \frac{1}{2}\bar I \omega_1^2 + mg(y_2 - y_1) + M(\theta_2 - \theta_1) = \frac{1}{2}m \bar v_2^2 + \frac{1}{2}\bar I \omega_2^2[/tex]

    Hope this helps.

  9. Sep 3, 2009 #8
    CS, yes, this is very helpful!

    Just to clarify - in my equation I was assuming very small "virtual" displacements, which do very small amounts of "virtual" work, so although all of my forces and moments are time-varying, I can use the same approach that you did when computing work (assuming a constant moment...). But we can ignore all of this and just look at your equations.

    So really my questions is related to the work done by the moment. You wrote
    [tex]W_M = M \left( \theta_2 - \theta_1 \right)[/tex]

    So the moment isn't doing any work in the x-direction (since the wheel is generating a horizontal force)? i.e. it would be incorrect to write
    [tex]W_M = M \left( \theta_2 - \theta_1 \right) + \frac{M}{r} \left( x_2 - x_1 \right)[/tex]

    So the change in kinetic energy would be (assuming we start from rest and there is no change in elevation)
    [tex]M(\theta_2 - \theta_1) = \frac{1}{2}m \bar v_2^2 + \frac{1}{2}\bar I \omega_2^2[/tex]
    and since v_bar and omega are related by v_bar = omega * r,
    [tex]M(\theta_2 - \theta_1) = \frac{1}{2}m \bar v_2^2 + \frac{1}{2}\bar I \left( \frac{\bar v_2}{r} \right)^2 [/tex]

    Is this correct?

    Thanks for your help,

  10. Sep 4, 2009 #9


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    The moment is not doing any work in the x direction. It has been fully described by the angular displacement. So there is no need for anything else to be accounted for regarding the moment. Note that if the moment is not constant then you will need to integrate from theta_1 to theta_2.

    Also note that the energy approach gives you an answer at two points only (and not what happened in between). It is simpler and faster but is limited by the fact that you only know what is happening at point 1 and 2.

    There is a change in elevation. The CoM of the body is changing elevation as it rotates. For example if you chose the center of the "wheel" portion of the body as a datum, then the CoM is at an elevation of d_cg (assuming point 1 is with the body in a vertical position).

    So if it is initially at rest (T_1 = 0) you get:

    [tex] 0 + mg(y_2 - y_1) + M(\theta_2 - \theta_1) = \frac{1}{2}m \bar v_2^2 + \frac{1}{2}\bar I \omega_2^2 [/tex]

    Hope this helps.

  11. Sep 6, 2009 #10
    After studying this for another week, I've convinced myself that the moment is doing work in the x-direction (effectively).

    If we add another variable to the mix which represents the angle through which the wheel is rotated (in the picture, this would be [tex]\omega[/tex], but we'll call it [tex]\phi[/tex] to avoid confusion with rotation rates), the the moment is doing work in both rotations. So the work done would be
    [tex]W_M = M \left( \theta_2 - \theta_1 + \phi_2 - \phi_1 \right)[/tex]

    But the rotation of the wheel is related to displacement in the x-direction (if we assume no slipping) by [tex]\phi = x / r[/tex].

    Making this substitution we have
    [tex]W_M = M \left( \theta_2 - \theta_1 + \frac{x_2 - x_1}{r} \right)[/tex]
    [tex]W_M = M \left( \theta_2 - \theta_1 \right) + \frac{M}{r} \left( x_2 - x_1 \right)[/tex]

    Is this incorrect?

    Thanks for the help!

  12. Sep 8, 2009 #11


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    I see the misunderstanding now. I've assumed all along the angle through which the wheel is rotated is theta. So the only work the moment is doing is [tex] M \left( \theta_2 - \theta_1 \right) [/tex]. However, it appears that your original problem was intended for that to be the initial angle from the vertical at position 1.

    Regardless, the work done by the moment is still [tex] M \left( \theta_2 - \theta_1 \right) [/tex] if the central angle is labeled theta. Or, if you prefer to call it phi then that is fine too.

    Using the energy approach, theta will not enter into the equation, unless you need it to find the height of the CoM for the potential energy at some point.

  13. Sep 8, 2009 #12
    I've been thinking that both angles are important - consider these examples:

    1. The angle of the pendulum relative to vertical is unchanged, but a moment is applied to the wheel, which rotates, moving the wheel and pendulum in the x-direction. Thus the work done is the moment times the angular displacement of the wheel.

    2. The wheel remains fixed (x is unchanged), but a moment is applied to the wheel, which causes the pendulum to rotate. Thus the work done is the moment times the angular displacement of the pendulum.

    This is the thought process that lead me to conclude that the work done by the moment is the moment times the angular displacement of the pendulum relative to the wheel. This also makes sense to me since one end of the motor is fixed to the wheel and the other end is fixed to the pendulum. The distance that the motor rotates (or the distance that the stator rotates relative to the rotor) is delta phi - delta theta, so the work done by the motor (in terms of x and theta) is
    [tex]W_M = M \left( \theta_1 - \theta_2 \right) + \frac{M}{r} \left( x_2 - x_1 \right)[/tex]

    I think the key thing here is (yes - this equation is different from the one in my previous post) that a positive motor torque rotates the wheel clockwise, moving positive in x. The same positive moment drives the pendulum counter-clockwise. So the work is delta phi - delta theta, not delta phi + delta theta.

    I'm pretty sure this is correct now, but please do correct me if I am wrong.

    Thanks for all of your help.

  14. Sep 8, 2009 #13


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    I'm not sure how else to describe it. The only work being done is by the moment and by gravity. There is no need to account for the displacement in the x direction since this is the same thing, just stated another way, as the moment times the angular displacement.

    Sorry I couldn't help more!

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