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Forces of inverted pendulum

  1. Jan 8, 2012 #1
    1. The problem statement, all variables and given/known data
    I am trying to understand where the forces in the inverted pendulum comes from starting with this horizontal force equation:
    http://www.engin.umich.edu/group/ctm/examples/pend/inveq2.GIF
    It is associated with this picture http://www.engin.umich.edu/group/ctm/examples/pend/invFBD.GIF
    from this website
    The variables are:
    m = mass of pendulum
    l = length to pendulum center of mass
    θ = angle
    [itex]\dot{θ}[/itex] = angular velocity
    [itex]\ddot{θ}[/itex] = angular acceleration

    2. Relevant equations
    http://www.engin.umich.edu/group/ctm/examples/pend/inveq2.GIF
    I understand the first term is the transitional motion, but I don't know why there are two rotational motion terms, which looks the same but are different in the diagram, looks like it is 90 degrees apart. I understand why the 2 terms on the left are opposite of each other in sign, but i don't know why one uses [itex]\ddot{θ}[/itex] while the other uses [itex]\dot{θ}[/itex][itex]^{2}[/itex]
    3. The attempt at a solution
    I first thought this had to do with parallel axis theorem, but this equation is calculating the sum of horizontal forces, not inertia. In the end, the two term breaks down to ([itex]\frac{dθ}{dt}[/itex])[itex]^{2}[/itex] and [itex]\frac{d^{2}θ}{dt^{2}}[/itex]. What is the difference, if any between these two? i know one is angular velocity squared and the other one is angular acceleration, but i don't know where that angular velocity comes from.

    thanks for all the answers!
     
  2. jcsd
  3. Jan 8, 2012 #2

    Simon Bridge

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    The force making the rod fall (gravity) has a perpendicular component, leading to the angular acceleration and an axial component which tries to translate it through the center of mass. Both of these have been derived from the angular contribution, and both contribute to the net horizontal force.

    Note: [itex]I\ddot{\theta}/L[/itex] is the torque-force and [itex]\frac{1}{2}I\dot{\theta}^2[/itex] would be kinetic energy. Looks like L in the diagram is replaced with l in the math.

    I don't think I'd analyze it that way - it looks like an engineer using centrifugal force.
     
  4. Jan 8, 2012 #3
    In the equation i posted he's calculating force, he doesn't use I until later in the equations in the website.

    so you're saying [itex]\frac{1}{2}I\dot{\theta}^2[/itex] comes from angular acceleration due to gravity? but then shouldn't the arrow be pointed in the bottom left direction?
     
  5. Jan 8, 2012 #4

    Simon Bridge

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    That's how I'd analyze it all right... which is why I wondered if he's using centrifugal force - which would point the other way - so he can treat it like statics. You'll notice that the torque is opposite the direction of the only applied force too.

    The arrow pointing out the end looks like it should be a force but it has units of energy.
    May have to face the possibility of typos in the diagram.

    Construct your own free-body diagrams and compare.
     
  6. Jan 9, 2012 #5
    Isn't the torque from the second term be due to gravity, while the centrifugal force from the third term also due to gravity. So gravity is acting on both? I am having trouble creating my own free-body diagram for inverted pendulum on a cart, since i dont even really get this example.
     
  7. Jan 9, 2012 #6

    Simon Bridge

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    The diagrams have too much on them.
    Clean slate - draw each separately ... on the pendulum part, you have only gravity and two reaction forces.

    The cart has received a small impulse - so it is moving left to right, but slowing down - there are only the reaction forces and friction.
     
  8. Jan 9, 2012 #7
    ok so i drew my own and actually figured out where his calculations come from, but i am not sure if i drew the forces right, and I am wondering why the website is missing the 1/3 from the inertia of a rod rotating on its end. I also don't know why my signs are different.
    Ar8xK.png
    (i left out the trig cos and sine)
     
  9. Jan 10, 2012 #8
    I am looking at this equation, but you mentioned that the centrifugal force and the torque is due to gravity, but how come there's no mg or g term in the equation?
     
  10. Jan 10, 2012 #9

    Simon Bridge

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    Ftorque=mgsinθ=Iα

    If these forces do not come from gravity, then where else?
     
  11. Jan 10, 2012 #10
    OH so [itex]\ddot{θ}[/itex] is equal to gsinθ??? (i am used to seeing gravity terms as just g)
     
  12. Jan 10, 2012 #11

    Simon Bridge

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    Not quite - [itex]I\ddot{\theta} = mg(l/2)\sin{\theta}[/itex] - in your case, [itex]I=\frac{1}{3}ml^2[/itex] so that makes: [itex] 2l\ddot{\theta}=3g\sin\theta[/itex]

    You can resolve your forces against any coordinate system you like, it's just that you are used to having axis lined up with gravity so it only has one component. If you have a box sliding down a slope, then the accelerating force is mgsinθ (where θ is the angle of the slope) and the friction is -μmgcosθ. Here it would be useful to have axes lined up with the rod.

    Your mgcosθ acts down the length of the rod - the vertical reaction force at the pivot must be equal to the vertical component of this, but the falling rod will be pushing the cart to the right so the horizontal reaction force does not cancel it out.

    You'll have to play around with different representations before you get the hang of it - there is no one "right way" - but experienced people will intuitively pick the easier approaches. To gain that experience you have to stumble around a bit. After a while you'll start to see why the text books pick one way rather than another.
     
  13. Jan 10, 2012 #12
    I understand [itex]Fl=I\ddot{θ}[/itex] but where is the 1/2 from?

    oh so you're saying this would be substituted into the main force equation, which would be how gravity is accounted for?

    since the force acted on by gravity going down the rod is [itex]mgcosθ[/itex], the horizontal force on the cart would be [itex]mgcosθsinθ[/itex]?

    how come that term isn't in any of the equations in the free body diagram modeling?
     
  14. Jan 11, 2012 #13

    Simon Bridge

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    Sorry: I used [itex]l[/itex] = length of the rod - so the moment arm for the torque is [itex]l/2[/itex] - so [itex]Fl/2 = I\ddot{\theta}[/itex] ...

    If the length of the rod is L and [itex]l=L/2[/itex] then the moment of inertial should be adjusted to [itex]I=\frac{1}{3}m(2l)^2 = \frac{4}{3}ml^2[/itex]

    yep - and the vertical force would be mgcos2θ ... the vertical force has to be balanced by the upwards reaction force, but the horizontal force does not have to be.

    Because the person doing that analysis was using a different representation of the forces. All his components go in the opposite direction for example... part of going through all this is to try to figure out how things work by means that you already understand and so, hopefully, come to realize what he was doing.

    If it turns out you've missed something out, it will likely show up during the analysis.

    I'm going to tell you this carefully ... I'm only signposting: you are doing the analysis. You will, almost certainly, need to fiddle around a bit before hitting on a good approach. It is very common in science to set out on a course of action expecting to fail - but you do it for the understanding you gain as a result. And you never know - you may succeed :)

    When you find success comes as a surprise, but you feel good anyway, you've probably got the right mindset for science :)
     
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