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Thinking about 3D rotating

  1. Nov 1, 2015 #1
    Hi folks!

    First of all, English is not my native language so I hope there is not much misleading spelling mistakes, discrepancies and inaccuracy.

    I’m currently working with a hobby-project of a flying drone/UAV with two adjustable angled rotors. I’m able to make it fly more or less in a way it is supposed but the physics behind all is partly blur.

    More precisely angular momentum, conservation of it, precession, etc…

    I have an example which I use to try to understand things correctly. It includes some assumptions and simplifications(no friction, no drag) since the very basics are what I’m into at this moment.

    The system has a horizontal rod which midpoint is attached with bearing to a base below. So the rod is able to rotate about the base. Main coordinate is fixed and attached to the base. Propellers are mounted to the both ends of that rod. Behind the link there is a picture of the initial situation => http://kuva.termiitti.com/v.php?img=38304

    http://kuva.termiitti.com/image/38304.jpg [Broken]

    At the initial state the rotor A is rotating about z-axis with angular velocity of 100 rad/s (clockwise from top view). Rotor B rotates at same amount of angular velocity, but to opposite direction. Rod is still without angular velocity to any direction.

    Mass moment of inertia of both rotors is 0.001 kgm^2.

    So now angular momentums of rotors are,
    Rotor A: Lza = -0.10 Nms
    Rotor B: Lzb = 0.10 Nms

    Since the rod is still, angular momentum of the whole system about z-axis is zero.

    I think this is correct so far?

    At the second state rotor A is turned 60 degrees about the x-axis pointing now mainly to right http://kuva.termiitti.com/v.php?img=38305.

    http://kuva.termiitti.com/image/38305.jpg [Broken]

    During this turn, system’s angular momentum about z-axis is about to chance. In this case, about to differ from zero.

    Since angular momentum has a tendency to remain the same, rod now has to start rotating clockwise. The amount of angular momentum rod have to compensate is half of the one rotors angular momentum (angle was 60 deg), being -0.05 Nms.

    I hope I’m still on the beam…?

    System’s mass moment of inertia is 0.02 kgm^2 when it is rotating around it’s mid point. (for simplicity, inertia and centre of mass of the system won’t chance even the rotor has moved)

    This means, that angular velocity of the rod will be -0.05 Nms / 0.02 kgm^2 = -2.5 rad/s.

    Is this correct?

    But now, because of the new position of rotor A, there is also angular momentum parallel to the rod. Direction of this momentum in a fixed main coordinate of course chance over time, but the amount of it remains the same being sin(60 deg)*0.10 Nms = 0.087 Nms.

    What torque generated this? I guess the torque that turned the axis of rotor A to new angle.

    Let’s say axis of rotor A turns that 60 degrees in 0.1 seconds. That is about 10 rad/s. How is this related to other velocities and momentums?

    The power needed to turn rotor A axis is unknown and also the inertia of rotor/rotor’s axis -system is unknown when it rotates around the pivot point at the and of the rod so I can’t say what is the force or torque which is needed. How should I think this?

    Because physics book says that angular momentum = torque*time, I am forced to think that since the amount of angular momentum of the system is increased by 0.087 Nms, the torque turning rotor A’s axis must be 0.087 Nms / 0.1 s = 0.87 Nm. (I’m pretty sure this is wrong thinking)

    Maybe I stop thinking and reasoning for now and wait if you have something to say. Is there something that I have understood totally wrong? Is there something else I should take into consideration?

    edit: pictures added
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Nov 1, 2015 #2


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    I don't think it can be that, because that torque has no component in the Y direction.

    I think what happens is that, as the rotor is pushed downwards - by a torque in the negative X direction, conservation of angular momentum makes the rotor shaft exert a torque in the Y direction on the rod, which is in turn transmitted to the base. If the base is held stationary then the torque that holds it stationary is the one that imparts the angular momentum in the positive Y direction. If it's not held stationary, the whole assembly will turn, so that its angular momentum in the Y direction is conserved.
  4. Nov 2, 2015 #3
    Hey, thanks for the reply!

    I forgot that there is of course a torque affecting to the base also. In this example it is stationary. Actually with this kind of assembly I’m measuring different effects when something moves/turns and of course currents of motors etc.

    If I would like to simulate this kind of scenario, am I able to know the angular momentum and velocity of the rod around the base without knowing the amount of torque which turns the rotor’s shaft?

    I mean if I know the total amount of angular momentum at the initial state and also know the angle that shaft turns and the time it takes to turn?

    Of course as long this is totally lossless system, the torque that turns the shaft can not exist no longer than the duration of the turn. Or is this right? Would the position of rotor keep that 60 deg angle also after the torque/force disappears? My understanding is that yes it would and will.

    I now actually think that what I pondered in a next to last paragraph of the first post is true. Increasing in the total amount of systems angular momentum comes from the torque turning the shaft. New amount of angular momentum can be calculated by the angles and directions and actually in the and the duration of the turn determines the torque. So that is what should be known. I guess...
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