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Load rotating device - will it slide?

  1. Dec 4, 2015 #1

    rst

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    1. The problem statement, all variables and given/known data

    Bez_tytu_u.png
    I'm designing a device for changing a load position from vertical to horizontal. It has a wheeled frame (2) which allows an operator to transport the load after reorientation.

    The combined centre of gravity of a rotating platform (1) and the load moves forward due to rotation, which may cause the frame (and thus the entire device) to move back, if a friction force between wheels and surface is insufficient. Brakes are engaged during the reorientation, therefore we consider sliding friction, not rolling.

    I want to make sure that the frame stays in place. My calculations include determining a force which would push it back if there was no friction (conservation of momentum). If it outweighs the frictional force, then the device will slide.

    Given data:

    masses m1 (platform+load), m2 (frame)
    angular velocty ω
    CoG rotation radius R
    initial CoG angle α0 (the angle varies from α0 to α0 +90 deg)
    coefficient of sliding friction (wheels to surface) μ

    2. Relevant equations
    ##
    F_2 = \frac{dp_2}{dt} \\
    p_1 + p_2 = 0\\
    F_{fmax} = μ \cdot F_n = μ (m_1 +m_2)g\\
    ##

    3. The attempt at a solution

    1) Device standing on a frictionless surface
    ##
    V = \omega \cdot R\\
    V_1 = V \cdot cos(\alpha_0 + \omega t)\\
    m_1 (V_1 - V_2) – m_2 V_2 = 0\\
    V_2 = \frac{m_1 V_1}{m_1 + m_2}\\
    p_2 = m_2 V_2 = \frac{m_1 m_2 V_1}{m_1 + m_2} = \frac{m_1 m_2 V cos(\alpha_0 + \omega t)}{m_1 + m_2}\\
    F_2 = \frac{dp_2}{dt} = - \frac{m_1 m_2 V \omega}{m_1 + m_2}sin(\alpha_0 + \omega t)\\
    |F_{2max}| = \frac{m_1 m_2 V \omega}{m_1 + m_2} = \frac{m_1 m_2 \omega^2 R}{m_1 + m_2} \\
    ##

    2) The device will not move back if the following condition is satisfied:
    ## |F_{2max}| < μ (m_1 +m_2)g##

    Question: is my reasoning valid?
     
  2. jcsd
  3. Dec 4, 2015 #2

    haruspex

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    Since rotating the platform raises the CoG, a force is required. That force may add to the normal force, so you need to specify where and at what angle that force will be applied.
    Since you require that there is no slipping, you should use the coefficient of static friction.
     
  4. Dec 5, 2015 #3

    rst

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    Yes, I was thinking of the static coefficient. Sorry I didn't put it clearly.

    Rotation is achieved by means of a torque applied to the platform's axis of rotation. The torque is generated through a worm gearbox, there is no force acting directly on the platform. Should I decompose the torque to (radius of CoG) x (force acting on CoG, perpendicular to R) and then include the vertical component in the friction equation?
     
  5. Dec 5, 2015 #4

    haruspex

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    Ok, so I'm guessing there is a motor mounted on the platform to generate this torque.
    I don't think it is valid to calculate momentum based on conservation law then try to apply its rate of change at some instant to the frictional case. Friction would have been acting up to that time, changing the momentum. It might happen to be right, but I wouldn't trust it.

    You mention a rate of rotation, but the only force that implies is a centripetal one. When the rotation starts, there must also be an angular acceleration. That might be more significant for the sliding question.

    Another concern could be toppling.
     
  6. Dec 7, 2015 #5

    rst

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    Let's assume that in the beginning the platform moves with an angular acceleration ε to achieve the rotation rate ω. Therefore, the CoG moves with a tangential acceleration a = ε*R. Horizontal component: ax = a * cosα. Vertical component: ax = a * sinα.
    The friction force decreases, because there is an additional force, a * sinα * m1.
    The CoG is moved forward by a horizontal force: a * cosα * m1.
    But I don't really see how does it affect the frame slipping. Should I simply apply the third law of motion? The slipping force is then a * cosα * m1?
    If I'm not right, please give me a hint. I've run out of ideas.

    Correct me if I'm wrong: toppling will occur if the following happens:
    1. Common centre of gravity of the platform and load moves beyond the front wheels.
    2. A moment of m1 weight about a front wheels' axis is greater then a moment of m2 weight about that axis.

    As long as the CoG stays between the rear and front wheels, the device is safe from toppling.
     
    Last edited: Dec 7, 2015
  7. Dec 7, 2015 #6

    haruspex

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    That force is upwards on the rotating platform, yes? So the reaction is downwards on the trolley, increasing the normal force reaction from the ground. That would increase the maximum frictional force.
    Yes, it applies a horizontal force that needs to be opposed by friction.
    But later on, at steady rotation ω, there are still forces of interest. What are the horizontal and vertical forces on the axle from the platform at that time?
    yes, but...
    that does not follow. In the initial acceleration phase, the torque accelerating the platform's rotation leads to a reactive torque on the trolley. If large enough, it would topple trolley even though (while the wheels are on the ground) the CoM is vertically between the sets of wheels.
    Likewise, after steady rotation is reached, there is still in principle the possibility of toppling.

    Finally, there is a deceleration phase. That could create the risk of toppling in the other direction.

    Do all four wheels have brakes? If not, you need to discriminate normal forces on the braked wheels from the normal forces on the others.
     
  8. Dec 9, 2015 #7

    rst

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    Thanks a lot! Your question has made me aware of some issues. I've decided to change the conditions. Now, the rate of rotation varies, but instead the motor's torque (T) is constant. An updated force diagram looks like this:
    219uo01.png
    ***EDIT***
    FT should be pointed in the opposite direction - my bad.

    FR = m*g*cosα
    FT = T/R - m*g*sinα
    FC = m12*R
    where:
    ## \omega = \int \varepsilon dt ##
    ## \varepsilon = F_T m_1 R ##

    FR & FC as a single force:
    Fw = m*g*cosα - m12*R

    Components of a total force:
    Fx = Fw*sinα + FT*cosα
    Fy = FT*cosα - Fw*sinα

    Fx is responsible for the trolley slipping, while Fy affects the normal force (it reduces or increases friction, depending on its orientation). Additionally, a moment of FT about the rotation axis (FT*R) may lead to toppling.

    Is the above correct?

    The front wheels do not have brakes. Should I assume that only the rear normal forces create friction opposing the frame motion? The front ones are then responsible just for a (negligible) rolling resistance?
     
    Last edited: Dec 9, 2015
  9. Dec 10, 2015 #8

    rst

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    In my previous post, in some equations there is a m mass. It should be m1 instead, I cannot edit the post for some reason.
     
  10. Dec 10, 2015 #9

    haruspex

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    Only one pair of wheels having brakes certainly opens up a much greater risk of sliding - depending on which wheels they are.
    But looking at the diagram, the front wheels are on the right, yes? When the table is rotating anticlockwise, the normal force will increase on those and decrease on the others. So the bigger sliding risk may be when lowering the table.

    In your force equations in post #7, these are forces acting on the platform, right? The reaction on the trolley will be equal and opposite.
    Think again about how centripetal force fits in. Remember, it is a resultant force, not an applied force.
     
  11. Dec 10, 2015 #10

    rst

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    Yes, the front is on the right. So, if the rolling resistance gets outweighed by the reaction on the trolley, the device slides even though a friction force on the braked wheels is greater then the reaction? It's counterintuitive to me.

    Yes, that's right.

    Then, if I understand it correctly, Fc will act only on the table and load. The reaction force on the trolley will be caused solely by FT and FR (?)
     
    Last edited: Dec 10, 2015
  12. Dec 10, 2015 #11

    haruspex

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    That's not what I'm saying.
    Forget rolling resistance, it's insignificant here.
    When the platform is rising, it will shift load towards the front wheels, increasing the available frictional force, so I doubt you have any tendency to slide in that phase. What it might do, if raised too suddenly, is shift all the load onto the front wheels. If that happens it will be in danger of toppling over.
    I presume that at some point the table will be lowered. I thought at first that might mean too much weight would go to the back wheels, creating the possibility of sliding. But I overlooked that centripetal force will be in the same direction as on rising.
    Centripetal force does not act. It is the resultant force the trolley must supply to achieve the rotational motion of the table. It will be more intuitive to think in terms of centrifugal force instead. Which ever way the table is rotating through its quadrant, centrifugal force will pull the trolley down and to the right.
     
  13. Dec 11, 2015 #12

    rst

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    The centrifugal force is equal and opposite to the centripetal one. Therefore, including the centripetal force into the reaction (which is wrong) would arithmetically give the same result as calculating the reaction based on FT and FR and then adding the centrifugal force. So, it is "just" a matter of addressing the problem in a physically adequate way, isn't it?
     
  14. Dec 11, 2015 #13

    haruspex

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    Same magnitude, opposite direction.
    The centripetal force needed is up, left on the table. That means the trolley needs to exert a force up, left to supply it, in addition to the up, left force to counter the weight of the table. By action and reaction, the table exerts this total down right on the trolley.
    In the centrifugal view, Fc acts down right on the table, so adds to the down right component of the weight of the table, and this total acts down, right on the trolley.
    In your post #7, you had Fc acting up, left on the table, thereby having the opposite sign to the radial component of the weight.
    In short, you had ##mg\cos(\alpha)-mR\omega^2## where you should have had ##mg\cos(\alpha)+mR\omega^2##.
     
  15. Dec 13, 2015 #14

    rst

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    It is getting clearer now. I have one more question. Why do we calculate friction on the wheels without brakes? I thought that the braked ones are responsible for slipping resistance.
     
  16. Dec 13, 2015 #15

    haruspex

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    I don't know, why would you? Did I suggest you should? Which post?
     
  17. Dec 13, 2015 #16

    rst

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    In post #7:

     
  18. Dec 13, 2015 #17

    haruspex

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    For some reason I misread your post #7 as saying only the front wheels have brakes. This puts a very different slant on things.
    Why are the brakes on only the back wheels? Is it so that it won't tip over when braking during forward motion? The downside is that this reduces the available frictional force, both while braking forward motion and while raising the table.
     
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