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Rolling Cylinders across a grid of bars

  1. Aug 13, 2017 #1
    1. The problem statement, all variables and given/known data
    A cylinder is placed on bars.The separation of each bar is l.The cylinder is then pulled with a force F through its center.Find the terminal velocity of the object.Assume l<<R and the ball rolls without slipping.

    2. Relevant equations
    1/2 mv^2 +1/2 Iω^2 = kinetic energy
    Work is F*distance

    3. The attempt at a solution
    My thought is that when the work done by the force is equal to the energy loss due to the change of axis of rotation ,the velocity will be constant.
    Thus,if its velocity before the collision with the new ridge is Vo then after the collision it will be V'.The distance it travels after every ridge is l/2.So the work is F*L/2.This work is equal to the kinetic energy loss,but I cannot figure out what V' is ? Can anyone give me advice or if my idea of this problem is correct ? Thanks in Advance !
     
  2. jcsd
  3. Aug 13, 2017 #2

    berkeman

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    I'm having a hard time visualizing the situation. Does this problem come with a figure that you could scan and Upload? Also, the thread title "Rolling Cylinders through Bar" is pretty confusing -- Can I change it for you to make it less confusing?

    EDIT/ADD -- Thread title updated. :smile:

    Cylinder Rolling in a Bar...
    http://2.bp.blogspot.com/-9dtRWNy-_2s/Vol6CqB8DHI/AAAAAAAAHVo/BOX0lm92N2U/s1600/barrel+track+2016.png
    barrel%2Btrack%2B2016.png
     
    Last edited: Aug 13, 2017
  4. Aug 13, 2017 #3

    haruspex

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    Here's my guess at the setup:
    There is a horizontal grid of parallel, equally spaced bars of negligible thickness, like a cattle grid.
    The cylinder is being dragged across them, so it keeps rolling over one bar and crashing onto the next. The work done by the force at each interval is lost in the inelastic collision with the next bar.
    Is that it?
     
  5. Aug 13, 2017 #4
    Yes that is it.The cylinder is dragged on grids of round bars( negligible in size) with the axis of rotation being parallel to the bars.Sorry if the explanation is confusing;I have difficulty finding the word to describe it.
     
    Last edited: Aug 13, 2017
  6. Aug 13, 2017 #5

    haruspex

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    If the velocity when on top of a bar is v0, what will it be just before it reaches the low point?
    When it hits the next bar, what is conserved?
     
  7. Aug 13, 2017 #6
    On the point of contact,is the angular momentum conserved ? But there is a force creating torque around the point.For the velocity ,here is what i think.The cyllinder will rotate around the bar so kinetic energy will be 1/2 I ω2 where I is moment of inertia around point of contact. So the difference in total kinetic energy is F*l/2 ?(assuming the ball drops only a small amount) and use the relation v =ωr to find the speed in the bottom.
     
    Last edited: Aug 13, 2017
  8. Aug 13, 2017 #7

    haruspex

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    The impact on the next bar takes negligible time, so the gravitational force, the normal and frictional forces from the previous bar, and the force from the string contribute negligible angular momentum during it. The only force of interest during the impact is the collision with the new bar. You can eliminate that by taking moments about that point of contact.
     
  9. Aug 13, 2017 #8
    ο=Iω with I being around the pont of contact.This equation doesnt really make sense considering the inertia is the same around the contact points,therefore the omega is constant (?)
     
  10. Aug 13, 2017 #9

    haruspex

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    The impact changes the direction of motion of the centre of the cylinder. Consider the angular momentum about the impact point as made of two components, the rotation of the cylinder about its centre, plus the moment of the linear momentum of the cylinder.
    Just after impact, the linear motion is orthogonal to the line from impact point to centre of cylinder. Just before impact it is not.

    By the way, it occurs to me it makes more sense to consider the interval from one impact to the next, rather than from the high point to an impact. This avoids having to think about gravity.
     
  11. Aug 13, 2017 #10
    I am quite confused,sir.What do we need to know is he loss in energy after every collision right ? So what is the next step I should do ?
     
  12. Aug 13, 2017 #11

    haruspex

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    Suppose the cylinder is rotating at rate ω just before an impact. Let the two consecutive bars subtend an angle 2θ at the cylinder's centre.
    Regard the cylinder's motion as the sum of the linear motion of its centre and a rotation ω about that centre.
    What is the linear velocity of the cylinder's centre? What direction is it in? What is the angular momentum from that about the bar it is about to strike?
     
  13. Aug 14, 2017 #12
    Before the collision ,The linear velocity of the cylinder should be perpendicular to the bar on the left.I imagine as if the cylinder rotates around an axis that is the bar,therefore the speed is or is this wrong ?.If i consider it as a normal cylinder motion then the cylinder will have V=wrcosθ,but I dont think that is the case.
     
  14. Aug 14, 2017 #13

    haruspex

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    Yes... is what?
    That is one component of it. Which component? If the mass of the cylinder is m, what angular momentum does that velocity component give about the new bar?
     
  15. Aug 14, 2017 #14
    Oopa i thought I had typed it. What I meant to say was the speed will be ωR and the direction is perpendicular to the line connecting center of mass and left bar,with ω being rotation wrt to the bar.(Now that i think of it ,it seems incorrect )
    After the collision only the horizontal component of the speed is conserved so wRcosΘ is conserved so the speed loss is WR-Wrcosθ ?(the angular momentum conservation would give Iω+MωRcosθRcosθ = I ω'+MRω'cosθRcosθ) but this would yield nothing (ithink). what part of the speed would be conserved after the collision ? Is it the horizontal component ,sir? (Sorry if I am a bit slow)
     
  16. Aug 14, 2017 #15

    haruspex

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    That is correct. Why do you doubt it?
    The collision impulse will have a horizontal component. The only thing that is necessarily conserved is angular momentum about the impact point.
     
  17. Aug 15, 2017 at 2:17 AM #16
    Angular momentum before the collision will be Iω+MVRcosΘ=Iω'+MvRcosΘ where v is total linear speed. What should i do next ?
     
  18. Aug 15, 2017 at 4:37 AM #17

    haruspex

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    Not quite. Look at how I defined θ in post #11.
    Also, write v in terms of ω, and I in terms of m and r.
     
  19. Aug 15, 2017 at 9:31 AM #18
    Why wpuld it be wrong ? Please point it out, sir
     
  20. Aug 15, 2017 at 4:39 PM #19

    haruspex

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    Draw a diagram. Label the bar it was just rolling on A, the bar it is about to hit B, the centre of the cylinder O.
    I defined theta as half of the angle AOB.
    Draw the velocity direction vector from O. This is normal to AO. For the angular momentum about B you want the component of the velocity that is normal to OB. What is the angle between those two normals?
     
  21. Aug 16, 2017 at 2:26 AM #20
    Here is the image
     

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