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Rolling without slipping question

  1. Aug 21, 2012 #1
    We are given with an hollow thin cylinder that has a mass of M and radius R, and a full cylinder that has a mass of m and radius r. The full cylinder is glued to the bottom of the hollow cylinder
    (As in the picture - ignore the speed v0 that is drawn in the picture) and now the hollow cylinder can roll without slipping on the floor while the full cylinder is glued to it from the inside. In order for this system to move - they push slightly to the right the full cylinder so that its center has a speed of v in that direction.
    What is the horizontal speed of this system's center of mass just as it starts to move?

    I really don't have any idea how to solve this problem... I read something about using the Instant centre of rotation in order to solve this problem but I don't know what it is... untill now I alwyas could solve pure roll questions using the fact that the axis of rotation went through the center of mass of the pure rolling body. I'd like to know how can I solve this problem using only the actual axis of rotation, starting with where does it go through... And if there's no choice but to solve this problem using the Instant centre of rotation concept, I'd like to know how does it come to hand for this question...

    Thanks a lot!
     

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  2. jcsd
  3. Aug 21, 2012 #2

    ehild

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    The instantaneous axis is at the bottom of the outer cylinder, where it touches the ground (the red point in the drawing). It is in rest for an instant, and as the cylinder rolls, the instantaneous axis of rotation advances with the same velocity as the centre of the big cylinder.

    For an instant you can consider the whole body rotating around that axis : all points move along a circle of radius equal to the distance from the axis.

    hint: try it with a mug or some cylinder shaped thing stuck a piece of gum inside:smile:


    ehild
     

    Attached Files:

  4. Aug 21, 2012 #3
    And did you get the same answer as the answer written in the bottom left corner of the picture I uploaded... because I got almost the same, but instead of the R-r/R element I got R-r/r...
     
  5. Aug 21, 2012 #4

    ehild

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    You mean (R-r)/R and (R-r)/r I guess. What you wrote R-r/r = R-1 which is entirely wrong.

    Yes, I got t the same result as in the figure. If you show your work in detail I could find out where you went wrong.

    ehild
     
  6. Aug 21, 2012 #5
    Calculation of the center of mass with respect to the contact point with the floor:

    y(cm) = (M*R+m*r)/(M+m)

    Since this system performs pure roll - all of the points move in the same angular speed with respect to the instantaneous axis. If we know the speed of the full small cylinder which is v, we can find v(cm) using the ratio between the distances from the contact point:

    v/r = v(cm)/y(cm) -> v/r = (M+m)/(M*R+m*r)*v(cm)
    and so

    v(cm) = v * (M*R+m*r)/((M+m)r)

    The anser according to the solution is:

    v*[1-(m/(M+m))*((R-r)/R)] -> v* (M*R+m*r)/((M+m)R)
     
  7. Aug 21, 2012 #6

    ehild

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    The centre of the big cylinder moves with speed v.

    ehild
     
  8. Aug 21, 2012 #7
    but the problem says that the center of the full cylinder moves with speed v... not the big hollow culinder
     
  9. Aug 21, 2012 #8

    ehild

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    Yes, you are right, the full cylinder was meant the small one, but I understood "full" as "whole" and I think, the writer of the problem wanted to say something that. The small full cylinder is inside the big one, how can you push it? Have you translated the original text?
    If the centre of the big cylinder moves with horizontal velocity v at an instant, the result is the same as given in the problem text.

    Anyway, I hope you have understood what the initial axis of rotation is, haven't you?


    ehild
     
  10. Aug 21, 2012 #9
    I kind of have, but I don't understand why this problem couldn't be solved without it... I mean untill now in pure roll problems, I solved them using the fact that the axis of rotation went through the center of mass (a cylinder rolling down hill, a pulley with a rope passing through it, a ball that a rope is binded on it etc.). Couldn't I solve this problem without the instantanious axis of rotation concept?
     
  11. Aug 21, 2012 #10

    ehild

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    You can solve the problem by putting the axis of rotation also into the centre of the big cylinder. In that case the rolling condition holds that the velocity of the centre is equal to the speed of the perimeter, that is, v=ωR. Note that v is the velocity of the big cylinder. The CM rotates around the centre of the big cylinder, while it moves with linear velocity v. The relative velocity of the CM with respect to the centre is -Dω, where D is the distance of the CM from the centre, so its velocity is v-Dω with respect to the ground. You will get the same result as with the other method.

    ehild
     
  12. Aug 22, 2012 #11
    Thanks a lot man!
     
  13. Aug 22, 2012 #12

    ehild

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    You are welcome. Have you got it now?

    ehild
     
  14. Aug 22, 2012 #13
    For this question yes, time will tell if the concept really got fixed in my head :)
     
  15. Aug 22, 2012 #14

    ehild

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    all right then:smile:

    ehild
     
  16. Aug 22, 2012 #15
    By the way, are we dealing here with two different angular velocities in this question: one from the circular motion of m inside M with radius of R-r, and another from the rotation of m in respect to its center?
     
  17. Aug 23, 2012 #16

    ehild

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    No, it is the same angular velocity.

    During the time the big cylinder rotates once, the centre of the small one makes a compete revolution around the centre of the big cylinder, and at the same time, it rotates once around its own centre. Like the Moon orbiting around the Earth.

    ehild
     
  18. Aug 23, 2012 #17
    I think I should have mentioned that my question about the 2 angular velocities reffers to another section of this question, where M (the big cylinder) is glued to the floor, and m performs pure roll inside of M (like a body inside a rail). They say that they give to the center of m a start velocity vo and they ask what's the minimum value of vo for which m could complete a full circyle inside M.
    In this situation I reffered to the angular velocity of m as it rotated around it's axis (vo/r in the start point) and to the angular velocity of m as it performes circular motion around the center of M (vo/(R-r) in the start point)
     
  19. Aug 23, 2012 #18

    ehild

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    You did not mention that you asked about a different problem: the small cylinder rolls inside the big one. There are two angular velocities then.

    ehild
     
  20. Aug 23, 2012 #19
    Yeah I forgot, thanks!
     
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