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Rolling motion with mass attached to the rim

  1. Oct 11, 2012 #1
    Hello everyone ,
    I recently encountered a question in my Russian physics book .
    I did a lot of brainstorming with this question , I couldn't reduce the eqauation in one variable though .
    The question is

    1. The problem statement, all variables and given/known data
    The problem is marked yellow in the attachment . Please note the figure for the problem is also marked yellow .

    The attempt at a solution

    I first deduced an expression for energy conservation .
    Found the velocity of mass A with respect to ground .
    And finally formed an equation in which I equated the total weight to net centripetal force in that direction .

    Any help would be appreciated .
    Thank you
  2. jcsd
  3. Oct 11, 2012 #2
    This is the attachement

    Attached Files:

  4. Oct 11, 2012 #3
  5. Oct 12, 2012 #4


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    Kushan, you know the rules. Show your attempt. First show what is the minimum Vo when the ring can turn over and roll forward instead moving to and fro like an inverted pendulum.
    I think, that is what the problem asks. That "bounce" can be just a wrong interpretation of the original Russian expression, as it is said that the ring moves without slipping. It is not called "rolling" when the ring detaches from the ground and flies over it. It can happen that the speed is so high that the condition for rolling can not be fulfilled, that is, negative normal force is needed. So I think you have a minimum and maximum Vo for a rolling motion.

    Your previous "solution" that the mv2/r=mg holds for the small mass was wrong as the small mass is fixed to the rim, and the force from the rim keeps it moving together with the rim along the circle.
    The ring and the small object on the rim is a "rigid body" You need to treat it according to the laws valid for rigid bodies.

  6. Oct 12, 2012 #5
    This is from Irodov's book of problems, and I assure you the translation is correct. The problem is indeed about motion without slipping, but this is possible only up to a certain speed, after which the point mass moves too fast to be constrained to circular motion by the weight of itself and the hoop.
  7. Oct 12, 2012 #6


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    Well, the point mass moves along a cycloid, and the motion is quite complicated, with time-dependent linear and angular velocity.

    I tried to perform some experiment, sticking a pebble into an empty cylinder at the rim, but I haven't seen that the cylinder jumped up. Do you think, it really happens? I can not exclude it happens if the coefficient of static friction is big enough.

  8. Oct 12, 2012 #7
    It has to. The force required to keep it rotating increases infinitely with the velocity. The point mass is unbalanced, so there is no force of the same magnitude acting in the opposite direction, as in a uniform disk or hoop, that cancels it. As soon as it exceeds the weight of the entire system, it will have to lift up.

    Note, by the way, that if such a hoop is rolling only inertially, its linear motion will be anything but uniform for the same reason.
  9. Oct 12, 2012 #8
    Ok , here are the attachments which show my incomplete soultion .

    Attached Files:

  10. Oct 12, 2012 #9
    this is the last part

    Attached Files:

  11. Oct 12, 2012 #10
    In these attachments , I have taken two cases , or you can call two different assumptions .
    In case I , I considered , the axis of rotation as the com of ring and only one centripetal force will act that is on the mass attached to its rim .
    In case II , I took to the instantaneous axis as my primary axis , and from that perspective I equated weight with centripetal force of both the masses .

    Now my problems ,
    a) What after the final equations after respective cases .
    And if you say find v0 max wrt to theta ,(how to find its maxima)

    b) why don't both the cases resolve to same equations or answers?

    c) What will be the trajectory of the system after we exceed the minimum v0 .
    will it jump as said by book and voko , or will it slip or something (ehild) .

    Thank you all .
  12. Oct 12, 2012 #11
    I found after integrating the acceleration that velocity is not uniform , it is somewhat falls exponentially wrt to theta .
    Last edited: Oct 12, 2012
  13. Oct 12, 2012 #12
    You seem to be solving a more general problem than you have in the book. The book only asks for the maximum velocity without bouncing. That is easy to find, the condition is that the centripetal force when the mass is at the top of the hoop equals the weight of the mass and the hoop.

    What exactly is the problem you are solving?
  14. Oct 12, 2012 #13
    How do you know centripetal force will max out at top ,
    As component of centripetal force does increase will increasing theta , but velocity goes down .
  15. Oct 12, 2012 #14


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    You need a minimum velocity Vo to ensure that the ring turns round.
    If you exceed a maximum possible velocity the ring might hoop (I am not sure) but will not roll.

    Do not forget that the normal force from the ground acts to the whole system, at the rim of the ring. If you want to treat the motion of the particle separately you need to calculate with the force the ring acts on it -entirely unknown.

    Last edited: Oct 12, 2012
  16. Oct 12, 2012 #15
    how do i find it ?
    and which case should i take?
  17. Oct 12, 2012 #16


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    First find the minimum velocity Vo which is needed to turn the ring round. It needs energy considerations only. I can not see what you got from your handwriting.
    As for the other limit of velocity, treat the whole system as a rigid body. What are the laws of motion for a rigid body?

  18. Oct 12, 2012 #17
    ok thanks
  19. Oct 12, 2012 #18


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    You know that the centre of mass moves in such a way as if all the external forces acted to it. The general problem of describing the motion is quite complicated. Find the condition of rolling (maximum speed) for the case when the particle is at the top of the circle. From that, you get the maximum Vo from energy conservation. And do not forget, that the centre of the ring moves along a horizontal line. Forget the instantaneous axis.

  20. Oct 12, 2012 #19


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    The first part is correct, so you got the first condition of rolling: the speed must be greater than zero even when the point mass is at the top: that is v2(top)=(Vo2-2gr)/3≥0, Vo2≥2gr. Good job!

    As for the other part, the planar motion of a rigid body can be described as motion around a fixed axis or the combined motion of translation of the CM and rotation about it. While rolling, the CM will move around the centre of the ring, and translate at the same time. Its motion is governed by the external forces: gravity, normal force (both vertical) and static friction (horizontal). You can not say that the vertical component of the centripetal force is equal to the sum of the vertical forces, as the tangential component of acceleration also has vertical component at angle theta except when theta is either zero or pi.

    Last edited: Oct 13, 2012
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