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Velocity question with moment of Inertia

  1. Nov 8, 2006 #1
    Referring to the attached figure; the inner hoop rolls without slipping inside the larger hoop. The large one moves freely on the frictionless table. The mass of the smaller hoop is m while the larger one has a mass of 3m. They start from rest in the position shown and the inner hoop rolls down to the bottom of the large hoop with negligible energy loss for the system as a whole. How fast, relative to the table, is the center of the large hoop moving when the small hoop has its center directly below the center of the large hoop? Is the large hoop rolling clockwise or counterclockwise?

    If b=radius of the small hoop and 3b is the radius of the large hoop, then the center of mass of the system is -b/2. The moment of inertia (I) of the small hoop I2= m*b^2, and I1=27*m*b^2 for the large one. The moment of inertia of the large hoop about the center of mass of the system is Il=27*m*b^2 + 3*m*(-b/2)^2 ......from the parallel axis theorem. Therefore Il=111*m*b^2/4. The moment of inertia of the small hoop about the center of mass of the system is Is=13*m*b^2/4....parallel axis theorem. The total moment about the center of mass is It = Is + Il = 124*m*b^2/4 = 31*m*b^2

    m*g*2*b = .5*4*m*v^2 + .5*It*w^2
    => 2*m*g*b = 2*m*v^2 + 2*It*v^2/b^2. Therefore
    m*g*b = m*v^2 + 31*m*v^2. Therefore
    g*b = 32*v^2 => v = (g*b/32)^.5. And v for the large hoop is:
    vl=6*(g*b/32)^.5
    My approach to this question may be wrong as the correct answer
    is vl= 6*(g*b/199)^.5. Any help would be welcome.
     
  2. jcsd
  3. Nov 8, 2006 #2

    OlderDan

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    The hoops are not connected, so there is no total moment of inertia. There is a system center of mass, and although I have not seen the diagram I believe you have found that correctly. Each hoop has its own center of mass and its own moment of inertia. The only external force is vertical. That tells you something about how the center of mass of the system can move. The initial potential energy you found looks OK. The final kinetic energy will have 4 contributions, two from each hoop, with each of those having a linear contribution and a rotational contribution. It appears you have left out a conservation principle that will enable you to relate the velocities of the two hoops.
     
  4. Nov 9, 2006 #3
    I must say OlderDan you are pretty good and to be able to give me the correct direction for my calculations considering you did not see the diagram, which is the same diagram as for my last topic on this forum. Thanks very much for your help.
     
  5. Nov 9, 2006 #4

    OlderDan

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    I did recall that other problem. In fact at first I thought you were doing that one again. I hope that one worked out for you too.
     
  6. Nov 13, 2006 #5
    The conservation laws are:
    0 = m1*v1 + m2*v2 ...(i) Where m1 = mass of small hoop
    0 = I1*w1 + I2*w2 ...(ii) and m2 = mass of large hoop.

    m1*g*2*b = .5*m1*v2^2 + .5*I1*w1^2 + .5*m2*v2^2 + .5*I2*w2^2 ...(iii)The problem is: it seems rather arbitrary in the substitutions you can make, e.g.,(and certainly cannot be)

    4*m1*g*b = m1*(-m2*v2/m1)^2 + I1*(I2*w2/I1)^2 + m2*v2^2 + I2*w2^2.

    4*m*g*b = m*(-3*m*v2/m)^2 + m*b^2*(-27m*b^2/(m*b^2)*w2)^2 + 3*m*v2^2 +27*m*b^2*w2^2.
    Where m*b^2 = I1 = moment of inertia of the small hoop and 3*m*(3*b)^2 = 27*m*b^2 =I2 is the moment of inertia of the large hoop. Thus we get

    4*m*g*b = 9*m*v2^2 + 729*m*b^2*w2^2 + 3*m*v2^2 + 27*m*b^2*w2^2 View attachment Doc1.doc . Therefore
    4*g*b = 12*v2^2 + 756*(v2/3/b)^2*b^2.
    4*g*b = 864*v2^2. Therefore v2=6*(gb/864)^.5
     
  7. Nov 13, 2006 #6
    We are told that the system as a whole dosen't lose energy implying that there is no friction between the small hoop and the large hoop as the small hoop rolls down.
     
  8. Nov 13, 2006 #7

    OlderDan

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    It's not that there is no friction between the hoops. It is that there is no slipping. Without slipping the frictional forces do no work on the system, so mechanical energy is conserved. My guess is that there has to be some slipping of the outer hoop on the table, although I have not actually done the calculation, but the table is frictionless, so there is no work done there either. I have not looked carefully at your calculation in the previous post. Are you comfortable with it?
     
  9. Nov 13, 2006 #8

    OlderDan

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    See the annotations in the quote.
     
    Last edited: Nov 14, 2006
  10. Nov 14, 2006 #9
    I think that I am comfortable enough with the answer I got in the last post. In fact I mixed up the answer of the last topic/post with what I posted at the beginning of this post, see #1 in this post It is my answer to the last topic I posted. Thanks again for your help and time.
     
  11. Nov 14, 2006 #10

    OlderDan

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    Without using your second equation (which is not valid), you get a different answer. See annotations and see my revised earlier post.

     
  12. Nov 15, 2006 #11
    I got that answer too of v2 = (g*b/6)^.5, I though it wrong because the answer in the book is 6*(g*b/199)^.5. Thanks for the help.
     
  13. Nov 15, 2006 #12

    OlderDan

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    Those answers are not very different, and theirs is about 4% bigger than ours which means their kinetic energy would be about 8.5% bigger. I can't see any way to add energy to the system. I don't believe their result. I wonder if they assumed the outer hoop rolled [edit: without slipping] on the table. That is a harder problem, but it could potentially account for an additional force (friction) accelerating the outer hoop.
     
    Last edited: Nov 15, 2006
  14. Nov 15, 2006 #13
    I think they assumed it rolled on the table because (i) they ask How fast, relative to the table is the center of the large hoop moving when the small hoop has its center directly below the center of the large hoop: and (ii) v2 = 0 if the outer hoop was slipping.
     
  15. Nov 15, 2006 #14

    OlderDan

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    The center of mass of the large hoop moves to conserve momentum if there is no table friction, so there is a v2 in any case. What I should have said is maybe they assumed it rolled without slipping on the table. That is not necessarily the same thing, and if rolling without slipping requires a frictional force, the assumption of horizontal momentum conservation is no longer valid. It seems to me at first glance that is a much more difficult problem than the one we did. I think that is why they specified a frictionless table in the problem.
     
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