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Rigid body dynamics

  1. May 29, 2012 #1
    A perfectly rough rod is gently placed with one end
    upon another rod of equal mass and in the same vertical
    plane, moving with the velocity √2gc on a smooth table. If
    the initial inclination of the first rod to the horizon be α, and
    its length 2a, shew that it will just rise to a vertical position
    if
    2a (1 -sin α) (5 + 3 cos ^2 α) = 3c sin ^2 a.
    i have used the fact that moment of momentum just after putting rough rod will be zero about point of contact to determine initial angular velocity.after that I used equation of energy and conservation of momentum in horizontal direction for the system.but i am not sure about the end condition, i used that reaction will become zero in vertical position and also may be angular velocity will become zero.none of these have yielded answers.at least i want to know the end condition.so please help.
     
  2. jcsd
  3. May 29, 2012 #2

    haruspex

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    Not sure I've understood the set-up. The second rod is horizontal, moving in the same vertical plane, horizontally, towards the first rod - is that it? I assume that's √(2gc), where g is gravitational acceleration. The thickness of the rods can be ignored, so it's the end away from the second rod that's raised.
    Yes, if it just reaches the vertical then the rotation will be zero at that point. But the rotation will also drop to zero if it doesn't reach the vertical, so that won't be enough in itself. The condition you need to look for is whether rotation = 0 has solutions. You want the boundary case, where it only just manages to have a solution.
    Please provide the working you have.
     
  4. May 30, 2012 #3
    the working is just too long to write here, so I have not provided it here.you have misunderstood the problem.the rod which is rough put upon another rod at certain angle
    2a (1 -sin α) (5 + 3 cos ^2 α) = 3c sin ^2 (alpha) ,perhaps that looks like 'a' there.
    the first rod which is moving is supposed to move horizontally on the table.
     
  5. May 30, 2012 #4

    haruspex

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    You could at least outline your approach, including the initial equations.
    The interpretation of the set up which I wrote in my previous post is consistent with "the rod which is rough put upon another rod at certain angle". If the interpretation is wrong you need to point out where.
    Maybe you didn't understand what I wrote. Try this:
    A rod AB is sliding in direction AB. A second rod CD is placed with C on B. ABCD are in the same vertical plane. Angle ABD is 180 - alpha.
     
  6. May 31, 2012 #5
    what does that mean.I am not saying i can not do that but what I have gotten is this
    32a(1-sinα)=9csin^2α(1+cos^2α)
    which is not what is required.I have written in the post what procedure I have followed to get the result ,angular velocity is zero in vertical position.it can not be zero in between because when you put dθ/dt=0 you get a linear eqn in θ.so it is not possible to make it zero in between.
     
  7. Jun 1, 2012 #6

    haruspex

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    andrien, I have tried hard to figure out what the set-up is. Your description is far from clear. Can you provide a diagram? Can you comment on my guess in post #4?
    One thing I struggle with in particular is the notion of placing a rough (presumably this means no sliding will occur) object gently on a moving one. What does this mean? That there is no energy lost? Surely the placed rod will have to go from stationary to moving (effectively) instantaneously, implying an inelastic collision.
    Yes, but I asked for your initial equations based on those conservation laws.
    If the initial velocity of the one rod is not enough to make the other reach the vertical then at some point before the vertical dθ/dt will be zero.
     
  8. Jun 1, 2012 #7
    I have done the problem.there was a mistake I was doing in determining the initial angular velocity,and yes the rod is placed gently that means both moment of momentum about the point of contact and linear momentum is conserved.I was assuming that momentum distribution takes place after the rod was put.but there will also be a momentum redistribution just after putting the rod of both rods and I was right with my calculation.nevertheless the problem is from old days math tripos and yes it is hard one.In fact I was not having any more information then I have written.so it was just my guess to find out what can be the case and how the figure will look like.the situation is just what you have guessed but it is placed GENTLY.that is the main word about which I was not caring earlier much.
    However thanks for seeing this problem,no one else has seen it.in fact it was my second post of the same problem here and you are the only one who responded to it.so thanks.
     
  9. Jun 1, 2012 #8

    haruspex

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    OK, thanks for letting me know. I can stop worrying about it ;-).
    Good luck!
     
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