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Rotating rod

  1. May 9, 2017 #1
    1. The problem statement, all variables and given/known data
    A uniform rigid scale is held horizontally with one of its end at the edge of the table and the other supported by hand. Some coins of negligible mass are kept on the meter scale.
    rod.jpg
    As the hand supporting the scale is removed, the scale starts rotating about its edge on the table and the coins start moving. Which of the following would look closest to the situation if a photograph of the scale is taken soon after
    rod.jpg
    2. Relevant equations

    Torque = ##r \times F##
    ##L = r \times p##
    Torque = ## \frac {dL} {dt}##
    3. The attempt at a solution
    This problem looked trivial until I tried to reason why the different options given would not be correct. This is how I thought about. Initially coins are at rest supported by reaction force provided by the hand. Now when the hand is removed, there is going to be a moment about the hinged part as a result of change in angular momentum. The torque is highest at the ends due to the length of the arm and lowest at points closer to the hinge. Also I think it is fair to assume that friction is greater than the force pulling it to the right on the coins closer to the hinge as it is about the situation immediately after the hand is removed and therefore I can conclude the coins are going to stick to the rod. However I am not able to come to conclusion regarding the coins at the right end especially trying to reason why option C and B could or could not be possible. Also I could not get my reasoning correct as to why the coins should separate from the right end as shown,
     
  2. jcsd
  3. May 9, 2017 #2
    if the coins are of negligible mass then they remain to hang in the space along straight horizontal line after the rod has fallen down
     
  4. May 9, 2017 #3
    Why should it remain in the air? Also there are three options that show coins in the air so which one among them?
     
  5. May 9, 2017 #4
    The correct formulation of this problem should be as follows. A uniform rod of mass ##M## and length ##l## is held horizontally by its right end while the left end can rotate in a perfect hinge. There is a chain of small identical coins on the rod. The mass of the whole chain is ##m##. Find acceleration of the points of the rod right after its right end has been released. The rod is smooth.
     
  6. May 9, 2017 #5
    I do not understand how finding acceleration is relevant to the problem. Nevertheless, for the rod I can write
    ##Mv^2/l =Mgsin\theta ##
    Not quite sure how to model the equation for chain.
     
  7. May 9, 2017 #6
    I think , all such coins lose contact with the stick where the underlying contact surface on the stick move with an acceleration greater than 'g' ( acc. due to gravity ) after the hand is removed .

    An important/interesting exercise for you would be to

    1) Calculate the acceleration of the tip of the stick of mass M and length L just after hand is removed ? Note that coins are very light , so you can neglect their effect .

    2) Now , find a point on the stick whose acceleration is 'g' ?

    After you have solved the above two you would get length of part of the stick whose linear acceleration is greater than 'g' and part whose acceleration is less than 'g' .

    This should enable you to spot the best option .
     
  8. May 9, 2017 #7
    81672b680811.png

    Let ##\rho=m/l## be the density of the chain of coins; see post #4. Let ##dm=\rho dx## be infinitesimal element of this chain. By ##N(x)dm## denote the reaction force that acts on ##dm## from the rod.
    By ##J=Ml^2/3## denote moment of inertia of the rod about ##O##; ##\epsilon## is angular acceleration of the rod.
    So we have
    $$J\epsilon=-Mgl/2-\int_0^XxN(x)\rho dx,\quad a(x)dm=N(x)dm-gdm;\qquad (*)$$
    here ##a(x)## is the vertical part of acceleration of the rod's point with coordinate ##x,\quad a(x)=\epsilon x##; ##X## is ##x-##coordinate of the point at which chain leaves the rod. The second equation of (*) holds for ##0\le x\le X##.
    From the second equation of (*) we have ##N(x)=\epsilon x+g## and thus ##X=-g/\epsilon.##
    Substituting these formulas to the first equation of (*) we have the following cubic equation on ##\epsilon##
    $$J\epsilon^3+\frac{1}{2}Mgl\epsilon^2+\frac{1}{6}\rho g^3=0.$$ It is easy to see that this equation has a unique real root ##\epsilon_*## and ##\epsilon_*<0##.
     
    Last edited: May 9, 2017
  9. May 9, 2017 #8
    Okay I can calculate the tangential acceleration as follows,
    ## a_t = r\alpha## and torque ##\tau = I\alpha##
    ##\alpha = \tau/I##
    ##\tau = r \times F = LMg/2## (uniform rod, weight approximated to act at centre half way along)
    ##I = ML^2/3##
    ##\alpha = \frac{3g}{2L}##
    ## a_t = L \frac{3g}{2L} = 3/2g##

    From the above calculation, I can say ##a_t = g## is at ##\frac{2L}{3}##

    So two third of the length of rod from the hinge has linear acceleration less than g and one third has it greater than g.
    Going by that, option B should be the correct one.

    However I have got a doubt, assuming that coins have negligible mass, wouldn't all the coins experience an acceleration the same as all the points on the rod? So why would some be left hanging in the air as opposed to going along with the rod?
    .
     
  10. May 9, 2017 #9

    haruspex

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    Right.
    Does that change their acceleration in free fall?
    Consider the forces on one of those coins on the right. Is it possible that these forces will cause the coins to accelerate as fast as the rod?
     
  11. May 10, 2017 #10
    Why ?
    They are not hanging in air . Rather all those coins not in contact with stick I.e those in a horizontal line , are in free fall . They are accelerating downwards with acceleration 'g' :smile:
     
  12. May 10, 2017 #11
    The coin experiences gravity force mg. If we neglect m that is m=0 then for such a coin there is no gravity, it will hang in air. The statement of the problem is incorrect.
     
  13. May 10, 2017 #12
    I agree .
     
  14. May 10, 2017 #13
    Coins are of negligible mass , not massless .
     
  15. May 10, 2017 #14
    So considering a coin at the right end,
    ##N-mgcos\theta = -ma##
    ##N = m (g cos\theta-a)##
    therefore ##g'=gcos\theta-a## not sure if this is right but it tells me the coins experience a lesser acceleration and thereby I can conclude they experience a lesser force?
     
  16. May 10, 2017 #15

    haruspex

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    As in you analysis in post #8, you only need to consider the initial acceleration, so θ=0.
    What is g'? You already have a for the acceleration the coins experience.
     
  17. May 10, 2017 #16
    Those are just words. Write exact equations then neglect some terms in the equations then obtain approximate equations then it will be matter for discussion
     
  18. May 10, 2017 #17
    Okay yes, so the equation will be N=m(g-a)

    g' is from the normal force on the coin, (mg').
     
  19. May 10, 2017 #18

    haruspex

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    There's nothing to discuss. It would have been clearer if the question had said the mass of the coins is negligible compared with that of the rod, but Vibhor's reading is obviously correct.
     
  20. May 10, 2017 #19

    haruspex

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    Yes, but you are interested in the value of a, so what's a better way to write that?
    What limits can you place on N?
     
  21. May 10, 2017 #20
    for acceleration to be g, N should be zero but that does not convince me. Putting a limit on N, we make it a case for free fall rather than it being a case of free fall as a result of the formulations.
     
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