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Velocity of end points of rectangular frames

  1. May 4, 2013 #1
    1. The problem statement, all variables and given/known data
    (see attachment, ignore those marks done with the pen)


    2. Relevant equations



    3. The attempt at a solution
    I think this has to do with calculating torques and forces but I don't know about which point to calculate torque about. I know this is a very less attempt towards the problem but I can't really figure out the point about which to calculate torque about. :confused:
     

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  2. jcsd
  3. May 4, 2013 #2

    tiny-tim

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    Hi Pranav-Arora! :smile:
    Can't you simply use conservation of energy?
     
  4. May 4, 2013 #3
    Hi tiny-tim!

    Initial energy=##2mgb\cos(\theta/2)## where m is the mass of one frame. (U=0 on the smooth horizontal surface)
    Final energy=##mv_{cm}^2##

    Equating them doesn't give the right answer. :confused:
     
  5. May 4, 2013 #4

    tiny-tim

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    nooo …

    isn't it also rotating? :wink:
     
  6. May 4, 2013 #5
    If using conservation of energy, one has to demonstrate that the system is conservative, not just take that for granted.
     
  7. May 4, 2013 #6
    Oh yes, I was wondering as to why they gave the value c. :P

    The moment of inertia came out to be ##20m/3## (Do I have to calculate the moment of Inertia when the frames lie on the ground? This is what I have done to calculate 20m/3.)

    [tex]2mgb\cos(\theta/2)=mv_{cm}^2+I\omega^2[/tex]
    Replacing ##\omega## with ##v_{cm}/b##, I don't end up with a nice answer. Is there a mistake in my calculated moment of inertia or in my arithmetic? :confused:

    How to do that? I have never encountered this type of question before where you need to show that system is conservative.
     
  8. May 4, 2013 #7

    tiny-tim

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    how did you get that? :confused:

    (and the moment of inertia, and rotational energy, must be calculated about the centre of mass)
    sorry, i disagree …

    the question says the system is frictionless (and there's no sharp collision), so it obviously intends conservation of energy to be assumed
     
  9. May 4, 2013 #8
    You can't apply conservation of energy unless you know energy is conserved.

    This is done by considering all the forces acting on the system. Either they must be potential, or their work must always be zero.
     
  10. May 4, 2013 #9
    Yes, I know it should be calculated about CM but I am not sure about which axis should I do it. Should it be about the axis shown by a dotted line in the question?
     
  11. May 4, 2013 #10

    tiny-tim

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    yes :smile:

    but i don't see what you think the other choices might be :confused:

    the axis must be parallel to the angular velocity (which is clearly parallel to that line), and there's only one line parallel to a particular line through a particular point

    (and i assume you're treating each half separately?)
     
  12. May 4, 2013 #11
    See attachment for the sketch.

    The moment of inertia of AB and CD about the axis is ##4mb^2/12=mb^2/3##
    The moment of inertia of BC and AD is ##mb^2##

    Hence, the moment of inertia of one rectangular frame is ##13mb^2/12##. Is this correct?
     

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    Last edited: May 4, 2013
  13. May 4, 2013 #12

    tiny-tim

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    yes :smile:
    i'd be happier if you used ρ for line density, rather than m …

    i'm not sure what your m is supposed to be :confused:

    anyway, can you please show your working
     
  14. May 4, 2013 #13

    tiny-tim

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    ooh, just occurred to me … ω isn't vc.o.m/b :wink:

    (and now i've got to go and watch doctor who on telly, and then i'm going out for the rest of the evening)
     
  15. May 4, 2013 #14
    Sorry I did a lot of mistakes in my previous post. :redface:

    Let ##ρ## be the line density.
    Moment of inertia of AB and CD: ##(2b\rho )4b^2/12=2b^3\rho/3##
    Moment of inertia of BC and AD: ##(c\rho )b^2##

    Hence the moment of inertia of the rectangular frame is ##\displaystyle \frac{2b^3\rho}{3}+\frac{2b^3\rho}{3}+c\rho b^2+c\rho b^2=\frac{4b^3\rho}{3}+2c\rho b^2=\frac{2b^2\rho (2b+3c)}{3}##

    where ##\rho=\frac{m}{2(2b+c)}## and m is the mass of one rectangular frame.

    Why? What is it?
     
  16. May 4, 2013 #15

    tiny-tim

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    (just got back :biggrin:)
    sorry, i was in a hurry and i got confused :redface:

    at the instant the frame hits the ground, one end of the frame is stationary, so ω = vc.o.m/b does work
     
  17. May 5, 2013 #16
    Sorry for the late reply, I had a test.

    We had
    [tex]2mgb\cos(\theta/2)=mv_{cm}^2+I\omega^2[/tex]
    Substituting the value of I and replacing ##\omega## with v_(cm)/b
    [tex]2mgb\cos(\theta/2)=mv_{cm}^2+\frac{2b^2\rho (2b+3c)}{3}\frac{v_{cm}^2}{b^2}[/tex]
    [tex]\Rightarrow 2mgb\cos(\theta/2)=mv_{cm}^2+\frac{m(2b+3c)}{3(2b+c)}v_{cm}^2[/tex]
    [tex]\Rightarrow 2\times 10 \times \frac{4}{5}=\frac{3}{2}v_{cm}^2[/tex]

    This doesn't give me a nice value for v_(cm). :confused:
     
  18. May 5, 2013 #17

    tiny-tim

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    (and v = 2vcm)

    hmm, it doesn't match any of the choices :confused:

    sorry, i can't see the mistake
     
  19. May 5, 2013 #18
    I have uploaded the solution presented in the solution booklet. I couldn't understand it so I had to post the question here. Can you please explain this? :)
     

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  20. May 5, 2013 #19

    tiny-tim

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    ah, i see it now …
    you left out the b (= 3/2) on the LHS :wink:
     
  21. May 5, 2013 #20
    Thanks a bunch tiny-tim! :smile:

    I should have taken care. :redface:

    The answer comes out to be 8 m/s.
     
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