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Rotational Motion Problem - 7

  1. Dec 16, 2013 #1
    Hi friends,
    Please help me in solving this problem, I'll appreciate the help.

    The problem is as:


    https://fbcdn-sphotos-f-a.akamaihd.net/hphotos-ak-ash3/q71/s720x720/575434_1461727284054377_268945819_n.jpg

    Attempt -

    https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-ash4/1501673_1461727677387671_1024341158_n.jpg


    Thank you all in advance.
     
  2. jcsd
  3. Dec 16, 2013 #2

    haruspex

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    As with your problem 4, choose a fixed reference point for angular momentum.
     
  4. Dec 17, 2013 #3
    haruspex: Can you please tell me why it is important to take any external reference point while conserving angular momentum?
     
  5. Dec 17, 2013 #4

    haruspex

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    If your reference point can accelerate then that clearly has the potential to give a wrong result. Consider a bar floating in space, struck perpendicularly at one end. If we take the point of the bar that was struck as the reference point, the impulse has no moment about that point, so the bar should acquire no angular momentum about it. But from the perspective of that end of the bar, just after impact, the entire of the rest of the bar is rotating about it (in the opposite direction to the impulse).
    If, instead, we use the fixed point in space where that end of the bar had been, we find that the part of the bar near that point acquires a moment one way, but the far end of the bar acquires an opposing moment, the two cancelling out.
     
  6. Dec 17, 2013 #5
    Ok, first let me clear Problem - 4 and all the things behind that, then I'll again try to solve this problem. For any confusion beyond that, I'll again come to you. Thanks a lot in advance.
     
  7. Dec 17, 2013 #6
    Please don't mind
     
  8. Dec 18, 2013 #7
    https://fbcdn-sphotos-e-a.akamaihd.net/hphotos-ak-prn2/1480736_1462448153982290_1020254380_n.jpg
    In this case if I take the reference point A, which is momentary at rest after collision,

    Conserving angular momentum, mv(2R) = [2MR2(ring) + MR2(particle)] ω

    Which gives the result, ω = 2v/3R

    Still not the correct one.
     
  9. Dec 18, 2013 #8

    haruspex

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    Looks like you mean that point of the hoop/loop. How can you be sure that any point of the hoop is momentarily at rest after the collision? Try taking a point fixed in the plane. I would try the point where the hoop's centre was when at rest.
     
  10. Dec 18, 2013 #9
    I think that point would be point C, the centre of line AB. And if i'll conserve angular momentum then post 1 again comes in front of me.
     
  11. Dec 18, 2013 #10
    Can you calculate the new Center of Mass of the system ,after the bullet gets embedded in the loop ?

    You may consider any fixed point in the plane .Two good choices are 1) center of the loop 2) new position of the center of mass in the plane . Both will yield correct answer .

    Another thing you need to remember is that - The angular momentum of the system(bullet+loop) about a fixed point,say C, is the sum of angular momentum of the CM about C + angular momentum of the system around the CM
     
    Last edited: Dec 18, 2013
  12. Dec 18, 2013 #11

    haruspex

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    In the OP, you took moments about the centre of the hoop as it moved. Taking instead a fixed point, the angular momentum of the particle will be greater. It is not only rotating about the centre of the hoop, but also moving forwards with the hoop.
     
  13. Dec 19, 2013 #12
    Yes it will be at the distance R/2 from the centre of the loop.
     
  14. Dec 19, 2013 #13
    So, will this be the right approach?

    conserving linear momentum,

    m(v) = 2m v' => v' = v/2

    conserving angular momentum about centre of the loop,

    mv(R) = mR2(loop) + {mR2(particle)}. ω + (2m). (v/2).(R/2)(particle +loop com)
    it gives the answer, ω = v/4R
     
  15. Dec 19, 2013 #14
    So, will this be the right approach?

    conserving linear momentum,

    m(v) = 2m v' => v' = v/2

    conserving angular momentum about centre of the loop,

    mv(R) = mR2(loop) + {mR2(particle)}. ω + (2m). (v/2).(R/2)(particle +loop com)
    it gives the answer, ω = v/4R
     
  16. Dec 19, 2013 #15
    Now I got where I was doing mistake.

    conserving linear momentum,

    m(v) = 2m v' => v' = v/2

    conserving angular momentum about centre of the loop,

    mv(R/2) = mR2 + m(R2)/4(loop) + {mR2/4(particle)}. ω + (m). (v/2).(R/2) - (m). (v/2).(R/2)(particle +loop com)
    it gives the answer, ω = v/3R
     
  17. Dec 19, 2013 #16
    Why mv(R/2) in the LHS ? Shouldn't the angular momentum of the bullet about the center of the loop be mvR ?
     
  18. Dec 19, 2013 #17
    I conserve A.m. about the center of mass of the system.
     
  19. Dec 19, 2013 #18

    haruspex

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    You can also do it about the point where the centre of the hoop started:
    initial moment of particle = mvr
    final moment of hoop = mr2ω
    final linear velocity of system = v/2
    final moment of particle = mr2ω + (mv/2)r
    The particle has moment mr2ω in consequence of the hoop's rotation, and another (mv/2)r on account of the linear motion of the system.
    mvr = (mv/2)r + 2mr2ω
     
  20. Dec 19, 2013 #19
    But this will give the answer ω = v/ 4r, and answer is v/ 3r
     
    Last edited: Dec 19, 2013
  21. Dec 20, 2013 #20
    You have used MI of the system about the center of the hoop ,whereas it should be MI about the new CM .

    The item in red should mvr= 2m(v/2)(r/2) + (3/2)mr2ω
     
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