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Rotational Motion Problem - 4

  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-e-a.akamaihd.net/hphotos-ak-frc1/q71/s720x720/601231_1461726974054408_338858428_n.jpg

    Attempt-
    https://fbcdn-sphotos-c-a.akamaihd.net/hphotos-ak-frc3/q71/s720x720/1509289_1461726967387742_825364419_n.jpg


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

    haruspex

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    For angular momentum, you need to be careful how you pick your reference point. You can use the mass centre or any fixed point in the inertial frame. Your choice appears to be the middle of the bar, which is neither.
     
  4. Dec 17, 2013 #3
    So according to you which point should I take here for conserving the angular momentum, Bottom or Top?
     
  5. Dec 17, 2013 #4

    haruspex

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    Neither. You can use the mass centre of the system (messy) or any fixed point. How about the location of the rod's centre before it moves?
     
  6. Dec 17, 2013 #5
    Well in my attempt, I conserve angular momentum about the center of the rod.
     
  7. Dec 17, 2013 #6

    haruspex

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    Yes, but you must not use that in the sense of wherever the rod is at any given time, as it moves. You can use that fixed point in space where the middle of the rod was at the start.
     
  8. Dec 17, 2013 #7
    Ok Let me try for that.
     
  9. Dec 17, 2013 #8
    https://fbcdn-sphotos-b-a.akamaihd.net/hphotos-ak-prn2/1513192_1462156994011406_1703815557_n.jpg
     
  10. Dec 17, 2013 #9

    haruspex

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    I just reread the OP. A mistake in your diagram misled me. You drew it as though the masses are m and 2m, with both velocities v. It's the other way about: both masses are m and the velocities are v and 2v.
    This is easier. The middle of the bar will be the mass centre, so you can use your original method.
     
  11. Dec 17, 2013 #10
    https://fbcdn-sphotos-d-a.akamaihd.net/hphotos-ak-prn2/1471173_1462372860656486_905776386_n.jpg

    Still not getting the perfect result.
     
  12. Dec 17, 2013 #11

    haruspex

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    I'm having some trouble telling your v's from your b's. It looks like you may have done too. How did you get the amplitudes on the trig functions?
     
  13. Dec 18, 2013 #12
    Well, there I think I have done wrong calculation for the distances.
    What I have found is that, Now the rod has translational speed as well as the angular speed.
    Translational velocity => v/2
    and angular speed => ω = (3v)/b
    After time t angular displacement of the rod, θ = ωt => θ = (3v)t/b
    and due to angular motion,the top particle would be moving in the direction towards ivth quadrant[+x, -y] if θ < 900 (assumed at some time.)
    So its speed would be (b/2)ω = (3v)/2 due to rotation.
    Taking its component in x axis and y - axis
    X component, (3v)/2. cosθ => (3v)/2. cos[(3v)t/b]
    Y component, (3v)/2. sinθ => (3v)/2. sin[(3v)t/b]

    Then the net speed in the x direction would be, (v/2) + (3v)/2. cos[(3v)t/b]
    and in the y direction, (3v)/2. sin[(3v)t/b]
    Now after integrating it I got the answer, Thanks a lot.
     
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