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Rotational Motion Problem - 12

  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-b-a.akamaihd.net/hphotos-ak-prn1/q71/s720x720/1520824_1461728107387628_750491406_n.jpg

    Attempt -

    https://fbcdn-sphotos-b-a.akamaihd.net/hphotos-ak-frc3/q71/s720x720/1459220_1461728404054265_66267352_n.jpg


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

    haruspex

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    You don't need to break it into two stages. Just look at work conservation from moment of release to maximum extension.
     
  4. Dec 17, 2013 #3
    According to this, K.E. →Max→at mean position (I think it won't be the release point)
    and
    P.E.→Max→ at ext. position.

    1/2. mv2 + 1/2 Iω2 = 1/2. kx2 (since ω= v/r)

    => 1/2. mv2 + 1/2 (1/2.mr2)(v/r)2 = 1/2. kx2
    => 3/4. mv2 = 1/2. kx2
    x2 = (3/2)(mv2)/k
    Is that correct?
     
  5. Dec 17, 2013 #4

    haruspex

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    It starts at rest, and it will be instantaneously at rest at maximum extension, so there's no KE to consider. There's just the lost gravitational PE and the gained spring PE.
     
  6. Dec 17, 2013 #5
    https://fbcdn-sphotos-f-a.akamaihd.net/hphotos-ak-ash3/1488253_1462165574010548_1361638838_n.jpg
     
  7. Dec 17, 2013 #6

    haruspex

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    Well, well. Another error in the book! Your answer is correct.
     
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