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shubhamsrg

  • one year ago

A cube of side x is moving with velocity u on a horizontal smooth plane. There is a ridge at O. find angular speed of the cube after it hits O

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  1. shubhamsrg
    • one year ago
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    |dw:1357204606212:dw|

  2. shubhamsrg
    • one year ago
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    i know angular momentum will be conserved here but unable to apply it..

  3. Diwakar
    • one year ago
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    Apply the conservation of energy. Mu^2/2=Iw^2/2 where I is the moment of inertia about the vertical edge about which it will rotate.

  4. shubhamsrg
    • one year ago
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    so, mu^2 = 2/3 (mx^2) * w^2 w^2 = 3u^2 /2x^2 nopes , not the right ans why you used cons of energy ?

  5. Diwakar
    • one year ago
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    Are you sure you are getting the I right?

  6. shubhamsrg
    • one year ago
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    2/3 mx^2 ..i googled it! though i can derive that also..

  7. shubhamsrg
    • one year ago
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    :P

  8. Diwakar
    • one year ago
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    what is the correct answer? I will try to solve it at some other time?

  9. shubhamsrg
    • one year ago
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    though that was an option which came by this method, correct ans is 3u/2x

  10. yrelhan4
    • one year ago
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    answer is 3v/4x..

  11. Diwakar
    • one year ago
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    @yrelhan4 Did you use Mux/2=Iw

  12. yrelhan4
    • one year ago
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    yup.. I would be icm + ma^2/2

  13. Vincent-Lyon.Fr
    • one year ago
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    Use conservation of angular momentum, not of kinetic energy.

  14. Vincent-Lyon.Fr
    • one year ago
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    Moment of inertia about edge is correct.

  15. shubhamsrg
    • one year ago
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    ohh yes,,3u/4x..got it,,was easy..hmm

  16. shubhamsrg
    • one year ago
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    thanks all..

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