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heena Group Title

a thin uniform rod of length l and mass m is swinging freely about a horizontal axis passing through its end. its maximum angular speed is w(omega). its centre of mass rises to a max. height of?

  • 2 years ago
  • 2 years ago

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  1. heena Group Title
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    options are \[1) 1l ^{2}\omega ^{2}\div3g\] \[2) 1l \omega \div6g\] \[3) 1l ^{2}\omega ^{2}\div2g\] \[4) 1l ^{2}\omega ^{2}\div6g\]

    • 2 years ago
  2. heena Group Title
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    it comes in my xam and this qn move above from my head sso i didnt solve this :P and asking u guys how to solve

    • 2 years ago
  3. heena Group Title
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    eve i have solution too but need sm1 to explain may i post the solution?

    • 2 years ago
  4. hosein Group Title
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    i think this fig and think we use consevation of E

    • 2 years ago
  5. heena Group Title
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    yes

    • 2 years ago
  6. hosein Group Title
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    |dw:1329508279870:dw|

    • 2 years ago
  7. hosein Group Title
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    i get C.M point for source point of E

    • 2 years ago
  8. heena Group Title
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    this qn is so bad i hate such type qn really :/

    • 2 years ago
  9. hosein Group Title
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    i don't find solve from options

    • 2 years ago
  10. hosein Group Title
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    sory say that

    • 2 years ago
  11. heena Group Title
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    np :)

    • 2 years ago
  12. heena Group Title
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    loss in KE=gain in PE 1/2 Iw^2=mgh 1/2x Ml^3/3 x w^2=mgh h=l^2w^2/6g

    • 2 years ago
  13. heena Group Title
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    i dont get the lat 2 steps

    • 2 years ago
  14. hosein Group Title
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    this formula is right but i can't understand why don't get result \[0.5mv^2+0.5I \omega^2=mgl 0.5(1-\cos \theta)\]

    • 2 years ago
  15. hosein Group Title
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    ohhhhhhh thats right i did a big mistake

    • 2 years ago
  16. hosein Group Title
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    rod had a net rotational energy so the 2nd statement in left hand's of my equation is zero

    • 2 years ago
  17. hosein Group Title
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    and in right hand (mgl/2)cos theta=h and I =(1/3)ml^2 hence moment of inertia by this axis equal to\[I=I _{C.M}+Ma ^{2}\] where "a" is perpendicular distance between that axis and C.M axis I(C.M)for rod is (1/12)ml^2

    • 2 years ago
  18. hosein Group Title
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    do you get it?

    • 2 years ago
  19. hosein Group Title
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    I is moment of inertia around arbitrary axis.

    • 2 years ago
  20. hosein Group Title
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    you'r made mistake in typing last realtion,i think this:\[(1/2)I \omega^2=mgh\] in gain point rod has only PE, hence in that point "omega"=0

    • 2 years ago
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