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Angular acceleration problem - fishing reel

  1. Apr 21, 2006 #1
    A fish takes the bait and pulls on the line with a force of 2.2 N. The fishing reel, which rotates counterclockwise without friction, is a solid cylinder of radius 0.049 m and mass 0.82 kg.

    What is the angular acceleration of the fishing reel?

    I used angular acceleration = RT/I, and got 54.7536 rad/s^2, but the answer is not correct. Anyone know why? Thanks!
     
  2. jcsd
  3. Apr 21, 2006 #2

    nrqed

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    Two thoughts (I did not check your number). Are you sing Webassign? You have tried with fewer decimals? And what did you use for the moment of inertia?
     
  4. Apr 21, 2006 #3
    Yes, I am using WebAssign. I have not tried with fewer decimals. For the moment of inertia, I used 0.04018...mr^2 = (0.82)(0.049)^2.
     
  5. Apr 21, 2006 #4

    nrqed

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    I am confused.. what is the 0.04018???

    In any case, for a colid cylinder, I = 1/2 M R^2
     
  6. Apr 21, 2006 #5
    I never learned about that 1/2 thing. So (.5)(0.82)(0.049)^2 = 0.00098441
     
  7. Apr 21, 2006 #6

    nrqed

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    yes.

    Check in your book, you must have a list of moments of inertia for different objects around different axis of rotation. MR^2 works only for point masses or hollow cylinders.
     
  8. Apr 21, 2006 #7
    Oh okay, so I just solve RT/I, and I should get the answer...109.5072 rad/s^2?
     
    Last edited: Apr 21, 2006
  9. Apr 21, 2006 #8

    nrqed

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    ???? What did you use for R, T and I????
    Oh.. Ithink you corrected it as i was posting,

    That sounds right to me
     
  10. Apr 21, 2006 #9
    Okay cool!
     
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