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Homework Help: TORQUE A cylindrical fishing reel has a mass of 0.65 kg and a radius of 4.3 cm .

  1. Jan 9, 2008 #1
    TORQUE A cylindrical fishing reel has a mass of 0.65 kg and a radius of 4.3 cm.....

    1. The problem statement, all variables and given/known data

    A cylindrical fishing reel has a mass of 0.65 kg and a radius of 4.3 cm. A friction clutch in the reel exerts a restraining torque of 1.3 N·m if a fish pulls on the line. The fisherman gets a bite, and the reel begins to spin with an angular acceleration of 66 rad/s2.

    (a) Find the force of the fish on the line.
    ______N
    (b) Find the amount of line that unwinds in 0.50 s.
    ______m



    3. The attempt at a solution


    for the fishing reel, I=m*r^2

    Net torque equals I*α
    α=66 rad/s^2
    I will assume that the torque from the line is at the edge of the radius:
    F*4.3/100-1.3=I*66
    where F is the force of the fish
    F=32.1 N

    θ=.5*α*t^2

    θ=.5*66*.5^2
    8.25 radians
    d=2*π*r*θ
    =2*3.14*4.3*8.25/100
    2.23 m



    but it says those answers are wrong. what do i do?
     
  2. jcsd
  3. Jan 9, 2008 #2

    Doc Al

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    Staff: Mentor

    The rotational inertia of a cylinder is [itex](1/2)mr^2[/itex]. (Assuming that it's a solid cylinder.)
     
    Last edited: Jan 9, 2008
  4. Jan 11, 2008 #3
    but i dont understand why the second answer is wrong, either
     
  5. Jan 11, 2008 #4

    Doc Al

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    So far, so good.
    No, distance just equals r*θ. (Since theta is in radians.)
     
  6. Jan 11, 2008 #5
    i'm sorry, i still don't understand...
     
  7. Jan 11, 2008 #6

    Doc Al

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    Staff: Mentor

    Last edited by a moderator: Apr 23, 2017
  8. Jan 11, 2008 #7
    Last edited by a moderator: Apr 23, 2017
  9. Jan 11, 2008 #8

    Doc Al

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    I don't see anything wrong with the method you used in part a (except for that moment of inertia).

    What do you think you left out?
     
  10. Jan 11, 2008 #9
    I feel i need to rearrange the whole thing, according to the equation you gave me.
    So if I=.5mr^2, how would i set it up to that....
    I am really confused.
    Also i'm partly a perfectionist so i need to start from the beginning hah, just a habit i picked up..
     
  11. Jan 11, 2008 #10

    Doc Al

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    I would set it up exactly as you did:
    [tex]\tau_{net} = I \alpha[/tex]

    [tex]Fr -\tau_{rest} = I \alpha[/tex]

    [tex]Fr = I \alpha + \tau_{rest} = (0.5mr^2) \alpha + \tau_{rest}[/tex]

    And so on...
     
  12. Jan 11, 2008 #11
    using .5mr^2*a*Trest i get 1.34.......
     
  13. Jan 11, 2008 #12

    Doc Al

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    For F*r, I presume? Now solve for F.
     
  14. Jan 11, 2008 #13
    it worked out perfectly. thank you!!!!
     
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