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Tangential Force Required to Slow Cylinder to a rest?

  1. Nov 6, 2012 #1
    1. The problem statement, all variables and given/known data
    a)-The massive grinder for wood-pulp in a newsprint paper factory, which can render about a half Tonne of wood into a thin soup in less than a minute, is essentially a stone cylinder of radius 0.900 m and mass about 6.00 Tonnes. An enormous motor spins this cylinder on its axis while hydraulic cylinders press the pocket of wood against its rim. A typical rotational speed for the motor is 62.0 rpm (rotations per minute). Suppose that while the paper plant is in full operation there is a power outage due to an ice-storm. How much energy (in J) is available to grind wood just from the rotational kinetic energy of the grinding wheel?

    I HAVE ALREADY SOLVED PART a,
    NEED HELP WITH B:


    b)-If the grinding wheel comes to rest in 80.0 ms what is the tangential force (in N) that the wood presents to the grinding wheel if this force is constant during the slowing down?



    2. Relevant equations

    KE = 1/2Iw2
    F = ∆p/∆t ...?
    t = Fr
    ∆w/∆t = α

    3. The attempt at a solution
    honestly don't know where to start.
     
  2. jcsd
  3. Nov 6, 2012 #2

    tiny-tim

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    Hi ObviousManiac! :smile:
    First find the angular acceleration, using the standard constant acceleration equations but with θ ω and α instead of s v and a.

    Then find the torque that produces that angular acceleration, and so find the force at the rim. :wink:
     
  4. Nov 7, 2012 #3
    Silly question - what is my θ?
     
  5. Nov 7, 2012 #4

    tiny-tim

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    there's no θ in this particular question, only the angular equivalent of u v a and t :wink:
     
  6. Nov 7, 2012 #5
    to find α, could I not do this?:

    wf = wi + αt

    wf = 0, so

    -wi = αt

    (where wi = 62 rpm/60 sec * 2π rad = 6.49 rad/s)

    so

    - 6.49 = α(.08)

    α = -81.125 rad/s2
     
  7. Nov 7, 2012 #6

    tiny-tim

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    looks good! :smile:

    and so the tangential force is … ? :wink:
     
  8. Nov 7, 2012 #7
    T = mr2a

    = 6000(.92)(-81.125)

    = -394267.5

    rF = -394267.5

    .9F = -394267.5

    F = -394267.5/.9

    F = - 438075 N

    .... Is this correct?
     
  9. Nov 7, 2012 #8

    tiny-tim

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    No, that's the wrong I.

    Also, it would be easier (and quicker, and less likely to lead to mistakes) if you wrote all your equations at the start …

    in this case, Fr = τ = Iα, so F = Iα/r :wink:
     
  10. Nov 7, 2012 #9
    isn't I = mr2 ?

    So would it be

    F = Iα/r

    F = mr2α/r = mrα

    ?
     
  11. Nov 7, 2012 #10

    tiny-tim

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    yes :smile:, except I for a solid cylinder isn't mr2

    you need to learn the moment of inertia of every common geometrical shape :wink:
     
  12. Nov 7, 2012 #11
    OH

    I didn't even think about that... wow I'm really dropping the ball tonight.

    So:

    I = .5mr^2

    F = Ia/r

    f = .5mr^2a/r = .5mra

    = (6000)(.9)(-81.125)/2

    = -219037.5 N
     
  13. Nov 7, 2012 #12

    tiny-tim

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