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Torque question

  1. Feb 6, 2010 #1
    A 51.0 kg uniform solid cylinder has a radius of .392m. If the cylinder accelerates at 9.2*10^-2 rad/s^2 as it rotates about an axis through the center, how large is the torque acting on the cylinder? Answer in N*m.

    so I plugged into the equation:
    T=F*d
    so
    T=(5.1*(9.2^-2)*.392
    T=1.839

    that was wrong. If anyone has any ideas what I did wrong and feels like sharing their wisdom, that would be great.
     
  2. jcsd
  3. Feb 6, 2010 #2

    tiny-tim

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    Hi BoldKnight399! :smile:

    (have a tau: τ and an omega: ω and an alpha: α :wink:)
    uhh? :confused:

    But you don't know what F is.

    Use τ = Iα. :smile:
     
  4. Feb 6, 2010 #3
    But what is I? Is it:
    (1/12)mL^2
    and if so...what is my L? I only know the radius.
     
  5. Feb 6, 2010 #4

    tiny-tim

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    You need to learn every moment of inertia on this list … http://en.wikipedia.org/wiki/List_of_moments_of_inertia" [Broken] :wink:
     
    Last edited by a moderator: May 4, 2017
  6. Feb 6, 2010 #5
    ok wow this is really helpful except that I don't have the height of the cylinder. What do i do then?
     
  7. Feb 6, 2010 #6
    First of all, (5.1*(9.2^-2)*.392) is only 0.1839264
    However (51*(9.2*10^-2)*.392) = 1.839264
    The cylinder is solid. Rotational inertia is dependent on the shape of the object and its mass distribution.
    You should either look it up ( see http://hyperphysics.phy-astr.gsu.edu/Hbase/mi.html#cmi ), or compute it.
     
  8. Feb 6, 2010 #7
    You don't need the height. Moment of inertia is independent of it.
     
  9. Feb 6, 2010 #8
    Wait...i don't? o never mind. This all makes sense now. Thank you!
     
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