1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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:

    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


    User Avatar
    Science Advisor
    Homework Helper

    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:
    and if so...what is my L? I only know the radius.
  5. Feb 6, 2010 #4


    User Avatar
    Science Advisor
    Homework Helper

    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook