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Calculating a moment required, using radius of gyration and parallel axis theorem

  1. Oct 14, 2012 #1
    1. The problem statement, all variables and given/known data

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    2. Relevant equations

    radius of gyration:
    r = root (I/m)
    I = moment of inertia
    m = mass

    parallel axis theorem given above

    3. The attempt at a solution

    Okay, so I think the moment about CM is just m*0.24^2, but after that, I'm less sure.
    Is the moment about the hip just m*0.42^2 + m*0.24^2?

    If so, it's taking the acceleration and the distance to the toe into account that I'm having difficulty with. Is it just F = ma? If so, is the m taken from the moment of inertia at the hip, or the toe? I figure the distance to the toe must be significant, but I don't know how to account for it.

    Any help?
     
  2. jcsd
  3. Oct 14, 2012 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Yes, you've got it.

    How would you write the equivalent of F = ma for rotational motion?
    Given the tangential acceleration, how would you find the angular acceleration?
     
  4. Oct 14, 2012 #3
    Okay, maybe it's best to do the last two bits in reverse order;
    if the tangential acceleration is 18m/s^2, and the radius is 1m, is the angular acceleration also 18 radians/s^2?

    Then is the moment: angular acceleration * the moment about the hip?
    And that's the final answer?
     
  5. Oct 14, 2012 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Right!

    Right again.
     
  6. Oct 14, 2012 #5
    Awesome,
    thanks man, I really appreciate it.
     
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