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Masses and a Light Rod

  1. Jan 3, 2009 #1
    1. The problem statement, all variables and given/known data
    Two equal masses (m) are connected by a light rod of length L that is pivoted about its center. A downward Force F is applied to the rod at a distance L/4 from the pivot. If you ignore gravity, the linear acceleration of the mass is...

    O--------o--------O


    2. Relevant equations

    torque= Ia

    3. The attempt at a solution
    I don't know where to go with this one because gravity is ignored.
     
  2. jcsd
  3. Jan 3, 2009 #2
    What else is torque equal to?
     
  4. Jan 3, 2009 #3
    torque = r*F... torque= (L/4)(F) right?
     
  5. Jan 3, 2009 #4
    ya but torque doesn't equal to linear acceleration but angular acceleration. That might be what you mean. And can you find I for the object?
     
  6. Jan 4, 2009 #5
    I'm really lost at what to do in this question.
     
  7. Jan 4, 2009 #6

    Doc Al

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    Staff: Mentor

    Use the torque to find the angular acceleration of the rod. (What's the rotational inertia of this system?) Then relate angular acceleration to linear acceleration of the masses.
     
  8. Jan 4, 2009 #7
    a=torque/I == (F*(L/4))/ (2(m(L/2)^2)) = a= F/2mL ??? a(linear)= a(rot)*radius == F/2ml= a*(L/2) solve for a and thats the answer???
     
  9. Jan 4, 2009 #8

    Doc Al

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    Staff: Mentor

    Yes, that's all there is to it. Use different symbols for angular acceleration (α) versus linear acceleration (a) to avoid confusion.
     
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