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Torque 3

  1. Oct 20, 2011 #1
    1. The problem statement, all variables and given/known data
    A uniform horizontal rod of mass 2.3 kg and
    length 0.32 m is free to pivot about one end
    as shown. The moment of inertia of the rod
    about an axis perpendicular to the rod and
    through the center of mass is given by
    I = (ml^2)/12
    If a 4.6 N force at an angle of 76 to the hor-
    izontal acts on the rod as shown, what is the
    magnitude of the resulting angular accelera-
    tion about the pivot point? The acceleration
    of gravity is 9.8 m/s2 .
    Answer in units of rad/s2

    2. Relevant equations
    torque = r * F
    I = (ml^2)/12
    torque = I[itex]\alpha[/itex]

    3. The attempt at a solution
    I = ml^2/12
    =((2.3)(0.32)^2)/12
    =0.019626667

    torque = r * F
    = 0.32 * 4.6sin 76
    =1.428275309

    torque = I[itex]\alpha[/itex]
    1.428275309 = 0.01962667[itex]\alpha[/itex]
    [itex]\alpha[/itex]= 72.77216711
    a=r[itex]\alpha[/itex]
    =72.77216711 * 0.32
    =23.28709348
    BUT it's wrong...
     

    Attached Files:

  2. jcsd
  3. Oct 20, 2011 #2

    SteamKing

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    Staff Emeritus
    Science Advisor
    Homework Helper

    Hint: is the pivot point of the rod located at the center of mass?
     
  4. Oct 20, 2011 #3
    oh do i use parallel axis theorem here?
     
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