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Angular acceleration of a rod pivoted at an end, at angle below the horizontal

  1. Nov 10, 2009 #1
    1. The problem statement, all variables and given/known data

    Tgbeq.jpg

    The rod is at rest and is released, what is the angular acceleration immediately after release? The pivot is frictionless, gravity is 9.8 m/s^2.

    2. Relevant equations

    Icm = (1/12)ML^2 => Inertia at center of mass, this was given in the problem, i don't think it's necessary though

    Iend = (1/3)ML^2 => at pivot end, this was also given in the problem and i confirmed it with the parallel axis theorem

    t = rF*sin(A) =>torque, r is the radius(half the total length L), and F is the force of gravity(m*g)

    [tex]\alpha[/tex] = t/I =>ang accel


    3. The attempt at a solution
    t = rMg*sin(A)

    [tex]\alpha[/tex] = (rMg*sin(A)) / ((1/3)ML^2))
    since r is half the length of the total length L, i can say L = 2r

    [tex]\alpha[/tex] = (rMg*sin(A)) / ((1/3)M(2r)^2))
    [tex]\alpha[/tex] = (3rg*sin(A))/ (4r^2) => 'M's cancel out, (1/3) on bottom becomes 3 on top
    [tex]\alpha[/tex] = (3g*sin(A))/ (4r) =>simplify 'r's


    Is that the right process? I keep getting the wrong answer.

    The angular acceleration should be positive right? I know for translational motion, when an object is speeding up it's acceleration is positive, when an object is slowing down it has negative acceleration, and when the rate of change of the speed is constant, the acceleration is 0. So it should be the same for angular acceleration right?

    Btw: Is there a way to put LaTeX inside LaTeX code? I was trying to put the greek lower case letter "tau" inside fraction operation for alpha = tau / I but I kept getting an error.
     
    Last edited: Nov 10, 2009
  2. jcsd
  3. Nov 10, 2009 #2
    Hi, looks good but I don't understand why you use I=(1/3)MR2. Shouldn't you use the moment of inertia at the center of mass? Can we assume the rod to have uniform distribution of mass?
     
  4. Nov 10, 2009 #3
    It's rotating about the pivot though, that's why I used I=(1/3)MR^2.
     
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