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Torque of a Uniform Rod

  1. May 15, 2014 #1
    3. The attempt at a solution

    Hello, I'm having trouble with energy in rotational motion. The question asks:
    Consider a uniform rod of mass 12 kg and length 1.0 m. At its end, the rod is attached to a fixed, friction-free pivot. Initially the rod is balanced vertically above the pivot and begins to fall (from rest) as shown in the diagram. Determine, a.the angular acceleration of the rod as it passes through the horizontal at B.

    Ok. So, here's my logic. The answer is wrong, I'm aware, but I'd like to know why.
    Einitial=Efinal Einitial = mg(L)/2 =58.8J

    Efinal = (1/2)(I)(w^2) (I can use h=0 as a reference point in the middle of the motion, right?)
    =(1/6)mL^2 w^2
    w = 5.42rad/s


    Ok, so correct me if I'm wrong but this is the velocity as soon as the rod is horizontal. Let's continue -- using this equation:

    Where a = angular acceleration d = θ (in rad)
    Didn't the rod rotate 90 degrees (vertical to horizontal)? -- if so, its radians are 1.57

    w^2 = wo^2 + 2ad (5.42 ) /(2(1.57)) = 9.3rad/s^2
    The answer is something like 14.7rad/s^2... I feel like something's wrong with the theta.
     
    Last edited: May 15, 2014
  2. jcsd
  3. May 15, 2014 #2
    Your calculation for ω is correct but it also appears that it's irrelevant to the angular acceleration. When the rod is perfectly horizontal with the pivot (which I assume from the question even though you didn't attach a picture), the only thing we need to figure out is the net torque at that moment in time. And since you probably know that ##T_{net}/I=\alpha##...
     
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