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Rigid Body Rotational Motion

  1. Nov 30, 2005 #1
    "A uniform rod of mass m and length 2a stands vertically on a rough horizontal floor and is allowed to fall. Assuming that slipping has no occured, write the angular velocity of the rod as a function of the angle Theta the rod makes with the vertical."

    Ic = (ml^2)/3
    Torque = mgd
    d = asin(theta)
    Torque = angular accel * I

    mgasin(theta) = ang accel * (ml^2)/3
    l = 2a

    gasin(theta) = ang accel * (m(2a)^2)/3
    gsin(theta) = 4/3 * ang acell * a

    Ang Accel = 3/(4a) * gsin(theta)

    MAIN QUESTION: How do I substitute Angular Acceleration such that I can find an equation that solves angular velocity?
  2. jcsd
  3. Nov 30, 2005 #2


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    Staff: Mentor

    Angular acceleration [itex]\alpha\,=\,\dot\omega\,=\,\ddot\theta[/itex]

    Angular velocity [itex]\omega\,=\,\dot\theta[/itex]
  4. Nov 30, 2005 #3
    I considered using calc to solve for this. But the integral of Angular Acceleration is a function of time. I'm unsure how I can incorporate that together and produce an equation as a function of Theta.

    W = Integral [ 3/4 * 1/a * gsin(theta) ] dt

    Would I just slap on a variable T and give that as my answer (There is no initial angular velocity right)? Or can I somehow subtitute T as a function of Theta?
    Last edited: Nov 30, 2005
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