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Circular motion, kinematics

  1. Nov 13, 2008 #1
    this is my 1st excercise with changing tangent acceleration with circular motion, would appreciate if anyone can tell me where im going wrong and what to change

    a small body connected to a rod with a length of 2m, r=2m, is moving in a horizontal circular motion, with an angular acceleration of

    time is measured in seconds,
    at t=0s the body is at point A (at 12 o clock on a clock face) travelling anti clockwise, and at t=3s it is known v=0(the body stops).
    a) tangent acceleration (a)
    b) actual velovity (v)
    c) angular velocity ([tex]\omega[/tex])
    d) when will the body return to A(12 o clock position)
    e) when will the body reach B (6 o clock position)

    v=[tex]\int[/tex]adt=[tex]\int[/tex]=[-[tex]\frac{\Pi^{2}}{6}[/tex]sin[tex]\frac{\Pi}{6}[/tex]t]dt=cos[tex]\frac{\Pi}{6}[/tex]t + C
    v(t=3)=0=cos[tex]\frac{\Pi}{2}[/tex] + C ===>C=0


    now i get stuck, i realise that what i need to to is find the time, t, that the angle of the body relative to point A [tex]\vartheta[/tex], is equal to 0 or 2[tex]\Pi[/tex],



    does this mean that it will rach point A every 6 seconds??
    Point B
    similarily with B but comparing thetha with "pi" 180 degrees, but because of the limits of thetha, because it is limites by sinus' limits, it cannot pass 3, therefore cannot reach pi

    have i done something wrong, or is this true and the body will never reach B, ie it will start at A, reach C, at 9 o clock, after 3 seconds, stop, return to A, reach D, at 3 o clock, stop, A....... in a half circular motion
    Last edited by a moderator: Nov 13, 2008
  2. jcsd
  3. Nov 13, 2008 #2
    hope my terminology is correct, translated into english
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