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Angular Position

  1. Nov 5, 2008 #1
    This is the problem:
    During a certain period of time, the angular position of a swinging door is described by θ = 4.92 + 10.7t + 1.91t2, where θ is in radians and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the door at the following times.

    then I'm given for part a) t=0s and part b) is t=3.07s

    I know I have to use the equation for angular position s=r[tex]\theta[/tex] then derive that once for angular velocity ([tex]\omega[/tex]) then take the second derivative for angular acceleration ([tex]\alpha[/tex]). I just don't understand what the values for s and r are and where the come from within this problem. i have that [tex]\theta[/tex](0)=4.92 or .0859rad

    any help or guidance is much appreciated.
  2. jcsd
  3. Nov 5, 2008 #2
    s is the linear position.

    Theta is the angular position (as described in the question).

    That should get you started off.
  4. Nov 6, 2008 #3
    ok I understand that but how do I find out or determine the value of the linear position from the formula theta?
  5. Nov 7, 2008 #4
    The part of the question quoted only asks for the angular position (another term for angle), angular speed and angular acceleration. There'd be no need to use s = rtheta at all, just differentiate what you've been given in the question.
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