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Rotation physics homework

  1. Nov 28, 2003 #1
    Hey all--
    Can someone try to ifnd what I'm doing wrong with this question, I must have sort of brain lapse in the middle of a 4-day weekend and all:

    Question: A (very large) door has a mass of 44,000kg, and a rotational inertia about a vertical axis through its hingesof 8.7*10^4 kg*m^2, and has a front face of 2.4m. NEglecting friction, what steady force perpendicular to the door can move it from rest through an angle of 90 degrees in 30s?

    My incorrect solution: ("@" refers to the angle in radian measure), and "&" refers to angular acceleration)

    @ = 2(pi)*r/4 @=2.4(pi)/2 radians

    @ = .5(&)(t^2)
    1.2(pi) = .5(&)(30^2)
    & = .0084 rad/s^2

    I(&) = r x F = rFsin90
    (8.7*10^4)(.0084) = (2.4)(F)(1)
    F= 304.5 N

    However, the actual answer is apparently 130N. What did I do wrong?
  2. jcsd
  3. Nov 28, 2003 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Re: Rotation

    How many radians in 90 degrees?
  4. Nov 28, 2003 #3
    hah oops...yeah i knew i had a brain fart there (thinks all the way back to trig)
    of course...radians dont change proportionally to radius...duhhh
    Thanks Doc
    Last edited: Nov 28, 2003
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