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Angular acceleration

  1. Jun 3, 2005 #1
    More questions!
    A rotating door is made from four rectangular glass panes, as shown in the drawing. The mass of each pane is 91 kg. A person pushes on the outer edge of one pane with a force of F = 58 N that is directed perpendicular to the pane. Determine the magnitude of the door's angular acceleration.

    OK, so equations...

    F sub T = mass * alpha sub T,
    and alpha sub T = radius * angular accel

    So if I try to apply 58N as the tangential force and 364kg as the mass, and solve for alpha sub T, why can't I divide alpha sub T by the radius of the door and get the answer? :yuck:
     
  2. jcsd
  3. Jun 3, 2005 #2

    Pyrrhus

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    could you be more clear with your notation?
     
  4. Jun 3, 2005 #3
    Sorry, I'll give LaTeX a try.
    [tex]F_T=m\alpha_T[/tex], and [tex]\alpha_T=ra[/tex], so why doesn't [tex]F_T=mra[/tex]?

    If you were to do this, you'd get...

    58N=364kg*1.2m*a, right? That's .133, but it's wrong :grumpy:
     
  5. Jun 3, 2005 #4

    Pyrrhus

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    [tex] a = r \alpha [/tex]

    [tex] \sum_{i=1}^{n} F_{i} = ma [/tex]
     
  6. Jun 3, 2005 #5
    Something without integrals, please :) It's trig calc.
     
  7. Jun 3, 2005 #6
    Sorry, trig physics.
     
  8. Jun 3, 2005 #7

    Pyrrhus

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    That's the sigma notation for summation, that's not an integral... :uhh:
     
  9. Jun 3, 2005 #8
    Sorry, blonde moment. OK, so the only force being applied is the tangential force of 58N...right? So,

    [tex]a=\frac{58N}{364kg}[/tex], or 0.159. Applying this value to a, and applying to the other equation, [tex]0.159=1.2m\alpha[/tex];[tex]\alpha=0.1325[/tex]...but that isn't the answer. What have I missed?
     
  10. Jun 3, 2005 #9

    Pyrrhus

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    I won't know what's wrong without the image. So far your original equations were wrong, now it seems you're using "correctly" the right ones.
     
  11. Jun 3, 2005 #10
  12. Jun 3, 2005 #11
    bump, please :)
     
  13. Jun 3, 2005 #12
    I would tackle this from a torque perspective. You have a force of 58N acting at a radius of 1.2M. Find the center of mass of the panes and find that point's moment of inertia and used Newton's 2nd Law for torques to find alpha..
     
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