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Rotation about a fixed axis

  1. Nov 16, 2004 #1
    I am having a very hard time with this problem! Even my teacher had trouble with it(of course we are still expected to have it done on our homework!)

    2 doors are uniform and identical. Door A rotates about an axis to its left edge and door B rotates about an axis through the center. The same force F is applied perpendicular to each door at its right edge and the force remains perpendicular to as the door turns. Starting from rest, door A rotates through a certain angle in 3.00s. How long does it take door B to rotate through the same angle.

    So I got as far as 1/3ML^2(alpha)=1/12M(L/2)^2(alpha) and then I get stuck on how o make time fit in. I have been looking at all the equations for rotational motion, but I don't know which one????
     
  2. jcsd
  3. Nov 16, 2004 #2

    Doc Al

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    Staff: Mentor

    Use [itex]\tau = I \alpha[/itex] to find the [itex]\alpha[/itex] for each case. Then apply the kinematic equation that gives [itex]\theta[/itex] as a function of time.
     
  4. Nov 16, 2004 #3
    This is what I am coming up with:
    1/3ML^2(2[itex]\theta[/itex]/t^2)= 1/12ML/2^2(2[itex]\theta[/itex]/t^2)

    ML^2 cancel out and so does the 2[itex]\theta[/itex]
    and I am left with (1/3)(1/t^2)=(1/12)(1/4)(1/t^2)

    Is that right? Where do I go from there?
     
  5. Nov 16, 2004 #4

    Doc Al

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    Staff: Mentor

    The torque is different in each case.
     
  6. Nov 16, 2004 #5
    But the same force is applied to each door in the same spot?? If the torque is different then I am not sure what to do b/c the only known value is time.
     
  7. Nov 16, 2004 #6

    krab

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    Science Advisor

    A: torque is FL
    B: torque is FL/2
     
  8. Nov 16, 2004 #7
    That is what I had, but can't I set them equal to each other?
     
  9. Nov 16, 2004 #8

    Doc Al

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    No, since they are not equal! You can easily relate them though.
     
  10. Nov 16, 2004 #9
    You should use the information given to find out the angle for door A. Treat door A and B as separate cases, door A is there just to tell you the angle. Then you find out the [tex]\alpha[/tex] from the [tex]\tau and I[/tex] that you already have. If you can find out the angle of door A, then you shouldn't have problem working out time for door B because it's sort of working backward with the method you used for A.
     
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