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Oscillating Rectangular Plate

  1. May 30, 2003 #1
    I think I'm lost in the ugly algebra, but I want to make sure.

    A uniform rectangular plate is suspended at point P (top center of the rectangle), and swings in the plane of the paper about an axis through P. At what other point between P and O (center of the rectangle), along PO, could the plate be suspended to have the same period of oscillation as it has around P. O is the cm of the plate.
    The rectangle has a width and b length.

    The answer is (a^2b)/(3(a^2 + b^2))

    T = period.
    I = moment of inertia
    M = mass

    1) T = 2pi(I/mgd)^(1/2) I = (1/12)M(a^2 + b^2) + M(b/2)^2

    d = b/2

    2) T = 2pi[(8b^2 + 2a^2)/(12mgb)]^(1/2)

    I then substitute (b/2 - x) for d in equation 1 and set it equal to equation 2 and solve for x, but I'm not getting the correct answer. Is this the correct approach?

    Thanks.

    The perpetually confused discoverer02.
     
  2. jcsd
  3. May 30, 2003 #2
    I think you're on the right track, but it's not quite that "easy".

    The point you're looking for (let's call it Q) has a different moment of inertia, so the expression for the period when oscillating about Q will be different.

    Then, if x is the distance from cm to Q, I think the equation to solve is:
    IP/(b/2) = IQ/x

    but I haven't been able to get that answer. You're right, the algebra is uuuuugly.
     
  4. May 30, 2003 #3
    Where'd you get that answer?

    I get
    x = (a2+b2)/6b, where x is the distance from the center of mass to the new point Q.

    (Assuming my equation IP/(b/2) = IQ/x is correct.)
     
  5. May 30, 2003 #4
    The answer came from my trusty textbook.

    I did remember to substitute the new distance from cm to pivot point into the moment of inertia, but try as I may I don't come up with the same terms the book does.
     
  6. May 30, 2003 #5
    I was basically trying to solve the same equation except I was using b/2-x instead of x which makes the problem even uglier. You're solving for the distance from the pivot and I'm solving for the distance from the center of mass, but all in all both equations seem OK to me. I've posted my question on our class's bulletin board also. I'll let you know if there's a problem with the answer in the book.

    Thanks for your help.
     
  7. May 30, 2003 #6
    You're welcome.

    My answer seems to check out. If x = (a2+b2)/6b, then

    IQ = Icm + m((a2+b2)/6b)2
    IQ = (m/12)(a2+b2) + (m/36b2)(a2+b2)2

    and if you divide that mess by (a2+b2)/6b you get a worse mess that eventually simplifies to
    (ma2 + 4mb2)/6b

    And similarly
    IP = Icm + (mb2)/4
    IP = (m/12)(a2+b2) + (mb2)/4
    and if you divide that by 2b, it also becomes
    (ma2 + 4mb2)/6b

    So I think that proves that x = (a2+b2)/6b
     
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