I think I'm lost in the ugly algebra, but I want to make sure.(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Oscillating Rectangular Plate

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