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2015-10-05T18:22:10+05:30

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See Diagram.

   First we will find the Moment of Inertia of a rectangular sheet/plate of surface density  m units/square area.  Let the total mass be = M = m a b where a and b are the width and height of the plate.    M = 4 m s t.

Given,  P is midpoint of half of diagonal =>  OP = √(a²+b²)  / 4

Moment of Inertia of plate about X axis is :
I_X= \int\limits^{+\frac{b}{2}}_{-\frac{b}{2}} {r_y^2} \, dm \\\\= \int\limits^{\frac{b}{2}}_{-\frac{b}{2}} {y^2} \, ( m\ a\ dy)\\\\=\frac{1}{6}m\ a\ b^3=\frac{1}{6}M\ b^2\\\\Similarly,\ I_Y=\int\limits^{+\frac{a}{2}}_{-\frac{a}{2}} {r_x^2} \, dm \\\\= \int\limits^{\frac{a}{2}}_{-\frac{a}{2}} {x^2} \, (m\ b\ dx)\\\\=\frac{1}{6}m\ b\ a^3=\frac{1}{6}M\ a^2\\\\I_Z=I_X+I_Y=\frac{1}{6}M\ (a^2+b^2)\\\\I_{Z_P}=I_{Z_O}+M*OP^2\\\\=\frac{5}{12}M(a^2+b^2)=\frac{5}{6}M\ a^2,\ for\ a\ Square\\


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