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Establish the formula of moment of inertia of a rectangular rod along the axis passing through its center of mass and perpendicular to the length of the rod.

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Solid Rod of Rectangular Cross Section

Moment of Inertia about the axis perpendicular to the length and passing through the Center of Mass
A rectangular solid rod is shown in the figure, whose length is L, breadth B, width d and mass is M. We can assume that this is made up of various thin rectangular cross-sections. The total moment of inertia of the rod would be equal to the sum of the moment of inertia of all these rectangular cross-sections.

Suppose every rectangular cross-section’s mass is m. According to the figure a cross-section of dx breadth is at a distance x parallel to the YY’ axis. If mass of unit area of the strip is a then the whole cross-section mass is B dxσ. Therefore, the moment of inertia about the YY’ axis is = B dxσ x2

Fig: Moment of inertia of a solid rod of rectangular cross section about the axis perpendicular to the length and passing through center of mass

The moment of inertia of the total rectangular cross-section about the YY’ axis

Here, m = σBL is the mass rectangular cross-section.
Similarly, the moment of inertia about the XX’ axis is;
dIXX’ = \(\frac{m B^{2}}{12}\)
dIXX’ and dIYY’ are the moment of inertia of two perpendicular axis in the same plane. Figure shows the axis OZ perpendicular to the intersection point of these two axis and its inertia;
dIOZ = dIYY’ + dIXX’
\(=\frac{m L^{2}}{12}+\frac{m B^{2}}{12}=m\left[\frac{L^{2}+B^{2}}{12}\right]\)

The moment of inertia of such rectangular laminas will be the sum of all about the Z axis.
Hence, the moment of inertia about the OZ axis, which is perpendicular to length and passing through the center is;

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