Figure (a) shows a homogeneous slender rod of mass m and length L. Determine the moments of inertia of the rod about the x-, y-, and z-axes that pass through its mass center G.

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Accepted Answer

The term slender implies that the cross-sectional dimensions of the rod are negligible compared with its length. Therefore, the mass of the rod may be considered as being distributed along the x-axis, which means that its moment of inertia about that axis is negligible; that is,

[latex]I_x ≈ 0[/latex]

The differential mass dm chosen for integration is shown in Fig. (b). Because x is the perpendicular distance from both the y-axis and the z-axis to dm, the moments of inertia about these two axes are identical. Letting ρ be the (constant) mass of the rod per unit length, we have dm = ρ dx. The definition of moment of inertia gives

[latex]I_y = I_z = \int {x^2} dm = ρ \int_{−L/2}^{L/2}{x^2} ρ dx = \frac{ρ}{3} (\frac{L^3}{8} + \frac{L^3}{8}) = \frac{ρL^3}{12}[/latex]

Because the mass of the rod is m = ρL, this result may be written as

[latex]I_y = I_z = \frac{mL^2}{12}[/latex]

Question F.1: Figure (a) shows a homogeneous slender rod of mass m and len...

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