Physics · Ch 5 — Motion of System of Particles and Rigid Bodies

Moment of Inertia of a Uniform Rod

5.4.1

Moment of Inertia of a Uniform Rod

Consider a uniform rod of mass $M$ and length $\ell$ , and find its moment of inertia about the axis passing through its own center of mass, perpendicular to the rod (Figure 5.21). Choose the origin to coincide with the center of mass (the geometric center of the rod), with the rod lying along the x-axis, and consider an infinitesimally small mass element $dm$ , of width $dx$ , located at distance $x$ from this origin.

Since the mass is uniformly distributed, the linear mass density (mass per unit length) is $\lambda=\dfrac{M}{\ell}$ , so the mass of the small element is

dm=\lambda\,dx=\frac{M}{\ell}\,dx.

Its contribution to the moment of inertia is $dI=(dm)x^2=\dfrac{M}{\ell}x^2\,dx$ . Because the mass is distributed symmetrically on either side of the center, the integration limits run from $-\ell/2$ to $+\ell/2$ :

I=\int_{-\ell/2}^{\ell/2}\frac{M}{\ell}x^2\,dx=\frac{M}{\ell}\left[\frac{x^3}{3}\right]_{-\ell/2}^{\ell/2}=\frac{M}{\ell}\left(\frac{\ell^3}{24}+\frac{\ell^3}{24}\right)=\frac{M}{\ell}\cdot\frac{\ell^3}{12}.

\boxed{I=\frac{1}{12}M\ell^2}

Axis through one end. The very same integration technique, but with the origin now fixed at one end of the rod (rather than its center), and the limits of integration running from $0$ to $\ell$ instead of $-\ell/2$ to $\ell/2$ , gives

I=\int_0^\ell\frac{M}{\ell}x^2\,dx=\frac{M}{\ell}\left[\frac{x^3}{3}\right]_0^\ell=\frac{M}{\ell}\cdot\frac{\ell^3}{3},

$\boxed{I=\frac{1}{3}M\ell^2}.$ …

Figure 5.21Setting up the integral for the moment of inertia of a rod

What this figure shows. A uniform rod of length l lies along the x-axis with its origin O fixed at its own center (its center of mass and geometric center); an infinitesimally thin slice of the rod, of mass dm and width dx, is marked at a distance x from this origin, and the integration limits run symmetrically from minus l over 2 to plus l over 2 to cover the whole rod on both sides of the center. …

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