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==== Examples ==== {{main|List of moments of inertia}} [[File:Moment of inertia rod center.svg|thumb|right]] The moment of inertia of a '''compound pendulum''' constructed from a thin disc mounted at the end of a thin rod that oscillates around a pivot at the other end of the rod, begins with the calculation of the moment of inertia of the thin rod and thin disc about their respective centers of mass.<ref name="Beer"/> * The moment of inertia of a '''thin rod''' with constant cross-section <math>s</math> and density <math>\rho</math> and with length <math>\ell</math> about a perpendicular axis through its center of mass is determined by integration.<ref name="Beer"/>{{rp|p=1301}} Align the <math>x</math>-axis with the rod and locate the origin its center of mass at the center of the rod, then <math display="block"> I_{C, \text{rod}} = \iiint_Q \rho\,x^2 \, dV = \int_{-\frac{\ell}{2}}^\frac{\ell}{2} \rho\,x^2 s\, dx = \left. \rho s\frac{x^3}{3}\right|_{-\frac{\ell}{2}}^\frac{\ell}{2} = \frac{\rho s}{3} \left(\frac{\ell^3}{8} + \frac{\ell^3}{8}\right) = \frac{m\ell^2}{12}, </math> where <math>m = \rho s \ell</math> is the mass of the rod. * The moment of inertia of a '''thin disc''' of constant thickness <math>s</math>, radius <math>R</math>, and density <math>\rho</math> about an axis through its center and perpendicular to its face (parallel to its axis of [[rotational symmetry]]) is determined by integration.<ref name="Beer"/>{{rp|p=1301}}{{Failed verification|date=June 2019|reason=page 1301 is the index of the book. I assume someone made a mistake with the page number}} Align the <math>z</math>-axis with the axis of the disc and define a volume element as <math>dV = sr \, dr\, d\theta</math>, then <math display="block"> I_{C, \text{disc}} = \iiint_Q \rho \, r^2\, dV = \int_0^{2\pi} \int_0^R \rho r^2 s r\, dr\, d\theta = 2\pi \rho s \frac{R^4}{4} = \frac{1}{2}mR^2, </math> where <math>m = \pi R^2 \rho s</math> is its mass. * The moment of inertia of the compound pendulum is now obtained by adding the moment of inertia of the rod and the disc around the pivot point <math>P</math> as, <math display="block"> I_P = I_{C, \text{rod}} + M_\text{rod}\left(\frac{L}{2}\right)^2 + I_{C, \text{disc}} + M_\text{disc}(L + R)^2,</math> where <math>L</math> is the length of the pendulum. Notice that the parallel axis theorem is used to shift the moment of inertia from the center of mass to the pivot point of the pendulum. A [[list of moments of inertia]] formulas for standard body shapes provides a way to obtain the moment of inertia of a complex body as an assembly of simpler shaped bodies. The [[parallel axis theorem]] is used to shift the reference point of the individual bodies to the reference point of the assembly. [[File:Moment of inertia solid sphere.svg|right|thumb]] As one more example, consider the moment of inertia of a solid sphere of constant density about an axis through its center of mass. This is determined by summing the moments of inertia of the thin discs that can form the sphere whose centers are along the axis chosen for consideration. If the surface of the sphere is defined by the equation<ref name="Beer"/>{{rp|p=1301}} <math display="block"> x^2 + y^2 + z^2 = R^2,</math> then the square of the radius <math>r</math> of the disc at the cross-section <math>z</math> along the <math>z</math>-axis is <math display="block">r(z)^2 = x^2 + y^2 = R^2 - z^2.</math> Therefore, the moment of inertia of the sphere is the sum of the moments of inertia of the discs along the <math>z</math>-axis, <math display="block">\begin{align} I_{C, \text{sphere}} &= \int_{-R}^R \tfrac{1}{2} \pi \rho r(z)^4\, dz = \int_{-R}^R \tfrac{1}{2} \pi \rho \left(R^2 - z^2\right)^2\,dz \\[1ex] &= \tfrac{1}{2} \pi \rho \left[R^4z - \tfrac{2}{3} R^2 z^3 + \tfrac{1}{5} z^5\right]_{-R}^R \\[1ex] &= \pi \rho\left(1 - \tfrac{2}{3} + \tfrac{1}{5}\right)R^5 \\[1ex] &= \tfrac{2}{5} mR^2, \end{align}</math> where <math display="inline">m = \frac{4}{3}\pi R^3 \rho</math> is the mass of the sphere.
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