Finding Moments of Inertia

1. Find the polar moment of inertia \(I_{O}\) of a homogeneous lamina of mass density \(\rho \) in the shape of a region \(R\) in the \(xy\)-plane enclosed by the circle \(x^{2}+y^{2}=a^{2}\), \(a \gt 0\).
2. Use the polar moment to find the moments of inertia \(I_{x}\) and \(I_{y}\) of the lamina.

Solution (a) The region \(R\) is circular so we use polar coordinates \((r, \theta).\) Then \(0\leq r\leq a,\) \(0\leq \theta \leq 2\pi \), and \({\it dA}=r\,dr\, d\theta \). The polar moment of inertia is \[ \begin{eqnarray*} I_{O}&=& \displaystyle\iint\limits_{\kern-3ptR}(x^{2}+y^{2}) \rho (x,y) \,{\it dA} \underset{\underset{\color{#0066A7}{\hbox{\(x^{2}+y^{2}=r^{2}\)}}}{\color{#0066A7}{\uparrow }}}{=}\iint\limits_{\kern-3ptR}r^{2}\rho \,r\,{\it dr}\,d\theta \underset{\underset{\color{#0066A7}{\hbox{\(\rho\) is constant}}}{\color{#0066A7}{\uparrow }}}{=}\rho \int_{0}^{2\pi }\!\!\!\int_{0}^{a}r^{3}\,dr\,d\theta\\ &=&\rho \int_{0}^{2\pi }\dfrac{a^{4}}{4} \,d\theta =\dfrac{\pi a^{4}\rho }{2} \end{eqnarray*} \]

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(b) By symmetry, the moment of inertia about the \(x\)-axis, \(I_{x}\), equals the moment of inertia about the \(y\)-axis, \(I_{y}\). Then \[ I_{x}=I_{y}=\dfrac{1}{2}I_{O}=\dfrac{1}{2}\!\left( \dfrac{\pi a^{4}\rho }{2} \right) =\dfrac{\pi a^{4}\rho }{4} \]