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to derive the moment of inertia of a solid sphere of uniform density and radius R. [tex] dI = r_{\perp}^{2} dm = r_{\perp}^{2} \rho dV [/tex] â. where [itex] r_{\perp} [/itex] is the perpendicular distance to a point at r from the axis of rotation. Therefore, [itex] r_{\perp} = |\mathbf{r}|\sin{\theta} [/itex].
up vote 3 down vote favorite. Q: Find out the moment of inertia of a uniform circular disc of radius $r$ & mass $M$ & the axis passes through a point on the circumference. My attempt: Let the axis passes through $O$ on the circumference. Then I took a ring at $x$ unit from the axis and width $dx$. Now, area of the disc is$$dA =2\pi(r-x).dx$$ . Since, the disc is uniform, density $$ D = \dfrac{M}{\pi .r^2} .$$ Therefore, mass of the ring $$ dM = \dfrac{2M(r-x)dx}{r^2}$$ . Now, moment of inertia w.r.t. the axis $$dI = \dfrac{2M(r-x)x^2 .dx}{r^2}$$ .