# How to calculate the moment of inertia of the rigid body?

by copperboy
Tags: body, inertia, moment, rigid
 Sci Advisor HW Helper P: 3,147 Use the definition of moment of inertia: $$I = \int r^2 dm$$ In the case of the shell the element of mass is $dm = M {dA} /{4 \pi R^2}$ where $dA = R^2 \sin \theta d\theta d\phi[/tex]. The distance to a point on the shell from the z-axis is [itex]R^2 \sin^2 \theta$ so $$I = \frac {M}{4 \pi R^2} \int_{0}^{2 \pi} d\phi \int_{-\pi /2}^{\pi /2}R^4 \sin^3 \theta d\theta$$ from which the desired result follows. In the case of the solid sphere you will work with a volume integral.
 P: 8 Could you please explaim why $dA = R^2 \sin \theta d\theta d\phi[/tex] in detail? Sci Advisor HW Helper PF Gold P: 12,016 How to calculate the moment of inertia of the rigid body?  Quote by copperboy Could you please explaim why [itex]dA = R^2 \sin \theta d\theta d\phi[/tex] in detail? Consider an area segment dA on a sphere with radius R. We approximate this with a rectangle: a)Two of the sides are arclengths along great circles; the length of each of these is $$Rd\theta$$ b) The other two are arclengths in THE HORIZONTAL PLANE; the local radius there is $$R\sin\theta$$ Hence, the arclenth is $$R\sin\theta{d\phi}$$ c) Multiplying together, we get: $$dA=R^{2}\sin\theta{d\theta}d\phi$$ Sci Advisor HW Helper P: 3,147  Quote by copperboy Could you please explaim why [itex]dA = R^2 \sin \theta d\theta d\phi[/tex] in detail? dA is a differential element of area on a spherical surface using spherical coordinates and it represents, to lowest order in differentials, the area of a rectangle [itex]R \sin \theta d\phi$ high and $R d\theta$ units wide.