# Derivation of solid sphere moment of inertia

by jordanl122
Tags: derivation, inertia, moment, solid, sphere
 P: 14 I need to derive the moment of inertia for a solid sphere, but I'm having some trouble. I did the following. I=?r^2dm given density, p= m/V pV=m so pdV=dm and differentiating V wrt r, d(4/3?r^3)dr = 4?r^2 so p4?r^2dr=dm and plugging that in I get I=?r^2(p4?r^2)dr I pull the p4? out in front I=p4??r^4dr evaluating the integral I get I=(M/(4/3?r^3))4?(r^5/5) simplifying the terms I get I=3/5mr^2 which is a universe off from what the answer should be, if anyone can show me where I went wrong I would be very appreciative. Thanks.
 Sci Advisor HW Helper P: 3,147 Be careful! The r in your definition of the moment is the distance from the axis to the element of mass dm. It is NOT the radial variable in spherical coordinates.
 P: 579 $$\rho=\frac{m}{V}=\frac{m}{\frac{4}{3}\pi r^{3}}$$ $$\frac{dm}{dr}=4\rho\pi r^{2}$$ Also, $$I=\int{r^{2}}dm=\int{r^{2}4\rho\pi r^{2}dr}=4\rho\pi\int{r^{4}}dr=\frac{4}{5}\rho\pi r^{5}$$ Now factor something familiar... $$I=\frac{4}{5}\rho\pi r^{5}=\frac{4}{3}\rho\pi r^{3}\left(\frac{3}{5}r^{2}\right)=\frac{3}{5}mr^{2}$$
 Sci Advisor HW Helper PF Gold P: 12,016 Derivation of solid sphere moment of inertia Sirus: This is wrong; read Tide's post. The moment of inertia of a solid sphere is $$I=\frac{2}{5}mr^{2}$$ To derive this, we use that that the distance of a point to the axis of rotation (going through the center of the sphere) is $$r=\hat{r}\sin\phi$$ where $$\hat{r}$$ is the distance of a point to the center, and $$\phi$$ is the angle between the point's position vector $$\vec{r}$$(measured from the origin) and the rotaion axis. R is the radius of the sphere We have then: $$I=\int_{V}r^{2}dm=\int_{0}^{\pi}\int_{0}^{2\pi}\int_{0}^{R}\rho(\hat{r} \sin\phi)^{2}\hat{r}^{2}\sin\phi{d\hat{r}}d\theta{d}\phi=\frac{2\pi\rho }{5}R^{5}\int_{0}^{\pi}\sin^{3}\phi{d}\phi$$ Using the identity: $$\sin^{3}\theta=\sin\theta(1-\cos^{2}\theta)$$ we find that: $$\int_{0}^{\pi}\sin^{3}\phi{d}\phi=\frac{4}{3}$$
 Sci Advisor P: 905 Here's another way, that uses the moment of inertia of a disc. Let z be the coordinate along the axis. Then we can divide the sphere into discs along z. Each disc has mass $dm=\pi r^2\rho dz$ and moment of inertia $(dm)r^2/2$. But the disc radius r is given by $r^2=R^2-z^2$, where R is the sphere radius. Add them all together: $$I={\pi\over 2}\rho\int_{-R}^R (R^2-z^2)^2dz$$.
 P: 579 Thank you for the correction. Some of this stuff is obviously a little over my head.
 P: 1 I've likewise been having trouble with the moment of inertia of a sphere proof. I can do it using speherical coords without a problem but am out by a factor of 1/2 when I use cartesian coordinates. I don't know why the $(dm)r^2/2$ (as in Krab's post above). Why do we need to divide it by 2?
 Sci Advisor HW Helper P: 11,896 Because the moment of inertia of a disk wrt an axis perpendicular to the disk in the center is $$I=\frac{mR^{2}}{2}$$ Daniel.
 P: 1 but one more question. let's assume that we don't know the moment of inertia of a disc. and we start directly from the sphere.. which means we start from the formula integral (r ^2 * dm) dm = dV * p = A * dz * p = pi r^2 dz p r^2 = (R^2 - z^2) so the original formula then turns into: pi p integral ((R^2 - z^2)^2 dz distribute what's inside the parenthesis pi p integral ((R^4 - 2R^2 z^2 + z^4) dz integrate: pi p [R^4 z - 2R^2 1/3 z^3 + 1/5 z^5]from R to -R throw everything together 2 pi p [R^5- 2/3 R^5 + 1/5 R^5] <- R- (-R) = 2R pi p 16/15 R^5 M = V * p = 4/3 pi R^3 p I = 4/5 R^2 it's somehow 2 times the actual inertia. and.. i don't want to start with the inertia of the disk, which tells me to divide it by 2. Can anyone tell me where I got wrong? like where in my train of thoughts is wrong. don't refer to the disc's inertia please. thank you
 HW Helper P: 507 .. we start from the formula integral (r ^2 * dm) dm = dV * p = A * dz * p = pi r^2 dz p r^2 = (R^2 - z^2)... The whole mass dm is not at distance r from the axis of rotation, it is distributed over area A. (You have considerd it)