- #1

cepheid

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Hi,

So as not to lead anybody astray, I have decided to post a correction to my ill-fated attempt in this thread:

https://www.physicsforums.com/showthread.php?t=158832

to derive the moment of inertia of a solid sphere of uniform density and radius R.

So as not to lead anybody astray, I have decided to post a correction to my ill-fated attempt in this thread:

https://www.physicsforums.com/showthread.php?t=158832

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]

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]. Integrating over the volume,[tex] I = \int \! \! \! \int \! \! \! \int_V \rho r_{\perp}^2 \,dV = \rho \int_0^{2\pi} \int_0^{\pi} \int_0^R (r \sin{\theta})^2 r^2 \sin{\theta} \,dr\,d \theta\, d\phi[/tex]

[tex] = 2\pi \rho \int_0^{\pi} \sin^3{\theta}\, d\theta\, \int_0^R r^4\,dr [/tex]

[tex] = 2\pi \rho \int_0^{\pi} \sin^3{\theta}\, d\theta\, \int_0^R r^4\,dr [/tex]

Make the substitution u = cos[itex]\theta[/itex].

[tex]= 2\pi \rho \frac{R^5}{5} \int_{-1}^{1} (1 - u^2)\,du = 2\pi \rho \frac{R^5}{5} \left(1 - (-1) - \left[\frac{u^3}{3}\right]_{-1}^1 \right) [/tex]

[tex] = 2\pi\rho \frac{R^5}{5} \left(2 - \frac{2}{3}\right) = \underbrace{\frac{4}{3} \pi R^3 \rho}_{M} \left(\frac{2}{5}R^2\right) [/tex]

[tex] = \frac{2}{5} MR^2 [/tex]

[tex] = 2\pi\rho \frac{R^5}{5} \left(2 - \frac{2}{3}\right) = \underbrace{\frac{4}{3} \pi R^3 \rho}_{M} \left(\frac{2}{5}R^2\right) [/tex]

[tex] = \frac{2}{5} MR^2 [/tex]

where M is the total mass of the sphere.

Please let me know if there is anything wrong with

Thanks,

Cepheid

Please let me know if there is anything wrong with

*derivation.***this**Thanks,

Cepheid

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