- #1
- 2,810
- 605
We know,by symmetry,that the center of mass of a uniform sphere is at its center.So we expect the formula [itex] r_{com}=\frac{\int r \rho d\tau}{\int \rho d\tau} [/itex] to give us zero for this case.So let's see:
[itex]
r_{com}=\frac{\int_0^{R} \int_0^{\pi}\int_0^{2\pi} r^3 \sin{\theta} d\phi d\theta dr}{\int_0^{R} \int_0^{\pi}\int_0^{2\pi} r^2 \sin{\theta} d\phi d\theta dr}=\frac{\int_0^R r^3 dr \int_0^{2\pi} d\phi \int_0^{\pi} \sin{\theta} d\theta}{\int_0^R r^2 dr \int_0^{2\pi} d\phi \int_0^{\pi} \sin{\theta} d\theta}=\frac{\frac{1}{4}R^4\times 2\pi \times 2}{\frac{1}{3}R^3\times 2\pi \times 2}=\frac{3}{4}R
[/itex]!
What's going on?
[itex]
r_{com}=\frac{\int_0^{R} \int_0^{\pi}\int_0^{2\pi} r^3 \sin{\theta} d\phi d\theta dr}{\int_0^{R} \int_0^{\pi}\int_0^{2\pi} r^2 \sin{\theta} d\phi d\theta dr}=\frac{\int_0^R r^3 dr \int_0^{2\pi} d\phi \int_0^{\pi} \sin{\theta} d\theta}{\int_0^R r^2 dr \int_0^{2\pi} d\phi \int_0^{\pi} \sin{\theta} d\theta}=\frac{\frac{1}{4}R^4\times 2\pi \times 2}{\frac{1}{3}R^3\times 2\pi \times 2}=\frac{3}{4}R
[/itex]!
What's going on?