 You have two solid steel spheres. Sphere 2 has twice Maria Huey 2022-01-05 Answered
You have two solid steel spheres. Sphere 2 has twice the radius of sphere 1. By what factor does the moment of inertia $$\displaystyle{I}_{{2}}$$ of sphere 2 exceed the moment of inertia $$\displaystyle{I}_{{1}}$$ of sphere 1?

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Firstly, $$\displaystyle{I}\propto{m}{r}^{{{2}}}$$
$$\displaystyle{m}={\frac{{{4}}}{{{3}}}}\pi{r}^{{{3}}}\rho_{{{s}{t}{e}{e}{l}}}$$
As sphere 2 has twice the radius, same destiny< its mass is greater by a factor of $$\displaystyle{2}^{{{3}}}={8}$$
The extra mass is also distributed farther from the center $$\displaystyle{\left({r}_{{2}}={2}{r}_{{1}}\right)}$$, thus,
$$\displaystyle{I}\propto{m}{r}^{{{2}}}\Rightarrow{I}_{{2}}\propto{\left({8}{m}_{{1}}\right)}{\left({2}{r}_{{1}}\right)}^{{{2}}}={32}{I}_{{1}}$$