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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Expert Answer

Robert Pina
Answered 2022-01-06 Author has 1730 answers
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,
\(\displaystyle{I}\propto{m}{r}^{{{2}}}\Rightarrow{I}_{{2}}\propto{\left({8}{m}_{{1}}\right)}{\left({2}{r}_{{1}}\right)}^{{{2}}}={32}{I}_{{1}}\)
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