Maria Huey

2022-01-05

You have two solid steel spheres. Sphere 2 has twice the radius of sphere 1. By what factor does the moment of inertia $I}_{2$ of sphere 2 exceed the moment of inertia $I}_{1$ of sphere 1?

Robert Pina

Beginner2022-01-06Added 42 answers

Firstly, $I\propto m{r}^{2}$

$m=\frac{4}{3}\pi {r}^{3}{\rho}_{steel}$

As sphere 2 has twice the radius, same destiny< its mass is greater by a factor of${2}^{3}=8$

The extra mass is also distributed farther from the center$({r}_{2}=2{r}_{1})$ , thus,

$I\propto m{r}^{2}\Rightarrow {I}_{2}\propto \left(8{m}_{1}\right){\left(2{r}_{1}\right)}^{2}=32{I}_{1}$

As sphere 2 has twice the radius, same destiny< its mass is greater by a factor of

The extra mass is also distributed farther from the center

Eliza Beth13

Skilled2023-05-11Added 130 answers

Don Sumner

Skilled2023-05-11Added 184 answers

madeleinejames20

Skilled2023-05-11Added 165 answers

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