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

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

Eliza Beth13

To solve the problem, we need to find the factor by which the moment of inertia of sphere 2 exceeds the moment of inertia of sphere 1. Let's denote the radius of sphere 1 as ${r}_{1}$ and the radius of sphere 2 as ${r}_{2}$.
Given that sphere 2 has twice the radius of sphere 1, we have ${r}_{2}=2{r}_{1}$.
The moment of inertia of a solid sphere is given by the formula:
$I=\frac{2}{5}m{r}^{2}$,
where $m$ is the mass of the sphere and $r$ is its radius.
Since both spheres are made of solid steel, the masses of the spheres will be proportional to their volumes. The volume of a sphere is given by:
$V=\frac{4}{3}\pi {r}^{3}$.
So, the mass $m$ of a sphere will be proportional to ${r}^{3}$.
Let's assume the mass of sphere 1 is ${m}_{1}$, and the mass of sphere 2 is ${m}_{2}$. We can write the following equation:
$\frac{{m}_{2}}{{m}_{1}}={\left(\frac{{r}_{2}}{{r}_{1}}\right)}^{3}$.
Substituting ${r}_{2}=2{r}_{1}$ into the equation, we have:
$\frac{{m}_{2}}{{m}_{1}}={\left(\frac{2{r}_{1}}{{r}_{1}}\right)}^{3}={2}^{3}=8$.
So, the mass of sphere 2 is eight times the mass of sphere 1.
Now, let's find the ratio of the moments of inertia:
$\frac{{I}_{2}}{{I}_{1}}=\frac{\frac{2}{5}{m}_{2}{r}_{2}^{2}}{\frac{2}{5}{m}_{1}{r}_{1}^{2}}=\frac{{m}_{2}{r}_{2}^{2}}{{m}_{1}{r}_{1}^{2}}$.
Substituting ${m}_{2}=8{m}_{1}$ and ${r}_{2}=2{r}_{1}$ into the equation, we have:
$\frac{{I}_{2}}{{I}_{1}}=\frac{\left(8{m}_{1}\right)\left(2{r}_{1}{\right)}^{2}}{{m}_{1}{r}_{1}^{2}}=\frac{32{m}_{1}{r}_{1}^{2}}{{m}_{1}{r}_{1}^{2}}=32$.
Therefore, the moment of inertia ${I}_{2}$ of sphere 2 exceeds the moment of inertia ${I}_{1}$ of sphere 1 by a factor of 32.

Don Sumner

We are given two solid steel spheres, where Sphere 2 has twice the radius of Sphere 1. Let's denote the radius of Sphere 1 as ${r}_{1}$ and the radius of Sphere 2 as ${r}_{2}$.
The moment of inertia of a solid sphere is given by the formula:
$I=\frac{2}{5}m{r}^{2}$,
where $m$ is the mass of the sphere and $r$ is its radius.
Since both spheres are made of solid steel, their densities are the same. Hence, their masses will be proportional to their volumes. The volume of a sphere is given by:
$V=\frac{4}{3}\pi {r}^{3}$.
Thus, the mass $m$ of a sphere will be proportional to ${r}^{3}$.
Let's assume the mass of Sphere 1 is ${m}_{1}$, and the mass of Sphere 2 is ${m}_{2}$. We can write the following equation:
$\frac{{m}_{2}}{{m}_{1}}={\left(\frac{{r}_{2}}{{r}_{1}}\right)}^{3}$.
Since Sphere 2 has twice the radius of Sphere 1, we have ${r}_{2}=2{r}_{1}$. Substituting this into the equation, we get:
$\frac{{m}_{2}}{{m}_{1}}={\left(\frac{2{r}_{1}}{{r}_{1}}\right)}^{3}={2}^{3}=8$.
Thus, the mass of Sphere 2 is eight times the mass of Sphere 1.
Now, let's find the ratio of the moments of inertia:
$\frac{{I}_{2}}{{I}_{1}}=\frac{\frac{2}{5}{m}_{2}{r}_{2}^{2}}{\frac{2}{5}{m}_{1}{r}_{1}^{2}}=\frac{{m}_{2}{r}_{2}^{2}}{{m}_{1}{r}_{1}^{2}}$.
Substituting ${m}_{2}=8{m}_{1}$ and ${r}_{2}=2{r}_{1}$ into the equation, we have:
$\frac{{I}_{2}}{{I}_{1}}=\frac{\left(8{m}_{1}\right)\left(2{r}_{1}{\right)}^{2}}{{m}_{1}{r}_{1}^{2}}=\frac{32{m}_{1}{r}_{1}^{2}}{{m}_{1}{r}_{1}^{2}}=32$.
Therefore, the moment of inertia ${I}_{2}$ of Sphere 2 exceeds the moment of inertia ${I}_{1}$ of Sphere 1 by a factor of 32.

Explanation:
$I=\frac{2}{5}m{r}^{2}$
where $I$ is the moment of inertia, $m$ is the mass of the sphere, and $r$ is the radius of the sphere.
Let's assume the mass of both spheres is the same. Therefore, we can compare the moments of inertia by using the ratio:
$\frac{{I}_{2}}{{I}_{1}}=\frac{\frac{2}{5}m\left(2{r}_{1}{\right)}^{2}}{\frac{2}{5}m{r}_{1}^{2}}$
Simplifying the equation, we have:
$\frac{{I}_{2}}{{I}_{1}}=\frac{\frac{2}{5}\left(2{r}_{1}{\right)}^{2}}{\frac{2}{5}{r}_{1}^{2}}$
$\frac{{I}_{2}}{{I}_{1}}=\frac{\frac{2}{5}4{r}_{1}^{2}}{\frac{2}{5}{r}_{1}^{2}}$
$\frac{{I}_{2}}{{I}_{1}}=\frac{4{r}_{1}^{2}}{{r}_{1}^{2}}$
$\frac{{I}_{2}}{{I}_{1}}=4$
Therefore, the moment of inertia ${I}_{2}$ of sphere 2 exceeds the moment of inertia ${I}_{1}$ of sphere 1 by a factor of 4.

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