Certainly! Let's find the moment of inertia for each object given the provided conditions: a cylinder rotating about an axis through its center and through both of its circular faces, and a sphere rotating about an axis through its center.
### Cylinder
The moment of inertia [tex]\(I\)[/tex] for a rotating solid cylinder about its center through both of its faces is given by the formula:
[tex]\[ I_{\text{cylinder}} = \frac{1}{2} m r^2 \][/tex]
where:
- [tex]\(m\)[/tex] is the mass of the cylinder
- [tex]\(r\)[/tex] is the radius of the cylinder
Given:
- Mass, [tex]\(m = 5 \, \text{kg}\)[/tex]
- Radius, [tex]\(r = 0.2 \, \text{m}\)[/tex]
Substituting the values into the formula:
[tex]\[ I_{\text{cylinder}} = \frac{1}{2} \times 5 \, \text{kg} \times (0.2 \, \text{m})^2 \][/tex]
We can evaluate:
[tex]\[ I_{\text{cylinder}} = \frac{1}{2} \times 5 \times 0.04 \][/tex]
[tex]\[ I_{\text{cylinder}} = \frac{1}{2} \times 0.2 \][/tex]
[tex]\[ I_{\text{cylinder}} = 0.1 \, \text{kg}\cdot\text{m}^2 \][/tex]
### Sphere
The moment of inertia [tex]\(I\)[/tex] for a rotating solid sphere about its center is given by the formula:
[tex]\[ I_{\text{sphere}} = \frac{2}{5} m r^2 \][/tex]
where:
- [tex]\(m\)[/tex] is the mass of the sphere
- [tex]\(r\)[/tex] is the radius of the sphere
Given:
- Mass, [tex]\(m = 5 \, \text{kg}\)[/tex]
- Radius, [tex]\(r = 0.2 \, \text{m}\)[/tex]
Substituting the values into the formula:
[tex]\[ I_{\text{sphere}} = \frac{2}{5} \times 5 \, \text{kg} \times (0.2 \, \text{m})^2 \][/tex]
We can evaluate:
[tex]\[ I_{\text{sphere}} = \frac{2}{5} \times 5 \times 0.04 \][/tex]
[tex]\[ I_{\text{sphere}} = \frac{2}{5} \times 0.2 \][/tex]
[tex]\[ I_{\text{sphere}} = 0.08 \, \text{kg}\cdot\text{m}^2 \][/tex]
### Summary
The moments of inertia for the given objects are:
- Cylinder: [tex]\(0.1 \, \text{kg}\cdot\text{m}^2\)[/tex]
- Sphere: [tex]\(0.08 \, \text{kg}\cdot\text{m}^2\)[/tex]
