Answer: C) \( I_{\text{center}} = 0.5 \, \text{kg} \cdot \text{m}^2, I_{\text{edge}} = 1.0 \, \text{kg} \cdot \text{m}^2 \)
Step 1: Moment of Inertia about an axis through the center and perpendicular to its plane
The moment of inertia of a ring about an axis passing through its center and perpendicular to its plane is given by:
\[
I_{\text{center}} = M R^2
\]
Here, \( M = 2 \, \text{kg} \) and \( R = 50 \, \text{cm} = 0.5 \, \text{m} \):
\[
I_{\text{center}} = 2 \times (0.5)^2 = 2 \times 0.25 = 0.5 \, \text{kg} \cdot \text{m}^2
\]
Step 2: Moment of Inertia about a parallel axis through the edge
Using the parallel axis theorem, the moment of inertia about a parallel axis through the edge is:
\[
I_{\text{edge}} = I_{\text{center}} + M d^2
\]
Here, \( d = R = 0.5 \, \text{m} \):
\[
I_{\text{edge}} = 0.5 + 2 \times (0.5)^2
\]
Calculate \( 2 \times (0.5)^2 \):
\[
2 \times (0.5)^2 = 2 \times 0.25 = 0.5
\]
Now, add the values:
\[
I_{\text{edge}} = 0.5 + 0.5 = 1.0 \, \text{kg} \cdot \text{m}^2
\]
Final Answers:
- Moment of Inertia about the center: \( I_{\text{center}} = 0.5 \, \text{kg} \cdot \text{m}^2 \)
- Moment of Inertia about the edge: \( I_{\text{edge}} = 1.0 \, \text{kg} \cdot \text{m}^2 \)
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