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The problem involves calculating the moment of inertia of seven identical circular planar disks, each with mass \( M \) and radius \( R \), welded symmetrically as shown. We are tasked with finding the moment of inertia about an axis normal to the plane and passing through point \( P \).

### Step-by-Step Solution:

1. **Moment of Inertia for a Single Disk about its Center:**
   The moment of inertia of a single disk of mass \( M \) and radius \( R \) about an axis perpendicular to the plane and through its center is:
   \[
   I_{\text{center}} = \frac{1}{2} M R^2
   \]

2. **Using the Parallel Axis Theorem:**
   For disks that are not centered at the axis passing through point \( P \), we apply the parallel axis theorem. This theorem states that if the axis of rotation is shifted by a distance \( d \) from the center of mass, the new moment of inertia is:
   \[
   I = I_{\text{center}} + M d^2
   \]
   where \( d \) is the distance between the center of the disk and the axis passing through point \( P \).

3. **Symmetrical Arrangement:**
   - The central disk (labeled at point \( O \)) has its center at the origin, where the axis of rotation passes through. So, for the central disk:
     \[
     I_O = \frac{1}{2} M R^2
     \]
   - For the surrounding six disks, their centers are at a distance of \( 2R \) from point \( P \) (since they are arranged symmetrically and the distance from point \( O \) to \( P \) is \( 2R \)).

   Therefore, the moment of inertia for each surrounding disk is:
   \[
   I_{\text{surrounding}} = \frac{1}{2} M R^2 + M (2R)^2 = \frac{1}{2} M R^2 + 4 M R^2 = \frac{9}{2} M R^2
   \]

4. **Total Moment of Inertia:**
   There are 6 surrounding disks, each contributing \( \frac{9}{2} M R^2 \), and 1 central disk contributing \( \frac{1}{2} M R^2 \). The total moment of inertia is:
   \[
   I_{\text{total}} = I_O + 6 \times I_{\text{surrounding}} = \frac{1}{2} M R^2 + 6 \times \frac{9}{2} M R^2
   \]
   Simplifying this expression:
   \[
   I_{\text{total}} = \frac{1}{2} M R^2 + 27 M R^2 = \frac{1}{2} M R^2 + \frac{54}{2} M R^2 = \frac{55}{2} M R^2
   \]

Thus, the moment of inertia of the entire arrangement about the axis through point \( P \) is:

\[
I_{\text{total}} = \frac{55}{2} M R^2
\]

Would you like any further clarifications or additional details? Here are five related questions to expand on this topic:

1. How does the parallel axis theorem apply to other shapes, such as cylinders or spheres?
2. What would change in the calculation if the axis passed through a different point?
3. How do the moments of inertia of composite objects compare to simpler objects?
4. Can you derive the moment of inertia for a hollow disk using a similar approach?
5. How is the moment of inertia affected if the radius of the disks changes while the mass remains constant?

### Tip:
Always apply the parallel axis theorem when the axis of rotation is shifted from the center of mass, as it accounts for both the rotational inertia about the center and the effect of translating the axis.

Seven identical circular planar disks, each of mass M and radius R are welded symmetrically. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is to be determined.

Calculate the moment of inertia of seven identical circular disks symmetrically arranged, using the Parallel Axis Theorem. The problem involves determining the total moment of inertia about an axis normal to the plane and passing through a specified point. Suitable for advanced physics students in Grades 11-12.

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