Math Problem Statement

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.

Solution

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

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 MM and radius RR about an axis perpendicular to the plane and through its center is: Icenter=12MR2I_{\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 PP, we apply the parallel axis theorem. This theorem states that if the axis of rotation is shifted by a distance dd from the center of mass, the new moment of inertia is: I=Icenter+Md2I = I_{\text{center}} + M d^2 where dd is the distance between the center of the disk and the axis passing through point PP.

  3. Symmetrical Arrangement:

    • The central disk (labeled at point OO) has its center at the origin, where the axis of rotation passes through. So, for the central disk: IO=12MR2I_O = \frac{1}{2} M R^2
    • For the surrounding six disks, their centers are at a distance of 2R2R from point PP (since they are arranged symmetrically and the distance from point OO to PP is 2R2R).

    Therefore, the moment of inertia for each surrounding disk is: Isurrounding=12MR2+M(2R)2=12MR2+4MR2=92MR2I_{\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 92MR2\frac{9}{2} M R^2, and 1 central disk contributing 12MR2\frac{1}{2} M R^2. The total moment of inertia is: Itotal=IO+6×Isurrounding=12MR2+6×92MR2I_{\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: Itotal=12MR2+27MR2=12MR2+542MR2=552MR2I_{\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 PP is:

Itotal=552MR2I_{\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.

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Math Problem Analysis

Mathematical Concepts

Moment of Inertia
Parallel Axis Theorem
Symmetry

Formulas

Moment of inertia of a disk about its center: I_center = (1/2) M R^2
Parallel Axis Theorem: I = I_center + M d^2

Theorems

Parallel Axis Theorem

Suitable Grade Level

Grades 11-12 (Advanced Physics or Engineering Level)