Question

If moment of inertia of octagonal plate (Regular Octagon) is $I_0$ about axis passing through it's centre and perpendicular to it's plane (a)Find moment of inertia about AB

Answer 1

I_AB = (3I₀)/4

Answer 2

Solution Explanation

1. For a uniform regular octagon, symmetry implies that the inertia about any in‐plane axis through the center is the same. Using the perpendicular axis theorem, if
      I₀ = moment about the z–axis (through the center, perpendicular to the plane),
   then the two orthogonal in–plane moments are
      I_inplane = I₀/2.

2. Let AB be one of the edges. Its line does not pass through the centre; the perpendicular distance from the centre to the side (the apothem) is denoted by a.

3. By the parallel axis theorem the moment of inertia about AB (i.e. about an axis parallel to the in–plane axis through the centre) is:    I_AB = I_inplane + M·a²               (1)

4. For a regular octagon the geometry gives a relation between the apothem and the circumradius R. In fact,
      a = R cos(π/8).

5. It turns out that one may show (using the standard formula for the moment of inertia of a regular polygon) that    M·R² = 2I₀ sin²(π/8) so that    M·a² = M·R² cos²(π/8) = 2I₀ sin²(π/8) cos²(π/8).

6. Note that
      2 sin²(π/8) cos²(π/8) = (1/2) sin²(π/4) = (1/2)(1/2) = 1/4. Thus,
      M·a² = I₀/4.

7. Substituting into (1):
      I_AB = (I₀/2) + (I₀/4) = (3I₀)/4.