Given identical rings are arranged in a hexagonal plane pattern so as to touch each neighbouring ring as shown in figure. Each ring has mass M and radius R. The moment of inertia of the system of seven rings about an axis passing through the centre of central ring and normal to plane of all rings is
Show Hint
Use the parallel axis theorem for the six outer rings, each at distance 2R from the central axis.
Step 1: Understanding the Concept:
The figure shows one ring at the centre and six identical rings around it. Each outer ring touches the central ring and its two neighbours, so the centres of the outer rings lie on a regular hexagon around the central ring. The axis passes through the centre of the central ring, perpendicular to the plane.
Step 2: Key Formula or Approach:
1. A ring about its own axis: \(I_{cm} = MR^2\).
2. Parallel axis theorem: \(I = I_{cm} + Md^2\).
Step 3: Detailed Explanation:
Because the outer ring touches the central ring, the distance between their centres is \(R + R = 2R\), so \(d = 2R\).
Central ring: \(I_0 = MR^2\).
Each outer ring:
\[ I_1 = MR^2 + M(2R)^2 = MR^2 + 4MR^2 = 5MR^2 \]
Six outer rings:
\[ 6I_1 = 30MR^2 \]
Total:
\[ I = MR^2 + 30MR^2 = 31MR^2 \]
Option (B) 19 and (C) 11 would arise from using \(d = R\) or fewer rings. Option (D) 7 is the moment of inertia of seven rings with no shift.
Final Answer:
The moment of inertia of the system is \(31MR^2\), option (A).
\[ \boxed{31MR^2 \text{ (A)}} \]
