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

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TestHub · Accepted Answer

The moment of inertia of the central ring about the axis passing through its centre and perpendicular to its plane is:

$I_{0} = M R^{2}$

For each of the six surrounding rings, the distance from its centre to the central axis is $d = 2 R$ .

Using the parallel axis theorem, the moment of inertia of one surrounding ring about the central axis is:

$I_{1} = I_{c m} + M d^{2}$

$I_{1} = M R^{2} + M (2 R)^{2}$

$I_{1} = M R^{2} + 4 M R^{2} = 5 M R^{2}$

Since there are six identical surrounding rings, their total moment of inertia about the central axis is:

$I_{s u rro u n d in g} = 6 \times I_{1} = 6 \times 5 M R^{2} = 30 M R^{2}$

The total moment of inertia of the system of seven rings is the sum of the moment of inertia of the central ring and the surrounding rings:

$I_{t o t a l} = I_{0} + I_{s u rro u n d in g}$

$I_{t o t a l} = M R^{2} + 30 M R^{2} = 31 M R^{2}$

Answer: $31 M R^{2}$

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