Three rings each of mass $M$ and radius $R$ are arranged as shown in the figure. The moment of inertia of the system about the axis $YY'$ will be
Choose the correct answer from the options given below
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To avoid re-deriving the tangent inertia each time, memorize these two common orientations for a thin ring:
Tangent perpendicular to the ring's plane $= 2MR^2$. Tangent lying within the ring's plane $= \frac{3}{2}MR^2$.
Recognizing these standard orientations immediately simplifies multi-body system calculations.
Step 1: The arrangement.
Three equal rings, each of mass $M$ and radius $R$, are placed in one plane. One ring is on top and two rings sit below. The axis $YY'$ runs vertically through the middle. We add up the moment of inertia of all three about this axis.
Step 2: Inertia adds up.
Moment of inertia is a plain number for each body, so the total is just the sum:
\[ I_{\text{total}} = I_1 + I_2 + I_3 \]
Step 3: The top ring.
The axis $YY'$ passes right through the centre of the top ring, along a diameter. For a ring about its diameter,
\[ I_1 = \frac{1}{2}MR^2 \]
Step 4: The two lower rings.
For each lower ring the axis just touches its edge, that is, it is tangent in the plane of the ring. Using the parallel axis theorem with the diameter value and a shift $d = R$:
\[ I_2 = I_3 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2 \]
Step 5: Add them all.
\[ I_{\text{total}} = \frac{1}{2}MR^2 + \frac{3}{2}MR^2 + \frac{3}{2}MR^2 \]
Step 6: Combine the fractions.
\[ I_{\text{total}} = \frac{1 + 3 + 3}{2}MR^2 = \frac{7}{2}MR^2 \]
This is option (2).
\[ \boxed{I_{\text{total}} = \frac{7}{2}MR^2} \]
