Question

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

• A. $5\ MR^2$
• B. $\frac{7}{2}\ MR^2$
• C. $\frac{3}{2}\ MR^2$
• D. $3\ MR^2$

Hint:

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.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem presents a symmetric planar layout of three identical thin rings, each with mass $M$ and radius $R$. The top ring sits symmetrically balanced above two adjacent base rings. We need to find the total combined moment of inertia ($I_{\text{total}}$) of this three-body configuration about the central vertical axis line denoted as $YY'$.

Step 2: Key Formula or Approach:
Since moment of inertia is a scalar quantity, the total inertia is the sum of the individual moments of inertia of each of the three rings computed relative to that same central axis line: $$I_{\text{total}} = I_1 + I_2 + I_3$$ We will use standard rotational inertia values for a thin ring along with the

Parallel Axis Theorem: $$I = I_{\text{cm}} + Md^2$$

Step 3: Detailed Explanation:
Let's analyze the position of each ring relative to the vertical axis $YY'$: 1.

For the upper ring (Ring 1): The axis $YY'$ passes directly through the center of mass of the top ring, dividing it into two symmetrical halves within its plane. Therefore, the axis aligns with the ring's diameter. The moment of inertia of a ring about its diameter is: $$I_1 = \frac{1}{2}MR^2$$ 2.

For the two lower rings (Ring 2 and Ring 3): The vertical axis $YY'$ runs tangent to the outer edges of both base rings within their layout plane. The moment of inertia of a thin ring about a tangential axis lying in its own plane is calculated using the Parallel Axis Theorem, where $I_{\text{diameter}} = \frac{1}{2}MR^2$ and the shift distance to the edge is $d = R$: $$I_2 = I_3 = I_{\text{diameter}} + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$$ Now, sum the individual moments of inertia to find the total value for the entire system: $$I_{\text{total}} = I_1 + I_2 + I_3$$ $$I_{\text{total}} = \frac{1}{2}MR^2 + \frac{3}{2}MR^2 + \frac{3}{2}MR^2$$ Combine the fractions over a common denominator: $$I_{\text{total}} = \left(\frac{1 + 3 + 3}{2}\right)MR^2 = \frac{7}{2}MR^2$$

Step 4: Final Answer:
The total moment of inertia of the system about the axis $YY'$ is $\frac{7}{2}\ MR^2$, which corresponds to option (B).