A thin wire of length ‘L’ and linear mass density ‘m’ is bent into a circular ring (in x-y plane) with centre ‘C’ as shown in figure. The moment of inertia of the ring about an axis yy′ will be: ____.
Show Hint
For a ring, $I$ about an axis through center (perpendicular to plane) is $MR^2$. About a diameter, it's half of that ($\frac{1}{2}MR^2$). About a tangent in the plane, it's $\frac{3}{2}MR^2$.
Step 1: Understanding the Topic:
This problem falls under "Rotational Mechanics," specifically the calculation of the moment of inertia for continuous bodies. The moment of inertia depends not only on the mass but also on how that mass is distributed relative to the axis of rotation. For this problem, we must correctly identify the geometry and use the Parallel Axis Theorem. Step 2: Key Formulas and Approach:
Total mass $M = \text{Linear density } (m) \times \text{Length } (L)$.
Circumference $L = 2\pi R \implies R = L / (2\pi)$.
$I$ of a ring about a diameter: $I_d = \frac{1}{2} M R^2$.
Parallel Axis Theorem: $I_{axis} = I_{CM} + M \cdot d^2$.
Step 3: Detailed Explanation:
Find Mass and Radius: The total mass is $M = mL$. The radius of the ring formed from wire of length $L$ is $R = L / 2\pi$.
Identify the axis: The axis $yy'$ is a tangent to the ring lying in its own plane.
Apply Theorems: The moment of inertia about a diameter (which passes through the center) is $I_d = \frac{1}{2} MR^2$.
According to the parallel axis theorem, the MI about a tangent parallel to the diameter is:
\[ I_{yy'} = I_d + M \cdot R^2 = \frac{1}{2} MR^2 + MR^2 = \frac{3}{2} MR^2 \]
Substitute terms in terms of $L$ and $m$:
\[ I_{yy'} = \frac{3}{2} (mL) \cdot \left( \frac{L}{2\pi} \right)^2 \]
\[ I_{yy'} = \frac{3}{2} \cdot mL \cdot \frac{L^2}{4\pi^2} = \frac{3mL^3}{8\pi^2} \]
Step 4: Final Answer:
The moment of inertia about axis yy′ is 3mL³/8π².
