A thin wire of length \(L\) and uniform linear mass density \(\rho\) is bent into a circular loop with centre at \(O\) as shown. The moment of inertia of the loop about the axis \(XX'\) is:
center
center
Show Hint
For a ring:
\[
I_{\text{diameter}}=\frac12 MR^2
\]
and for tangent axis in the plane:
\[
I_{\text{tangent}}=\frac32 MR^2
\]
using Parallel Axis Theorem.
Step 1: Understand the axis. A wire is bent into a ring. The axis $XX'$ lies in the plane of the ring and just touches its edge (a tangent line). We want the moment of inertia about this tangent line.
Step 2: Recall the diameter result. For a ring, the moment of inertia about a diameter is \[ I_{dia} = \tfrac12 MR^2 \] Step 3: Shift to the tangent using parallel axis. The tangent is parallel to a diameter but moved out by $R$. The parallel axis rule adds $MR^2$. \[ I_{tan} = \tfrac12 MR^2 + MR^2 = \tfrac32 MR^2 \] Step 4: Express mass and radius from the wire. The wire of length $L$ has mass $M = \rho L$. When bent into a ring its length is the circumference, so \[ L = 2\pi R \quad\Rightarrow\quad R = \frac{L}{2\pi} \] Step 5: Substitute everything. \[ I = \tfrac32 (\rho L)\left(\frac{L}{2\pi}\right)^2 = \tfrac32 (\rho L)\frac{L^2}{4\pi^2} \] Step 6: Simplify the result. \[ I = \frac{3\rho L^3}{8\pi^2} \] \[ \boxed{\dfrac{3\rho L^3}{8\pi^2}} \]
