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' (tangent in the plane) will be :
$\frac{3mL^3}{8\pi^2}$
To find the moment of inertia of the ring about an axis \(yy'\) (tangent in the plane), we can use the parallel axis theorem and the known formula for the moment of inertia of a circular ring about its center and diameter.
- First, calculate the radius \(R\) of the ring. The circumference of the ring is equal to the length of the wire \(L\), so we have: \(L = 2\pi R \Rightarrow R = \frac{L}{2\pi}\)
- The moment of inertia of the ring about its center \(C\) (z-axis) is: \(I_C = mR^2 = m\left(\frac{L}{2\pi}\right)^2 = \frac{mL^2}{4\pi^2}\)
- The moment of inertia about any diameter (say x-axis) of the ring is half of \(I_C\): \(I_x = \frac{I_C}{2} = \frac{mL^2}{8\pi^2}\)
- Using the parallel axis theorem, the moment of inertia about axis \(yy'\) is: \(I_{yy'} = I_x + mR^2 = \frac{mL^2}{8\pi^2} + \frac{mL^2}{4\pi^2}\)
- Simplify the expression: \(I_{yy'} = \frac{mL^2}{8\pi^2} + \frac{2mL^2}{8\pi^2} = \frac{3mL^2}{8\pi^2}\)
- Finally, factor in the mass density \(m\) and length \(L\): \(I_{yy'} = \frac{3mL^3}{8\pi^2}\)
Thus, the moment of inertia of the ring about the tangent axis \(yy'\) is \(\frac{3mL^3}{8\pi^2}\).
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