Question

A thin wire of length L and uniform linear mass density p is bent into a circular loop with centre at O as shown in figure. The moment of inertia of the loop about the axis XY is

• A. $\frac{3p{L}^3}{8{\pi }^2}$
• B. $\frac{5p{L}^3}{16{\pi }^2}$
• C. $\frac{p{L}^3}{8{\pi }^2}$
• D. $\frac{p{L}^3}{16{\pi }^2}$

Answer 1

The correct option is A $\frac{3p{L}^3}{8{\pi }^2}$

The moment of inertia of a thin hoop about its diameter is $\frac{1}{2}{MR}^2$

here, $M=LP$ also, we have

$2\pi R=L$

$\Rightarrow R=\frac{L}{2\pi }$

So, we have

$I=\frac{1}{2}{MR}^2=\frac{1}{2}LP{\left(\frac{L}{2\pi }\right)}^2=\frac{L^{3}P}{{8\pi }^2}$

Now using parallel axis theorem we have

$L_{{xx}^1}=I_{cm}+{MR}^2=\frac{L^{3}P}{{8\pi }^2}+LP{\left(\frac{L}{2\pi }\right)}^2$

$=\frac{L^{3}P}{{8\pi }^2}+\frac{L^{3}P}{{4\pi }^2}=\frac{{3L}^{3}P}{{8\pi }^2}$