question_answer

A thin wire of length \[\ell \] and uniform linear mass density \[\rho \] is bent into a circular loop with centre \[o\] and radius r as shown in figure. The moment of inertia of the loop about the axis xx' is:

A)  \[\frac{3\rho {{\ell }^{3}}}{8{{\pi }^{2}}}\]              

B) \[\frac{\rho {{\ell }^{3}}}{16{{\pi }^{2}}}\]

C) \[\frac{3\rho {{\ell }^{3}}}{8{{\pi }^{2}}r}\]              

D) \[\frac{\rho {{\ell }^{3}}}{8{{\pi }^{2}}r}\]

Correct Answer: A

Solution :

[a] We have to calculate MOI about an axis \[x-{{x}^{1}}\] \[{{I}_{AB}}=\frac{M{{R}^{2}}}{2}\] Apply parallel axis theorem \[{{I}_{x-x1}}={{I}_{AB}}+M{{r}^{2}}\therefore r=R\] \[=\frac{M{{R}^{2}}}{2}+M{{R}^{2}}\] \[=\frac{3}{2}+M{{R}^{2}}\ell =2\pi R\] \[=\frac{3}{2}\rho \ell {{\left[ \frac{\ell }{2\pi } \right]}^{2}}R=\frac{\ell }{2\pi },M=\rho \ell \] \[=\frac{3}{2}\rho \ell \frac{{{\ell }^{2}}}{4{{\pi }^{2}}}\] \[=\frac{3}{2}\frac{\rho {{\ell }^{2}}}{{{\pi }^{2}}}\]