Question
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
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
A.
$$\frac{{\rho {L^3}}}{{8{\pi ^2}}}$$
B.
$$\frac{{\rho {L^3}}}{{16{\pi ^2}}}$$
C.
$$\frac{{5\rho {L^3}}}{{16{\pi ^2}}}$$
D.
$$\frac{{3\rho {L^3}}}{{8{\pi ^2}}}$$
Answer :
$$\frac{{3\rho {L^3}}}{{8{\pi ^2}}}$$
Solution :
Moment of inertia about the diameter of the circular loop (ring) $$ = \frac{1}{2}M{R^2}$$
Using parallel axis theorem
The moment of inertia of the loop about $$XX’$$ axis is
$${I_{XX'}} = \frac{{M{R^2}}}{2} + M{R^2} = \frac{3}{2}M{R^2}$$
Where $$M\,=$$ mass of the loop and $$R \,=$$ radius of the loop
Here $$M = L\rho $$ and $$R = \frac{L}{{2\pi }};$$
$$\therefore {I_{XX'}} = \frac{3}{2}\left( {L\rho } \right){\left( {\frac{L}{{2\pi }}} \right)^2} = \frac{{3{L^3}\rho }}{{8{\pi ^2}}}$$
Moment of inertia about the diameter of the circular loop (ring) $$ = \frac{1}{2}M{R^2}$$
Using parallel axis theorem
The moment of inertia of the loop about $$XX’$$ axis is
$${I_{XX'}} = \frac{{M{R^2}}}{2} + M{R^2} = \frac{3}{2}M{R^2}$$
Where $$M\,=$$ mass of the loop and $$R \,=$$ radius of the loop
Here $$M = L\rho $$ and $$R = \frac{L}{{2\pi }};$$
$$\therefore {I_{XX'}} = \frac{3}{2}\left( {L\rho } \right){\left( {\frac{L}{{2\pi }}} \right)^2} = \frac{{3{L^3}\rho }}{{8{\pi ^2}}}$$
