Question 739: Centre of Mass and Rotational Motion

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:

Accepted Answer

Mass of the ring $M=ho L$. Let $R$ be the radius of the ring, then
$$ L=2 \pi R \quad 	ext { or } \quad R=\frac{L}{2 \pi} $$
The moment of inertia about an axis passing through the centre $O$ and perpendicular to plane of the loop is $I_z=m R^2$. Using symmetry and perpendicular axis theorem, $I_x+I_y=I_z$, we get
$$ I_x=I_y=\frac{1}{2} I_z=\frac{1}{2} m R^2, $$
where $I_x$ is the moment of inertia about an axis passing through $O$ and lying in the plane of the loop. The centre of mass $O$ of the loop lies at a perpendicular distance $d=R$ from the axis $\mathrm{XX}^{\prime}$. The parallel axis theorem gives
$$ I_{X X^{\prime}}=\frac{1}{2} M R^2+M R^2=\frac{3}{2} M R^2 $$
Substituting values of $M$ and $R$
$$ I_{X X^{\prime}}=\frac{3}{2}(ho L)\left(\frac{L^2}{4 \pi^2}ight)=\frac{3 ho L^3}{8 \pi^2} $$

Answer

$\frac{\rho L^3}{8\pi^2}$

Answer

$\frac{\rho L^3}{16\pi^2}$

Answer

$\frac{5\rho L^3}{16\pi^2}$

Answer

$\frac{3\rho L^3}{8\pi^2}$

Centre of Mass and Rotational Motion

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:

Choose the correct answer: