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Q. A thin wire of length $l$ having a linear density $\rho$ is bent into a circular loop with $C$ as its centre as shown in the figure. The moment of inertia of the loop about the line $A B$ is
image

AP EAMCETAP EAMCET 2019

Solution:

Given, length of a thin wire $=l$
Linear density of the wire $=\rho$
Let the mass of the wire $=M$
Now, according to the question
image
Here, circular loop of the wire with centre, $C$.
$R$ is the radius of circular loop of wire.
$\therefore $ Moment of inertia axis passing through the centre and perpendicular to the plane $A B$. $I_{A B}=M R^{2}$ .......(i)
$\therefore $ Moment of inertia axis passing through the centre of circular loop about its diameter,
$I_{ CM }=\frac{1}{2} M R^{2} \ldots( ii)$
Now, the total moment of inertia for circular loop of wire. If $R$ is the distance between the axis, $I_{ CM }$ and $I_{A B}$ are the respective moments of inertia about these axies,
Then, $I=I_{ CM }+I_{A B}$
From Eqs. (i) and (ii), we get
$I=\frac{1}{2} M R^{2}+M R^{2} \Rightarrow I=\frac{3}{2} M R^{2} \ldots (iii)$
We know that,
Mass $=$ length $\times$ density
Mass of the wire, $M=\rho l $...(iv)
Thus, the length of the wire, $l=$ circumference of the circular loop of wire
$\therefore l=2 \pi R \text { or } R=\frac{l}{2 \pi}$ ....(v)
Now, from Eq. (iv) and (v) putting the value of $M$ and $R$ in Eq. (iii), we get
$I=\frac{3}{2}(\rho l) \times\left(\frac{l}{2 \pi}\right)^{2} \Rightarrow I=\frac{3 \rho l^{3}}{8 \pi^{2}}$
So, the moment of inertia of a circular loop of wire
is, $I=\frac{3 \rho l^{3}}{8 \pi^{2}}$.