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. The moment of inertia of the loop about the axis XY is

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

Answer 1

Answer: (A)

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{3 L ^{3} P }{8 \pi^{2}} $