First slide

Moment of intertia

Question

A thin wire of length L and uniform linear mass density p is bent into a circular loop with centre O as shown in fig. The moment of inertia of the loop about the axis XX' is

Moderate

A

$\frac{{ρL}^{3}}{8 π^{2}}$
B

$\frac{{ρL}^{3}}{16 π^{2}}$
C

$\frac{5 {ρL}^{3}}{16 π^{2}}$
D

$\frac{3 {ρL}^{3}}{8 π^{2}}$

Solution

The moment of inertia of the loop about the diameter = M R²/ 2. So, the moment of inertia of the loop about XX' is given by
$I_{XX} = \frac{{MR}^{2}}{2} + {MR}^{2} = \frac{3}{2} {MR}^{2}$
where M = mass of the loop and .R = radius of the loop Now
$M = Lρ and R = L / 2 π$
$∴ I_{{XX}^{'}} = \frac{3}{2} (Lρ) (L / 2 π)^{2} = 3 L^{3} ρ / 8 π^{2}$

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