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

Question

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(4)

$About the diameter of the circular loop (ring)$

$I = \frac{1}{2} M R^{2}$

$Using parallel axis theorem$

$Moment of inertia of the loop about X X' axis$

$I_{x x'} = \frac{M R^{2}}{2} + M R^{2} = \frac{3}{2} M R^{2}$

$Here mass M = L ρ and radius R = \frac{L}{2 π};$

$∴$ $I_{x x'} = \frac{3}{2} (L ρ) {(\frac{L}{2 π})}^{2} = \frac{3 L^{3} ρ}{8 π^{2}}$

Solution

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

1 $\frac{ρ L^{3}}{8 π^{2}}$

2 $\frac{ρ L^{3}}{16 π^{2}}$

3 $\frac{5 ρ L^{3}}{16 π^{2}}$

4 $\frac{3 ρ L^{3}}{8 π^{2}}$

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Solution

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

1 ρL38π2

2 ρL316π2

3 5ρL316π2

4 3ρL38π2