Question

A thin wire of length L and uniform linear mass density ρ is bent in to a circular loop with centre at O as shown in figure. The moment of inertia of the loop about the axis xx' is …. 

(A) (ρL² / 8π²) 

(B) (ρL³ / 16π²) 

(C) (5ρL³ / 16π²) 

(D) (3ρL³ / 8π²)

Answer 1

Correct option: (D) (3ρL³ / 8π²)

Explanation:

Length of wire = L

mass per unit length = ρ

hence mass of length L = ρL

when it is converted to loop then 2πr = L

r = (L / 2π)

MI of loop about axis passing through centre O is

I_o = [(mr²) / 2]

from parallel axes theorem,

MI about xx¹ = I = I_o + mr² = [(mr²) / 2] + mr² = (3/2)mr² 

hence I = (3/2) × ρL[L / (2π)]² = [(3ρL³) / (8π²)]