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Physics

Mechanics

Easy

Question

A thin wire of length L is bent into a circular wire of uniform linear density . When circular wire is in a vertical plane find the moment of inertia of loop about an axis BC, passing through centre of the loop and which makes an angle $theta$ with the tangent at the topmost point of the loop

$\frac{ρ L^{3}}{8 π^{2}}$
$\frac{ρ L^{3}}{2 \sqrt{2} π^{2}}$
$\frac{ρ L^{3}}{4 π^{2}}$
$\frac{ρ L^{3}}{3 π^{2}}$

The correct answer is: $fraction numerator rho L cubed over denominator 8 pi squared end fraction$

$table attributes columnalign left columnspacing 1em rowspacing 4 pt end attributes row cell M equals rho L semicolon space of 1em L equals 2 pi R space of 1em text so end text R equals L divided by 2 pi end cell row cell I subscript X X end subscript superscript straight prime equals 1 half M R squared space of 1em semicolon space of 1em I subscript straight capital pi equals 1 half M R squared end cell row cell I subscript 0 equals I subscript Z equals I subscript M plus I subscript straight capital pi T end subscript equals M R squared end cell end table table attributes columnalign left columnspacing 1em rowspacing 4 pt end attributes row cell text Now end text I subscript B C end subscript plus I subscript B C end subscript equals I subscript 0 equals I subscript Z equals M R squared end cell row cell text so end text I subscript B C end subscript equals fraction numerator M R squared over denominator 2 end fraction equals 1 half left parenthesis rho L right parenthesis fraction numerator L squared over denominator 4 pi squared end fraction equals fraction numerator rho L cubed over denominator 8 pi squared end fraction end cell end table$

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