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Physics

Mechanics

Easy

Question

A uniform rod of mass m and length 2l is rotating about an axis passing through its centre and perpendicular to its length. If is its angular velocity , its angular momentum about that axis is

$m l^{2} ω$
$\frac{m l^{2} ω}{3}$
$\frac{m l^{2} ω}{6}$
$\frac{m l^{2} ω}{2}$

The correct answer is: $fraction numerator m l squared omega over denominator 3 end fraction$

$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em rowspacing 3 pt end attributes row blank blank row blank cell text Angular momentum, end text L equals I omega equals fraction numerator m left parenthesis 2 h right parenthesis blank squared over denominator 12 end fraction cross times omega equals fraction numerator m l squared omega over denominator 3 end fraction end cell end table$

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