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Physics

Mechanics

Easy

Question

A uniform rod AB of mass m and length L is at rest on a smooth horizontal surface . An impulse J is applied to the end B perpendicular to the rod is horizontal direction . Speed of point P at distance $L over 6$ from the centre towards A of the rod after time $t equals fraction numerator pi m L over denominator 12 J end fraction$ is

$\frac{J}{m}$
$\frac{\sqrt{2} J}{m}$
$\frac{\sqrt{3} J}{m}$
$\frac{\sqrt{3} J}{2 m}$

The correct answer is: $fraction numerator square root of 2 J over denominator m end fraction$

$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em rowspacing 3 pt end attributes row blank cell text from end text theta equals omega t comma theta equals fraction numerator 6 J over denominator m L end fraction left parenthesis fraction numerator pi m L over denominator 12 J end fraction right parenthesis equals pi over 2 end cell row blank cell text Rod turns through end text 90 to the power of ring operator text speed of particle at end text end cell row blank cell P equals square root of v subscript c superscript 2 plus left parenthesis L over 6 omega right parenthesis squared end root equals fraction numerator square root of 2 J over denominator m end fraction end cell end table$

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