A uniform slender rod of mass $m$ and length $L$ is released from rest, with its lower end touching a frictionless horizontal floor. At the initial moment, the rod is inclined at an angle $θ = 30^{\circ}$ with the vertical. Then the initial acceleration of the lower end of the rod will be

g \sqrt{3} / 4

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g \sqrt{3} / 5

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3 g \sqrt{3} / 7

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None of these

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Solution

The correct option is C $3 g \sqrt{3} / 7$
$a_{P} = (O P) α$
$= (\frac{l}{2} cos 30^{\circ}) (\frac{12}{7} \frac{g}{l})$
$= \frac{3 \sqrt{3} g}{7}$

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A uniform slender rod of mass m and length L is released from rest, with its lower end touching a frictionless horizontal floor. At the initial moment, the rod is inclined at an angle θ=30∘ with the vertical. Then the initial acceleration of the lower end of the rod will be

The correct option is C 3g√3/7 aP=(OP)α=(l2cos30∘)(127gl)=3√3g7

A uniform slender rod of mass $m$ and length $L$ is released from rest, with its lower end touching a frictionless horizontal floor. At the initial moment, the rod is inclined at an angle $θ = 30^{\circ}$ with the vertical. Then the initial acceleration of the lower end of the rod will be

The correct option is C $3 g \sqrt{3} / 7$
$a_{P} = (O P) α$
$= (\frac{l}{2} cos 30^{\circ}) (\frac{12}{7} \frac{g}{l})$
$= \frac{3 \sqrt{3} g}{7}$