A linear rigid rod of mass m and length l, falls freely under gravity, at an angular position $α$ with vertical upward direction. The rod makes inelastic impact with the smooth ground at end A. Just after impact,

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The acceleration of end A is $[\frac{3 g \sin α \cos α}{1 + 3 \sin^{2} α}] (- \hat{i})$

The acceleration of end A is $[\frac{g \sin α \cos α}{1 + 3 \sin^{2} α}] (- \hat{i})$

Acceleration of centre of mass is $g [\frac{\sin^{2} α}{1 + 3 \sin^{2} α}] (- \hat{j})$

Acceleration of centre of mass $g [\frac{3 \sin^{2} α}{1 + 3 \sin^{2} α}] (- \hat{j})$

answer is A, C.

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Detailed Solution

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${\vec{a}}_{A}$ must be horizontal
$\Rightarrow a = \frac{L}{2} β \sin α$
${\vec{F}}_{n e t} = M {\vec{a}}_{c m} \Rightarrow m g - N = m a$
${\vec{S}}_{c m} = I_{c m} \vec{α} \Rightarrow N (\frac{L}{2} \sin α) = \frac{M L^{2} β}{12}$
on solving 1, 2 and 3

$\Rightarrow β = \frac{6 g \sin α}{L (1 + 3 \sin^{2} α)}$
${\vec{a}}_{A} = \frac{L}{2} β \cos α$
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