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A uniform rod of mass $M$ , length $L$ which can rotate freely in vertical $x - y$ plane about hinge point $O$ as shown is released from horizontal position. Find hinge reaction on rod just after releasing it.

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$M g$

$\frac{M g}{4}$

$\frac{M g}{2}$

$\frac{3 M g}{4}$

answer is B.

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Detailed Solution

By considering the FBD of the body,

$Mg - N = {Ma}_{cm}$ ...(i)

$I α = Mg . \frac{l}{2} \Rightarrow \frac{M l^{2}}{3} α = Mg . \frac{l}{2}$

$\Rightarrow α = \frac{3 g}{2 l}$

And the relation between $α$ and $a_{cm}$ will be $a_{c m} = \frac{l}{2} α$

So, $a_{cm} = \frac{3 g}{4}$

from (i) $N = M g - M (\frac{3 g}{4}) = \frac{M g}{4}$

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