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A thin and uniform rod of mass $M$ and length $L$ is held vertical on a floor with large friction. The rod is released from rest so that it falls by rotating about its contact-point with the floor without slipping. Which of the following statement($s$) is/are correct, when the rod makes an angle $60^{\circ}$ with vertical ? [ $g$ is the acceleration due to gravity]
$(1)$ The radial acceleration of the rod's center of mass will be $\frac{3 g }{4}$
$(2)$ The angular acceleration of the rod will be $\frac{2 g }{ L }$
$(3)$ The angular speed of the rod will be $\sqrt{\frac{3 g}{2 L}}$
$(4)$ The normal reaction force from the floor on the rod will be $\frac{ Mg }{16}$
$1,2,3$
$1,2,4$
$1,3,4$
$1,2$
Solution
We can treat contact point as hinged.
Applying work energy theorem
$W _{ g }=\Delta K . E .$
$mg \frac{\ell}{4}=\frac{1}{2}\left(\frac{ m \ell^2}{3}\right) \omega^2$
(image)
$\omega=\sqrt{\frac{3 g }{2 \ell}}$
radial acceleration of $C.M.$ of rod $=\left(\frac{\ell}{2}\right) \omega^2=\frac{3 g }{4}$
Using $\tau= I \alpha$ about contact point
$\frac{ mg \ell}{2} \sin 60^{\circ}=\frac{ m \ell^2}{3} \alpha$
$\Rightarrow \alpha=\frac{3 \sqrt{3}}{4 \ell} g$
Net vertical acceleration of $C.M.$ of rod
$a _2= a _{ r } \cos 60^{\circ}+ a _{ t } \cos 30^{\circ}$
$=\left(\frac{3 g }{4}\right)\left(\frac{1}{2}\right)+\left(\alpha \frac{\ell}{2}\right) \cos 30^{\circ}$
$=\frac{3 g }{8}+\frac{3 \sqrt{3} g }{4 \ell}\left(\frac{\ell}{2}\right)\left(\frac{\sqrt{3}}{2}\right)$
$=\frac{3 g }{8}+\frac{9 g }{16}=\frac{15}{16} g$
Applying $F_{\text {not }}= ma$ in vertical direction on rod as system
$mg – N = ma _{ v }= m \left(\frac{15}{16} g \right)$
$N =\frac{ mg }{16}$
(iamage)
