A thin and uniform rod of mass M and length L is held vertical on a floor with large friction. The rod is…

1. The radial acceleration of the rod's center of mass will be $\frac{3\mathrm{g}}{4}$
2. The angular acceleration of the rod will be $\frac{2\mathrm{g}}{\mathrm{L}}$
3. The angular speed of the rod will be $\sqrt{\frac{3\mathrm{g}}{2\mathrm{L}}}$
4. The normal reaction force from the floor on the rod will be $\frac{\mathrm{M}\mathrm{g}}{16}$

Solution

Using conservation of energy
$∆K+∆U=0$
$\frac{1}{2}{I}_0{\omega }^2=-∆U\left(I_0=\mathrm{m}\mathrm{o}\mathrm{m}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{ }\mathrm{a}\mathrm{b}\mathrm{o}\mathrm{u}\mathrm{t}\mathrm{ }\mathrm{H}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{e}\right)$
$\frac{1}{2}\frac{m{l}^2}{3}{\omega }^2=-\left(-mg\frac{l}{4}\right)$
$\omega =\sqrt{\frac{3g}{2l}}\Rightarrow \left(\mathrm{C}\right)$ is correct
$\Rightarrow a_{\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{l}}={\omega }^2.\frac{l}{2}=\frac{3g}{2l}\frac{l}{2}=\frac{3g}{4}\Rightarrow \left(\mathrm{A}\right)$ is correct
Now, $\tau =I.\alpha$
$\Rightarrow mg.\frac{l}{2}\sin 60=\frac{m{l}^2}{3}.\alpha \Rightarrow \alpha =\frac{3\sqrt{3}g}{4l}$
$\Rightarrow \left(\mathrm{B}\right)$ is incorrect
Acceleration in vertical direction $\Rightarrow a_v=\left(\alpha \frac{l}{2}\right)\sin {60}^{\mathrm{o}}+{\omega }^2\frac{l}{2}\cos {60}^{\mathrm{o}}$
$=a_v=\frac{3\sqrt{3}g}{8}\frac{\sqrt{3}}{2}+\frac{3g}{8}$
$a_v=\frac{9g}{16}+\frac{6g}{16}=\frac{15 g}{16}$
Now, using $\mathrm{N}\mathrm{L}\mathrm{M}\Rightarrow mg-N=m{a}_v\Rightarrow N=mg-m{a}_v=mg-\frac{15}{16}mg$
$\Rightarrow N=\frac{mg}{16}\Rightarrow \left(\mathrm{D}\right)$ is correct

Asked in: JEE Advanced 2019 (Paper 2)

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