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 ${60}^0$ with vertical? [g is the acceleration due to gravity]

The axis of rotation of rod is considered at the bottom  contact point in this case. Using work energy theorem, we have from the initial state of rod to the state when it makes an angle  ${60}^0$ with the vertical, it gives  $mg\frac{l}{4}=\frac{1}{2}\left(\frac{m{l}^2}{3}\right){\omega }^2$

$\Rightarrow \omega =\sqrt{\frac{3g}{2l}}$

Hence option (C) is correct.

Radial acceleration of centre of mass of rod is given as  $a_t=\left(\frac{l}{2}\right){\omega }^2=\frac{3g}{4}$

Hence option (A) is correct.

Using  $\tau =I\alpha$ about bottom contact point  gives

$\frac{Mgl}{2}\sin {60}^0=\frac{M{I}^2}{3}\alpha$

$\Rightarrow \alpha =\frac{3\sqrt{3}}{4l}g$

Hence option (B) is NOT correct.

Thus tangential acceleration of centre of mass of rod is given as  $a_t=\alpha \times \frac{1}{2}$

Total acceleration of  centre of mass of rod in vertical direction is given as

$\Rightarrow a_v=a_r\cos {60}^0+a_t\cos {30}^0$

$\Rightarrow a_v=\frac{3g}{8}+\frac{3\sqrt{3}g}{4l}\left(\frac{l}{2}\right)\left(\frac{\sqrt{3}}{2}\right)$

 $\Rightarrow a_v=\frac{3g}{8}+\frac{9g}{16}=\frac{15}{16}g$

In vertical direction on rod, we use

$Mg-N=M{a}_v=M\left(\frac{15}{16}g\right)$

$\Rightarrow N=\frac{Mg}{16}$

Hence option (D) is correct.