Question

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

• A. The angular acceleration of the rod will be $\frac{2g}{L}$
• B. The normal reaction force from the floor on the rod will be $\frac{Mg}{16}$
• C. The radial acceleration of the rod's center of mass will be $\frac{3g}{4}$
• D. The angular speed of the rod will be $\sqrt{\frac{3g}{2L}}$

Answer 1

Answer: (B), (C), (D)

The angular speed of the rod will be $\sqrt{\frac{3g}{2L}}$

When the rod has rotated by ${60}^o$, the centre of mass gets shifted vertically by $l/4$
Using conservation of energy,
$\Delta K+\Delta U=0$
$\frac{1}{2}{I}_0{\omega }^2=-\Delta U$
$\frac{1}{2}\frac{m{L}^2}{3}{\omega }^2=-(-mg\frac{L}{4})$
$\omega =\sqrt{\frac{3g}{2L}}$
Hence option D is correct.
$\Rightarrow a_{radial}={\omega }^2\frac{L}{2}=\frac{3g}{2L}\frac{L}{2}=\frac{3g}{4}\text{Hence option C is correct}\text{Since},\tau =I_0\alpha \left(\text{About hinge}\right)$
$\alpha =\frac{mg\frac{L}{2}\sin {60}^o}{m\frac{L^2}{3}}=\frac{3\sqrt{3}g}{4L}\Rightarrow a_v=\left(\alpha \frac{L}{2}\right)\sin {60}^o+{\omega }^2\frac{L}{2}\cos {60}^0$ where $a_v$ is the net vertical accelration .
$\Rightarrow a_v=\frac{3\sqrt{3}g}{8}\frac{\sqrt{3}}{2}+\frac{3g}{8}\Rightarrow a_v=\frac{9g}{16}+\frac{6g}{16}=\frac{15mg}{16}$
Let $N$ be the normal reaction at the point of contact of rod.
Net vertical forces = $m{a}_v$
$\Rightarrow mg-N=m{a}_v\Rightarrow N=\frac{mg}{16}$
Hence option B is also correct.