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

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

Answer 1

Answer: (A), (C), (D)

We can treat contact point as hinged.
Applying work energy theorem
$W_g=\Delta KE.$

$mg\frac{\ell }{4}=\frac{1}{2}\left(\frac{m{\ell }^2}{3}\right){\omega }^2$

$\omega =\sqrt{\frac{3g}{2\ell }}$

radial acceleration of C.M. of rod $=\left(\frac{\ell }{2}\right){\omega }^2=\frac{3g}{4}$

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

$\frac{mg\ell }{2}sin60^{\circ }=\frac{m{\ell }^2}{3}\alpha$

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

Net vertical acceleration of C.M. of rod

$a_v=a_{r}cos60^{\circ }+a_{t}cos30^{\circ }$

$=\left(\frac{3g}{4}\right)\left(\frac{1}{2}\right)+\left(\alpha \frac{\ell }{2}\right)cos30^{\circ }$

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

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

Applying $F_{net}$ = ma in vertical direction on rod as system

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

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