Question

A slender rod of mass $m$ and length $L$ is pivoted about a horizontal axis through one end and released from rest at an angle of ${30}^o$ above the horizontal. The force exerted by the pivot on the rod at the instant when the rod passes through a horizontal position is :

• A. $\sqrt{\frac{10}{4}}$ $mg$ along horizontal.
• B. $mg$ along vertical.
• C. $\frac{\sqrt{10}}{4}mg$ along a line making an angle of $ta{n}^{-1}\left(\frac{1}{3}\right)$ with the horizontal.
• D. $\frac{\sqrt{10}}{4}$ $mg$ along a line making an angle of $ta{n}^{-1}\left(3\right)$ with the horizontal.

Answer 1

Answer: (C)

$\frac{\sqrt{10}}{4}mg$ along a line making an angle of $ta{n}^{-1}\left(\frac{1}{3}\right)$ with the horizontal.

The angular velocity of the rod about the pivot when it passes through the horizontal position is given by

$mg\times \frac{L}{2}sin{30}^o=\frac{m{L}^2}{3}\times \frac{{\omega }^2}{2}$

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

Radial acceleration of the centre of mass (as centre of mass is moving in a circle of radius L/2) is given by

$a_r={\omega }^2\frac{L}{2}=\frac{3g}{4}$

Torque about pivot, in the horizontal position, is $\tau =mgL/2=I\alpha$

$\alpha =\frac{mgL/2}{m{L}^2/3}=\frac{3g}{2}$

Tangential acceleration of the centre of mass, $a_t=\frac{L}{2}\alpha =\frac{3g}{4}$

Draw the FBD of the rod an instant when it passes through the horizontal position. Use Newton's second law of equation.

$R_1=m{a}_r=\frac{3mg}{4}$

$mg-R_2=m\times a_t=\frac{3mg}{4}$

$R_2=\frac{mg}{4}$

So, reaction force by the pivot on the rod, $R=R_1^3+R_2^2=\sqrt{10}mg/4$ at an angle of $ta{n}^{-1}\left(R_2/R_1\right)[=ta{n}^{-1}(1/3\left)\right]$ with the horizontal.