Question

A uniform rod of length $l$ is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed$\omega$ the rod makes an angle $\theta$ with it (see figure). To find$\theta$ equate the rate of change of angular momentum (direction going into the paper) $\left(\raisebox{1ex}{$m{l}^2$}\!\left/ \!\raisebox{-1ex}{$12$}\right.\right){\omega }^2\sin \theta \cos \theta$ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces $F_H$ and $F_v$ about the CM. The value of $\theta$is then such that:

• A. $\cos \theta =\frac{2g}{3l{\omega }^2}$
• B. $\cos \theta =\frac{3g}{2l{\omega }^2}$
• C. $\cos \theta =\frac{g}{2l{\omega }^2}$
• D. $\cos \theta =\frac{g}{l{\omega }^2}$

Answer 1

The correct option is B

$\cos \theta =\frac{3g}{2l{\omega }^2}$

Step 1. Given data:

A uniform rod of length $l$

Angular speed is$\omega$ and the rod makes an angle $\theta$

The rate of change of angular momentum equals torque

$\tau =\left(\raisebox{1ex}{$m{l}^2$}\!\left/ \!\raisebox{-1ex}{$12$}\right.\right){\omega }^2\sin \theta \cos \theta$ $\left(1\right)$

Step 2. solving for $\theta$:

From the diagram,

The horizontal force is, $F_H=m{\omega }^2\frac{l}{2}\sin \theta$

The vertical force is, $F_V=-mg$

Where $g$ is the acceleration due to gravity

The net torque about the centre of mass is, ${\tau }_{net}=F_v\frac{l}{2}.\sin \theta -F_H\frac{l}{2}\cos \theta$ $\left(2\right)$

Equating equation $\left(1\right)$and $\left(2\right)$ , we get

$F_v\frac{l}{2}.\sin \theta -F_H\frac{l}{2}\cos \theta =\frac{m{l}^2}{12}{\omega }^2\sin \theta \cos \theta$

$\Rightarrow mg\frac{l}{2}.\sin \theta -m{\omega }^2\frac{l}{2}\sin \theta \frac{l}{2}\cos \theta =\frac{m{l}^2}{12}{\omega }^2\sin \theta \cos \theta$

$\Rightarrow g-\frac{l}{2}{\omega }^2\cos \theta =\frac{l}{6}{\omega }^2\cos \theta$

$\Rightarrow g=\left(\frac{l}{6}{\omega }^2+\frac{l}{2}{\omega }^2\right)\cos \theta$

$\Rightarrow g=\frac{2l{\omega }^2}{3}\cos \theta$

$\Rightarrow \cos \theta =\frac{3g}{2l{\omega }^2}$

Hence, option B is correct.