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) $\frac{ m \ell^{2}}{12} \omega^{2} \sin \theta \cos \theta$ 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. $\cos \theta=\frac{ g }{2 \ell \omega^{2}}$
• B. $\cos \theta=\frac{3 g}{2 \ell \omega^{2}}$
• C. $\cos \theta=\frac{2g }{3 \ell \omega^{2}}$
• D. $\cos \theta=\frac{ g }{\ell \omega^{2}}$

Answer 1

Answer: (B)

$F _{ V }= mg$
$F _{ H }= m \omega^{2} \frac{\ell}{2} \sin \,\theta$
$mg \frac{\ell}{2} \sin \theta- m \omega^{2} \frac{\ell}{2} \sin \theta \frac{\ell}{2} \cos \theta=\frac{ m \ell^{2}}{12} \omega^{2} \sin \theta \cos \theta$
$\cos \theta=\frac{3}{2} \frac{ g }{\omega^{2} \ell}$ ......(ii)