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. $\cos \theta =\frac{3g}{2\ell {\omega }^2}$
• 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)