A uniform rod of length â€˜$l$â€™ 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) ${{m{l^2}} \over {12}}{\omega ^2}\sin \theta \cos \theta $ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces FH and FV about the CM. The value of $\theta $ is then such that :

Question

A uniform rod of length â€˜$l$â€™ 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) ${{m{l^2}} \over {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 :

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

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

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

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

Solution

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