A uniform rod of length ' ℓ ^ ' is pivoted at one of its ends on a...

A uniform rod of length ' ℓ ^ ' is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft

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) $\dfrac{m{ℓ}^{2}}{12}{\omega}^{2}sin\theta$ 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:

Options

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

Correct answer

D. $cos\theta = \dfrac{3g}{2ℓ{\omega}^{2}}$

Step-by-step solution

Torque of centrifugal force ${τ}_{cf} = dm . xsin{\text{\theta \omega}}^{2}xcosϑ = \dfrac{m}{l}{\omega}^{2}sin\theta cos\theta{∫}_{0}^{l}{x}^{2}dx$

${τ}_{ef} = \dfrac{m{l}^{2}{\omega}^{2}sin \theta cos \theta}{3}$

${τ}_{mg} = {τ}_{cf}$

$mg . \dfrac{l}{2}sin\theta = \dfrac{m{l}^{2}{\omega}^{2}sin \theta cos\theta}{3}$

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

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