Tilting Rod

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Consider the rod in its vertical upright position. Treating the X-axis as the zero gravitational potential energy level, the PE of an elementary rod length $dy$ at a distance $y$ from the hinge is:

$dV = \left(\lambda \ dy\right)gy$

Here $\lambda = y$ is the mass per unit length of the rod. Integrating the above expression from $0$ to $L$ gives:

$V = \frac{gL^3}{3}$

Next, the moment of inertia of the rod about the hinge can be calculated as such:

$I = \int_{0}^{L} y^2\left(\lambda \ dy\right) = \frac{L^4}{4}$

So if the rod starts from rest from its vertically upward position and rotates till it becomes horizontal, its potential energy is converted to kinetic energy as such:

$V_{initial} + KE_{initial} = V_{final} + KE_{final}$ $\implies \frac{gL^3}{3} + 0 = 0 + \frac{1}{2}I\omega^2$

$\implies \boxed{\omega = \sqrt{\frac{8g}{3L}}}$

Mass of the rod is $\dfrac{L^2}{2}$ . Moment of inertia of the rod about one of it's ends is $\dfrac{L^4}{4}$ . Distance of the C.M. of the rod from the hinged end is $\dfrac{2L}{3}$ . So, initial P.E. of the rod is $\dfrac{L^2}{2}\times g\times \dfrac{2L}{3}=\dfrac{L^3g}{3}$ . Initial K.E. of the rod is zero. Final P.E. is zero and final K.E. is $\dfrac{1}{2}\times \dfrac{L^4}{4}\times \omega^2$ , where $\omega$ is the required angular velocity. Applying energy conservation law, we get $\omega =2\sqrt \dfrac{2g}{3L}$ . Substituting the given values of $g$ and $L$ we get $\omega\approx {\boxed {3.651}}$