Question

For the pivoted slender rod of length $l$ as shown in figure., find the angular velocity as the bar reaches the vertical position after being released in the horizontal position.

Answer 1

By parallel axis theorem, the moment of inertia of the rod about the hinge is

$\Rightarrow \frac{m{l}^2}{12}+m{\left(\frac{l}{4}\right)}^2=\frac{7m{l}^2}{48}$

When the rod reaches the vertical position the center of mass of the rod falls by $l4$

By conservation of energy,

$\Rightarrow \frac{mgl}{4}=\frac{1}{2}I{w}^2[I=\frac{{7ml}^2}{48}]$

$\therefore w={\left(\frac{24}{7}gl\right)}^{\frac{1}{2}}$

Hence, the answer is ${\left(\frac{24}{7}gl\right)}^{\frac{1}{2}}.$