The thin uniform rod in Fig. $\mathbf{10}\mathbf{-}\mathbf{53}\mathbf{}$has length $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{m}}\mathbf{}$and can pivot about a horizontal, frictionless pin through one end. It is released from rest at angle$\theta =40°$above the horizontal. Use the principle of conservation of energy to determine the angular speed of the rod as it passes through the horizontal position.

1. Length of the rod is$\text{L}=2.0\text{m}$
2. Angle is$\theta ={40}^0$

Use the concept of conservation of mechanical energy. First, the height of the centre of mass from the length of the rod and angle given. Then using the inertia of the rod about one end in the equation of conservation of mechanical energy, find the speed of the rod at horizontal position.

Formulae:

$\begin{array}{rcl}{\text{mgh}}_i+\frac{1}{2}\text{I}{\omega }_i^2& =& {\text{mgh}}_f+\frac{1}{2}\text{I}{\omega }_f^2\\ \text{I}& =& \frac{1}{3}{\text{ML}}^2\end{array}$

Angular speed of the rod as it passes through the horizontal position:

We know the centre of mass of the rod is at $\frac{\text{L}}{2}$, so initially, it will have gravitational potential energy; then at horizontal position, it will have kinetic energy.

Centre of mass will be at height, ${\text{h}}_i=\frac{\text{L}}{2}\text{sin}\theta$and we know moment of inertia of rod about one end is $\text{I}=\frac{1}{3}{\text{ML}}^2.$We get

$\text{Mg}\frac{L}{2}\text{sin}\theta +\frac{1}{2}\text{I}\left(0\right)=\text{mg}\left(0\right)+\frac{1}{2}\left(\frac{1}{3}{\text{ML}}^2\right){\omega }_f^2$

$\Rightarrow \text{Mg}\frac{\text{L}}{2}\text{sin}\theta =\frac{1}{2}\left(\frac{1}{3}{\text{ML}}^2\right){\omega }_f^2$

Mass is common on both sides, so it will be cancelled out. Plugging the values, we get

$\begin{array}{rcl}\left(9.8\right)\frac{2.0}{2}\text{sin}\left({40}^0\right)& =& \frac{1}{2}\left(\frac{1}{3}{2.0}^2\right){\omega }_f^2\\ & \Rightarrow & 6.2993=0.666{\omega }_f^2\\ & \Rightarrow & {\omega }_f^2=\frac{6.2993}{0.666}=9.4489\\ & \Rightarrow & {\omega }_f=3.07\\ & \Rightarrow & {\omega }_f=3.1\text{rad}/\text{s}\end{array}$