A uniform rod of length \(L\) is free to rotate in a vertical plane about a fixed horizontal axis through…
A uniform rod of length \(L\) is free to rotate in a vertical plane about a fixed horizontal axis through \(B\). The rod begins rotating from rest. The angular velocity \(\omega\) at angle \(\theta\) is given as:
$\sqrt{\frac{6 \mathrm{~g}}{l}} \sin \theta$
$\sqrt{\frac{6 g}{l}} \sin \frac{\theta}{2}$
$\sqrt{\frac{6 g}{l}} \cos \frac{\theta}{2}$
$\sqrt{\frac{6 \mathrm{~g}}{l}} \cos \theta$
Solution
The fall of centre of mass
$\begin{aligned}
& h=\frac{L}{2}(1-\cos \theta) \\
& mgh=mg \frac{L}{2}(1-\cos \theta)
\end{aligned}$
law of conservation of energy
$\begin{aligned}
& \frac{1}{2} I \omega^{2}=\frac{mgl}{2}(1-\cos \theta) \\
& \frac{1 mL^{2}}{2} \omega^{2}=\frac{mgL}{2}(1-\cos \theta) \omega^{2} \\
& \frac{6 g}{2 L}=(1-\cos \theta) \\
& \frac{6 g}{2 L} 2 \sin^{2} \frac{\theta}{2} \\
& \omega=\sqrt{\frac{6 g}{L} \sin^{2} \frac{\theta}{2}} \\
& \omega=\sqrt{\frac{6 g}{L}} \sin \frac{\theta}{2}
\end{aligned}$
