Question

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:

• A. $\sqrt{\left(\frac{6g}{L}\right)}\sin \left(\frac{\theta }{2}\right)$
• B. $\sqrt{\left(\frac{6g}{L}\right)}\cos \left(\frac{\theta }{2}\right)$
• C. $\sqrt{\left(\frac{6g}{L}\right)}\sin \theta$
• D. $\sqrt{\left(\frac{6g}{L}\right)}\cos \theta$

Answer 1

The correct option is A $\sqrt{\left(\frac{6g}{L}\right)}\sin \left(\frac{\theta }{2}\right)$

the fall of centre of mass

$h=\frac{L}{2}(1-cos\theta )$

$mgh=mg\frac{L}{2}(1-cos\theta )$

law of conservation of energy

$\frac{1}{2}I{\omega }^2=\frac{mgl}{2}(1-cos\theta )$

$\frac{1}{2}\frac{{mL}^2}{3}{\omega }^2=\frac{mgL}{2}(1-cos\theta ){\omega }^2$

$\frac{6g}{2L}=(1-cos\theta )$

$\frac{6g}{2L}2{sin}^2\frac{\theta }{2}$

$\omega =\sqrt{\frac{6g}{L}{sin}^2\frac{\theta }{2}}$

$\omega =\sqrt{\frac{6g}{L}}sin\frac{\theta }{2}$