Question

A uniform rod of length hangs vertically from a hinge passing through one end. The initial angular velocity ω that must be imparted to the rod so as to rotate it through 90° is

• A. $\sqrt{\frac{g}{\mathcal{l}}}$
• B. $\sqrt{\frac{2g}{\mathcal{l}}}$
• C. $\sqrt{\frac{3g}{\mathcal{l}}}$
• D. $\sqrt{\frac{6g}{\mathcal{l}}}$

Answer 1

Answer: (C)

$\sqrt{\frac{3g}{\mathcal{l}}}$

Answer 2

Initial K.E. of rotation = $\frac{1}{2}I\omega^{2} = \frac{1}{2}.\frac{M\mathcal{l}^{2}}{3}.\omega^{2}$

Finally, the total energy is P.E. = Mg$\frac{\mathcal{l}}{2}$

where $\frac{\mathcal{l}}{2}$ is the height through which c.m. of the rod is raised.

⇒ $\frac{1}{6}M\mathcal{l}^{2}\omega^{2} = \frac{1}{2}Mg\mathcal{l}$ ⇒ ω = $\sqrt{\frac{3g}{\mathcal{l}}}$

(3) is the correct