JEE MainPhysicsModerateRigid Body Dynamics

A uniform metre stick is held vertically with one end on the
floor and is allowed to fall. The speed of theother end when it hits the
floor assuming that the end at the floor does not slip

Show answer and solution

using energy conservation

mg\frac{\mathcal{l}}{2} = \frac{1}{2}I\omega^{2}

mg\frac{\mathcal{l}}{2} = \frac{1}{2} \cdot \frac{m\mathcal{l}^{2}}{3}\omega^{2}

= 1m

\omega = \sqrt{\frac{3g}{}}

V_{A} = \omega = \sqrt{3g} = (\sqrt{3g})

Source: Grass-G / P-07 Physics paper 17 October.docx

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