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Q.
A rod of length $l$ is hinged at one end and kept horizontal. It is allowed to fall. The velocity of the other end of the rod is
System of Particles and Rotational Motion
Solution:
As the mass is concentrated at the centre of the rod, therefore,
$m g \times \frac{l}{2}=\frac{1}{2} l \omega^{2}=\frac{1}{2}\left(\frac{m l^{2}}{3}\right) \omega^{2}$
$\Rightarrow \omega=\sqrt{\frac{3 g}{l}}$
Velocity of other end of the rod $v=\omega l=\sqrt{3 g l}$
