A metre stick of length 1m is held vertically with one end in contact of the floor and is then allowed to fall.  If the end touching the floor is not allowed to slip, the other end will hit the ground with a velocity of (g=9.8 m/s²)

Potential energy is converted into kinetic energy of rotation.

From law of conservation of energy, we have the potential energy of rod when it is vertical is converted to kinetic energy of rotation.

$\therefore \frac{\mathrm{M}}{2}\mathrm{gl}=\frac{1}{2}{\mathrm{I\omega }}^2$

Moment of inertia of a thin rod is $I=\frac{1}{3}M{l}^2$

$\frac{M}{2}gl=\frac{1}{2}\frac{M{l}^2}{3}{\omega }^2$

$\Rightarrow \omega =\sqrt{\frac{3g}{l}}$

Since l=1m, $\omega =\sqrt{3g}$

Also, v=rω

$\Rightarrow v=1\times \sqrt{3\times 9.8}=5.4 \mathrm{m}/\mathrm{s}$