Question

A uniform meter stick is held at the lower end and the other end is allowed to fall. The speed of the other end when it hits the floor assuming that the end at the floor does not slip

• A. $\sqrt{4g}$
• B. $\sqrt{3g}$
• C. $\sqrt{5g}$
• D. $\sqrt{g}$

Answer 1

Answer: (B)

$\sqrt{3g}$

The centre of mass of the stick comes down by $0.5m$ as the center of mass in the center of the stick.

Now, the stick rotates, so loss in potential energy is equal to gain in its kinetic energy.

$mg\left(0.5\right)=\frac{1}{2}I{\omega }^2$

$L=1m⟹I=\frac{m{L}^2}{3}=\frac{m}{3}$

$⟹\frac{mg}{2}=\frac{1}{2}\frac{m}{3}{\omega }^2$

$⟹{\omega }^2=3g$

$⟹\omega =\sqrt{3g}$

Speed of other end$=v=\omega \times L=\sqrt{3g}$ since $(L=1m)$

Answer-(B)