Question

A uniform meter stick is held vertically with one end on the floor and is allowed to topple such that the end at the floor does not slip. The speed of the other end, when it hits the floor, will be -

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

Answer 1

The correct option is B $\sqrt{3g}$


According to work-energy theorem,

$W_{\text{all forces}}=\Delta KE$

$\Rightarrow mgh=\frac{1}{2}I{\omega }^2$

$\Rightarrow mg\times \frac{L}{2}=\frac{1}{2}\left(\frac{m{L}^2}{3}\right){\omega }^2$

$\Rightarrow g=\left(\frac{L}{3}\right){\omega }^2$

$\Rightarrow \omega =\sqrt{\frac{3g}{L}}=\sqrt{\frac{3g}{1}}=\sqrt{3g}$

Now, speed of the other end,

$v=\omega R=\sqrt{3g}\times 1=\sqrt{3g}$

Hence, option $\left(B\right)$ is the correct answer.

Why this Question ? Tip: The potential energy of a rigid body is calculated by considering the position of its center of mass, which here is at $L/2$ from the floor. Caution : When the bar is horizontal, it won't have any translation kinetic energy, as the end attached to the floor does not slip. It will just have rotational kinetic energy.