# Question #638a4

Jun 16, 2016

$\text{option C}$

#### Explanation: $\text{it will gain a potential energy if object is brought at horizontal position.}$

${E}_{p} = m \cdot g \cdot h$

$m : \text{mass of object}$
$g : \text{acceleration of gravity}$
$h = \frac{l}{2}$

${E}_{p} = m \cdot g \cdot \frac{l}{2}$

$\text{it will turn and will gain a kinetic energy if object is released}$

${E}_{k} = \frac{1}{2} \cdot I \cdot {\omega}^{2}$

$I \text{ represents the moment of inertia of rod}$
$\omega \text{ represents angular velocity of object}$

$\text{we can write "E_p=E_k " because Energy is reserved.}$

$m \cdot g \cdot \frac{l}{\cancel{2}} = \frac{1}{\cancel{2}} \cdot I \cdot {\omega}^{2}$

$m \cdot g \cdot l = I \cdot {\omega}^{2}$

$\text{ so " I=1/3*m*l^2" (moment of inertia for a rod turning its one end)}$

$\text{we can write as:}$

$\cancel{m} \cdot g \cdot \cancel{l} = \frac{1}{3} \cdot \cancel{m} \cdot \cancel{{l}^{2}} \cdot {\omega}^{2}$

$g = \frac{1}{3} l \cdot {\omega}^{2}$

${\omega}^{2} = \frac{3 g}{l}$

$\sqrt{{\omega}^{2}} = \sqrt{\frac{3 g}{l}}$

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