Question

Two identical rods are joined as shown. The system is pivoted at point $O$ and is released from rest from the horizontal position. The speed of point $A$ when $OB$ become vertical is

• A. $6\sqrt{\frac{gl}{17}}$
• B. $3\sqrt{\frac{5gl}{17}}$
• C. $4\sqrt{\frac{3gl}{17}}$
• D. $2\sqrt{\frac{15gl}{17}}$

Answer 1

The correct option is B $3\sqrt{\frac{5gl}{17}}$

From conservation of mechanical energy at initial and final position. Considering horizontal line OB as the reference for $PE$.

Gain in kinetic energy = loss in gravitational potential energy,
$\frac{I{\omega }^2}{2}-0=3mg\frac{l}{2}$ (i)
Now, $I=I_1+I_2$
$I_1=\frac{m{l}^2}{3}$ and $I_2=\frac{m{l}^2}{12}+m{l}^2$
Putting the value of $I$ in equation (i)
$\Rightarrow \frac{17}{12}m{l}^2{\omega }^2=3mgl$

$\Rightarrow \frac{17}{12}l{\omega }^2=3g\Rightarrow \omega =\sqrt{\frac{36g}{17l}}$
$v_A=\omega \times OA=v_A=3\sqrt{\frac{5gl}{17}}$