Question

The rod is released. What is the angular speed when it turns by ${180}^o$?

• A. $\sqrt{\frac{46g}{7L}}$
• B. $\sqrt{\frac{45g}{7L}}$
• C. $\sqrt{\frac{47g}{7L}}$
• D. $\sqrt{\frac{47g}{7L}}$

Answer 1

The correct option is D

$\sqrt{\frac{47g}{7L}}$

Moment of inertia of rod about hinge point
$I=I_0+M\left(\right(\frac{L}{4}{)}^2$
$I=\frac{M{L}^2}{12}+\frac{M{L}^2}{16}=\frac{7}{48}M{L}^2$
$K.E_i=0;P.E_i=\frac{MgL}{4}$
$K.E_i=\frac{1}{2}I{\omega }^2;P.E_f=-\frac{MgL}{4}$
Applying conservation of energy, $\frac{MgL}{4}=\frac{1}{2}l{\omega }^2+(-\frac{MgL}{4})$
$\Rightarrow \frac{MgL}{4}=\frac{1}{2}\left(\frac{7}{48}M{L}^2\right){\omega }^2-\frac{MgL}{4}$
$\Rightarrow \omega =\sqrt{\frac{48g}{7L}}$