Question

A uniform rod is hinged as shown in the figure and is released from a horizontal position.The angular velocity of the rod as it passes the vertical position is:

• A. $\sqrt{\frac{12g}{3l}}$
• B. $\sqrt{\frac{2g}{3l}}$
• C. $\sqrt{\frac{24g}{7l}}$
• D. $\sqrt{\frac{3g}{7l}}$

Answer 1

The correct option is C $\sqrt{\frac{24g}{7l}}$

R.E.F image

By parallel axis Theorem, MOL of rod

about hinge is

$I_0=\frac{m{L}^2}{12}+m{\left(\frac{L}{4}\right)}^2=\frac{7m{L}^2}{48}$

When rod reaches the vertical position

the center of mass of rod falls by

$L/4$

By consecration of energy

$mg\left(\frac{L}{4}\right)=\frac{1}{2}\times I_0{w}^2$

$\frac{mgL}{4}=\frac{1}{2}\times \frac{7m{L}^2}{48}{w}^2$

$\therefore w=\sqrt{\frac{24g}{7L}}$