Question

A uniform thin rod of length $l$ and mass $m$ is hinged at a distance $l/4$ from one of the end released from horizontal position as shown in figure. The angular velocity of the rod as it passes the vertical position is

• A. $2\sqrt{\frac{5g}{7l}}$
• B. $2\sqrt{\frac{6g}{7l}}$
• C. $\sqrt{\frac{3g}{7l}}$
• D. $2\sqrt{\frac{g}{l}}$

Answer 1

The correct option is B $2\sqrt{\frac{6g}{7l}}$


Applying conservation of mechanical energy

Loss in gravitational P.E = Gain in rotational K.E

$\Rightarrow mg\left(\frac{l}{4}\right)=\frac{1}{2}I{\omega }^2.......\left(1\right)$

Here, using parallel theorem

$I=I_{cm}+m{d}^2$

$\Rightarrow I=\frac{m{l}^2}{12}+m{\left(\frac{l}{4}\right)}^2=\frac{m{l}^2}{12}+\frac{m{l}^2}{16}$

$\Rightarrow I=\frac{7m{l}^2}{48}$

Substituting in Eq. $\left(1\right)$

$mg\left(\frac{l}{4}\right)=\frac{1}{2}\times \frac{7m{l}^2}{48}{\omega }^2$

$\Rightarrow \omega =\sqrt{\frac{24g}{7l}}=2\sqrt{\frac{6g}{7l}}$

So, option $\left(b\right)$ is correct.

Why this question? Caution : while applying mechanical energy conservation, it should be ensured to use $I$ about the hinged point.