Question

A uniform rod of length $\frac{6}{10}\text{m}$ is suspended from one end such that it is free to rotate about an axis passing through that end and perpendicular to the length. The minimum speed $\text{m/s}$ must be imparted to the lower end so that the rod completes one full revolution, is (Integer only)

Answer 1

In case of rod to complete one full revolution the rod must reach atleast the top most point as the rod does not slack like thread.

Hence for one full revolution the increase in $P.E=MgL$. where $M$ is the mass of the rod.
Therefore, $P.E=\Delta K.E$,

$MgL=\frac{1}{2}I{\omega }^2=\frac{1}{2}\left(\frac{M{L}^2}{3}\right){\omega }^2$,
$($As the moment of inertia of the rod about an axis fixed to one end is $I=\frac{M{L}^2}{3})$

$\Rightarrow \omega =\sqrt{\frac{6g}{L}}$

Now, $v=r\omega =L\sqrt{\frac{6g}{L}}=\sqrt{6gL}$ $(\because r=L)$

Given, $L=\frac{6}{10}\text{m}$ and $g=10{\text{m/s}}^2$

$\therefore v=\sqrt{6\times 10\times \frac{6}{10}}=6\text{m/s}$

Correct answer$:6$