Question

A uniform rod smoothly pivoted at one of its ends is released from rest. If it swings in vertical plane , find the maximum speed of the end P of the rod:

Answer 1

$Vp=wR+Vcom$

As $wR=Vcom--\left(i\right)$

$Vp=QVcom--\left(ii\right)$

we know

$\Delta KE$ Total $=\frac{1}{2}m(Vcom{)}^2+\frac{(Icoa;)(w{)}^2}{2}---(iii)$

$\Delta PE=-mg\left(i/2\right)$

As only gravitational force i.e., cloing

work and it is a conservative force,

$\Delta KB$ roots $\Delta PE=0$

$\Delta KE$ Total $=-\Delta PE=-(-\frac{mg}{2}l)$

$\Rightarrow \Delta KE$ Total $=\frac{mgl}{2}---\left(iv\right)$

From $\left(i\right),\left(iii\right)$ and $\left(iv\right)$, we get

$\frac{1}{2}m(Vcom{)}^2+\left(\frac{M}{12}\right)\frac{(Vcom{)}^2}{2}=\frac{mg}{2}l$

$\Rightarrow (Vcom{)}^2=\frac{24}{13}\left(\frac{mgl}{2m}\right)$

$\Rightarrow Vcom=\sqrt{\frac{12}{13}gl}$

So, $Vp=2Vcom=4\sqrt{\frac{3}{13}ge}$