Question

A light rod carries three equal masses A, B and C as shown in figure. What will be velocity of B in vertical position of rod, if it is released from horizontal position as shown in figure?

Answer 1

$I=MI$ of rod about fixed point

$I=M\left(\frac{l}{3}\right)2+M\left[\left(\frac{2l}{3}\right)\right]2+M{l}^2$

$=M{l}^2[\left(\frac{1}{9}\right)+\left(\frac{4}{9}\right)+1]$

$=M{l}^2\cdot \left[\frac{14}{9}\right]$

$=\left[\frac{14}{9}\right]M{l}^2$ (1)

also when rod is released,

loss in PE = gain in KE

hence $Mg\left(\frac{l}{3}\right)+Mg\left[\left(\frac{2l}{3}\right)\right]+Mgl=\left(\frac{1}{2}\right)I{w}^2$

$2Mgl=\left(\frac{1}{2}\right)\times \left[\left(\frac{14}{9}\right)\right]M{l}^2{w}^2-----from\left(1\right)$

$2g=\left(\frac{14}{18}\right)l{w}^2$

hence$w^2=\left[\left(\frac{36g}{14l}\right)\right]$

$V=r\omega$\\ $VB=\left(\frac{2}{3}\right)l\cdot \sqrt{\left[\left(\frac{36g}{14l}\right)\right]}$

$=\sqrt{[\left(\frac{36g}{14l}\right)\times \frac{4}{9}{l}^2]}$

$=\sqrt{[l\cdot g\cdot \left(\frac{16}{14}\right)]}$

$=\sqrt{\left\{\left(\frac{8}{7}\right)gl\right\}}$