Question

Two sliders $A$ and $B$ connected by a light rigid rod $10\text{m}$ long, move in two frictionless shafts as shown in the figure. If $B$ starts from rest, determine the velocity of $B$ in $\text{m/sec}$ when $x=6\text{m}$. Assume $m_A=m_B=200\text{kg}$ and $m_C=100\text{kg}$. Also, rod is vertical initially. (Take $g=9.8{\text{ms}}^{-2}$)

Answer 1

$x^2+y^2=L^2$ $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$;
$\frac{xdx}{dt}=-\frac{ydy}{dt}$; $\frac{dx}{dt}=-\frac{y}{x}\frac{dy}{dt}$
$\therefore$ Velocity of $B=-\left(\frac{y}{x}\right)\times$ velocity of $A$. As when $x=6,y=8$,
$\therefore$ Velocity of $B=\frac{8}{6}{v}_A=\frac{4}{3}{v}_A$
Applying the law of conservation of energy, P.E. lost $=m_A.g.2+m_C.g.6$
Since when $B$ moves $6\text{m}$ from $O,A$ moves down $2\text{m}$ from its original position and $C$ moves down $6\text{m}$.
$\therefore$ P.E. lost $=200\times 2g+100\times 6g=1000g$
K.E.gained $=\frac{1}{2}{m}_A{v}_A^2+\frac{1}{2}{m}_B{v}_B^2+\frac{1}{2}{m}_C{v}_C^2$
$=\frac{1}{2}\times 200\times \frac{9}{16}{v}_B^2+\frac{1}{2}\times 200\times v_B^2+\frac{1}{2}\times 100\times v_B^2$
$=\frac{1}{2}\times 200\times v_B^2\{\frac{9}{16}+1+\frac{1}{2}\}=100\left\{\frac{33}{16}\right\}{v}_B^2$
$\therefore 100\times \frac{33}{16}{v}_B^2=1000\times 9.8$
$\therefore v_B^2=\frac{98\times 16}{33}$
$\therefore v_B=7\times 4\sqrt{\frac{2}{33}}=6.9\text{m/sec}$