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 $O$ and is at rest initially, determine the velocity of $B$ ( in $\text{m/s}$ upto two decimals) when $x=6\text{m}$. Assume $m_A=m_B=200\text{kg}$ and $m_c=100\text{kg},\sqrt{\frac{2}{33}}=0.25$.

Answer 1

From constraints,
$x^2+y^2=L^2$
Differentiating
$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$;
$\frac{xdx}{dt}=-\frac{ydy}{dt}$;
$\frac{dx}{dt}=-\frac{y}{x}\frac{dy}{dt}$
$\therefore$ 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 $2\text{m}$ downward from its original position and $C$ moves $6\text{m}$ downward.
$\therefore$ P.E. lost $=200\times 2g+100\times 6g=1000g$

As $v_B=v_C$
K.E. gained $=\frac{1}{2}{m}_A{v}_A^2+\frac{1}{2}{m}_B{v}_B^2+\frac{1}{2}{m}_C{v}_C^2$
$=\frac{1}{2}\times 200\times \frac{9}{16}{v}_B^2+\frac{1}{2}\times 200\times v_B^2+\frac{1}{2}\times 100\times v_B^2$
$=\frac{1}{2}\times 200\times v_B^2\{\frac{9}{16}+1+\frac{1}{2}\}$
$=100\left\{\frac{33}{16}\right\}{v}_B^2$

KE gained=PE lost
$\therefore 100\times \frac{33}{16}{v}_B^2=1000\times 9.8$
$\therefore v_B^2=\frac{98\times 16}{33}$
$\therefore v_B=7\times 4\sqrt{\frac{2}{33}}=7.00\text{m/sec}$