Question

Two particles $A$ and $B$ are connected by a rigid rod $AB$The rod slides along perpendicular Is as shown here. The velocity of $A$ to the left is $10\, m/s$. What is the velocity of $B$ when angle $\alpha$ = $ 60^\circ$

• A. 10 m/s
• B. 9.8 m/s
• C. 5.8 m/s
• D. 17.3 m/s

Answer 1

Answer: (C)

$V$ elocity $=10 m / s \alpha=60^{\circ}$
$L ^{2}= x ^{2}+ y ^{2}$
On differentiating we get,
$ 0=2 x \frac{d x}{d t}+2 y \frac{d y}{d t} $
The Velocity of $A =\frac{ dx }{ dt }=-10 m / s$
Thus the velocity of particle $B$ is:
$ \frac{d y}{d t}=\frac{-x}{y} \frac{d x}{d t}=\cot 60^{0} \times 10=\frac{1}{\sqrt{3}} \times 10=5.8 m / s $