Two particles A and B are connected by rigid | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$L ^{2}=x^{2}+y^{2}$

On differentiating we get,

$0=2 x \frac{ dx }{ dt }+2 y \frac{ dy }{ dt }$

The Velocity of $A =\frac{ dx }{ dt }=-10 m /

Vedclass · Accepted Answer

$5.8$

Vedclass · Answer

$L ^{2}=x^{2}+y^{2}$

On differentiating we get,

$0=2 x \frac{ dx }{ dt }+2 y \frac{ dy }{ dt }$

The Velocity of $A =\frac{ dx }{ dt }=-10 m / s$

The velocity of particle $B=\frac{ d y}{ dt }$

$\frac{ d y}{ dt }=\frac{- x }{ y } \frac{ dx }{ dt }=\cot 60^{0} 	imes 10=\frac{1}{\sqrt{3}} 	imes 10=5.8 m / s$

Two particles $A$ and $B$ are connected by rigid rod $A B$. The rod slides along perpendicular rails as shown here. The velocity of $A$ to the left is $10\; m / s$. What is the velocity of $B$(in $m/s$) when angle $\alpha=60^{\circ}$?

Similar Questions

In the figure shown the velocity of different blocks is shown. The velocity of $C$ is ......... $m/s$

For the given diagram when block $B$ is pulled with velocity $V$ then velocity of block $A$ will be :-

If pulleys shown in the diagram are smooth and massless and $a_1$ and $a_2$ are acceleration of blocks of mass $4 \,kg$ and $8 \,kg$ respectively, then

Two equal masses $A$ and $B$ are arranged as shown in the figure. Pulley and string are ideal and there is no friction. Block $A$ has a speed $u$ in the downward direction. The speed of the block $B$ is :-

In the given figure acceleration of wedge $'A'$ is $10\ m/s^2$ along the inclined plane. (There is no friction between $A$ $\&$ $B$ and $A$ $\&$ fixed inclined plane.) Then acceleration of block $'B'$ will ............ $m/s^2$