Question

A rod of length ${}^{\prime }{l}^{\prime }$ slides down along the inclined wall as shown in figure. At the instant shown in figure the speed of end $A$ is ${}^{\prime }{v}^{\prime }$, then the speed of $B$ will be:

• A. $\frac{v\sin \alpha }{\sin \beta }$
• B. $\frac{v\sin \beta }{\sin \alpha }$
• C. $\frac{v\cos \alpha }{\cos \beta }$
• D. $\frac{v\cos \beta }{\cos \alpha }$

Answer 1

The correct option is A $\frac{v\sin \alpha }{\sin \beta }$

velocity of point A is V and that of B is v' as there is no external torque, center of man (c) will follow only linear downward trajectory. Applying conservation of momentum about c,

$v\frac{l}{2}\sin \alpha =v^{\prime }\frac{l}{2}\sin \beta$

$\Rightarrow v^{\prime }=\frac{v\sin \alpha }{\sin \beta }$