Question

A rigid (AB) of length $L=\sqrt{2}$ is undergoing translational as well as rotational motion in the x-y plane (see the figure). The point A has the velocity $V_1=i+2jm/s$. The end B is contrained to move only along the x-direction . The magnitude of velocity $V_2$(in m/s) at the end B is.

Answer 1

Given, $V_1-\hat{i}+2\hat{j}$ m/s
$V_1=V_{x_1}\hat{i}+V_{y_1}\hat{j}$
So, $V_{x_1}=1m/sand{V}_{y_1}=2$ m/s
$tan(\theta +{45}^o)=\frac{V_{y_1}}{V_{x_1}}=\frac{2}{1}$
$\theta +{45}^o={63.435}^o={18.435}^o$
$\theta ={18.435}^o$
According to geometry,
$\theta +{45}^o+\alpha ={90}^o$
$\alpha ={90}^o-{45}^o-\theta$
$={26.565}^o$
In $\Delta ABO$
If $AB=\sqrt{2}m,thenAO=BO=1m$
in $\Delta OA{I}_{A{B}^{\prime }}$
$cos\alpha =\frac{AO}{A{I}_{AB}}\Rightarrow A{I}_{AB}$
$=\frac{AO}{cos26.565}=\frac{1}{cos26.565}$
$tan\alpha =\frac{O{I}_{AB}}{AO}$
$O{I}_{AB}=\left(tan\alpha \right)\left(AO\right)=1\times tan26.565$
$=0.4999\underset{–}{\sim }0.5$ m
So, $B{I}_{AB}=BO+O{I}_{AB}=1+0.5=1.5m$
So, for velocity of rod, $V_2$
$\frac{V_1}{A{I}_{AB}}=\frac{V_2}{B{I}_{AB}}$
$\sqrt{\frac{1+4}{1.118}}=\frac{V_2}{1.5}$
$V_2=\frac{\left(\sqrt{5}\right)\times 1.5}{1.118}=3$ m/s