Question

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

The magnitude of the velocity $V_2$ (in m/s) at the end B is

Answer 1

Method I:

$\sqrt{2}\sin {45}^{\circ }=1m$
$\tan \theta =\frac{1}{2}$
$V_A=i+2j=\sqrt{1+2^2}=\sqrt{5}m/s$
$\theta ={\tan }^{-1}0.5={26.5651}^{\circ }$
$\alpha ={90}^{\circ }-({45}^{\circ }+\theta )={18.4349}^{\circ }$
$\beta ={90}^{\circ }-({45}^{\circ }+\alpha )={26.5651}^{\circ }$
$\frac{V_A}{IA}=\frac{V_B}{IB}$
$V_B=\frac{V_A.IB}{IA}=\frac{\sqrt{5}.IB}{IA}$
$\cos \beta =\frac{1}{IA}$
$IA=\frac{1}{\cos 26.5651}=1.118m.$
$IC=\sqrt{(IA{)}^2-1^2}=\sqrt{(1.118{)}^2-1}$
$=0.5m$
$IB=(1+0.5)m$
$\frac{V_A}{IA}=\frac{V_B}{IB}$
$V_2=V_B=\frac{\sqrt{3}\times [1+0.5]}{1.118}=3m/s$
Method II:
As AB is rigid rod, velocity of point A and B along AB will be same.
Let ${\overrightarrow{V}}_B\text{be the velocity of B then}{\overrightarrow{V}}_B={\overrightarrow{V}}_B\hat{i}$
${\overrightarrow{V}}_A=\hat{i}+2\hat{j}$
Unit vector along,
$AB,\overrightarrow{u}=\frac{i+j}{\sqrt{2}}$
Now ${\overrightarrow{V}}_A.\overrightarrow{u}={\overrightarrow{V}}_B.\overrightarrow{u}$
$\Rightarrow (i+2j).\left(\frac{i+j}{\sqrt{2}}\right)=\left(V_B\hat{i}\right).\left(\frac{i+j}{\sqrt{2}}\right)$
$\Rightarrow \frac{3}{\sqrt{2}}=\frac{V_B}{\sqrt{2}}$
$V_B=3m/s$