Question

In the figure shown $v=2\text{m/s}$, $\omega =5\text{rad/s}$ and $\text{CP}=1\text{m}$.


In terms of $\hat{i}$ and $\hat{j}$ , find linear velocity of point $\text{P}$.

• A. $4\hat{i}-5\hat{j}$
  
• B. $5\hat{i}-4\hat{j}$
• C. $-5\hat{i}-4\hat{j}$
• D. $5\hat{i}-5\hat{j}$

Answer 1

The correct option is B $5\hat{i}-4\hat{j}$

Given,
$v=2\text{m/s}$, $\omega =5\text{rad/s}$ and $\text{CP}=1\text{m}$

For particle $\text{P}$,


$r=CP=1\text{m}$
So,
$r\omega =\left(1\right)\left(5\right)=5\text{m/s}$

Net velocity of particle $\text{P}$ is the vector sum of $v$ and $r\omega$ as shown in figure.

$V_P=2\hat{i}+(5\cos {53}^{\circ }\hat{i}-5\sin {53}^{\circ }\hat{j})$

$\Rightarrow V_P=2\hat{i}+3\hat{i}-4\hat{j}$

$\therefore V_P=(5\hat{i}-4\hat{j})\text{m/s}$

Hence, option (b) is the correct answer.