The linear velocity perpendicular to radius vector of a particle moving with angular velocity `omega=2hatK` at position vector `r=2hati+2hatj` is

Options

A. `4(hati-hatj)`
B. `4(hatj-hati)`
C. `4hati`
D. `-4hati`

Answer

To solve the problem, we need to find the linear velocity of a particle moving with a given angular velocity at a specified position vector. Let's break it down step by step.

### Step 1: Identify the given quantities
We have:
- Angular velocity, \(\omega = 2 \hat{k}\)
- Position vector, \(\mathbf{r} = 2 \hat{i} + 2 \hat{j}\)

### Step 2: Use the formula for linear velocity
The relationship between linear velocity \(\mathbf{v}\) and angular velocity \(\omega\) is given by the cross product:
\[ \mathbf{v} = \omega \times \mathbf{r} \]

### Step 3: Substitute the values into the equation
Substituting the values of \(\omega\) and \(\mathbf{r}\):
\[ \mathbf{v} = (2 \hat{k}) \times (2 \hat{i} + 2 \hat{j}) \]

### Step 4: Expand the cross product
Using the distributive property of the cross product:
\[ \mathbf{v} = 2 \hat{k} \times (2 \hat{i}) + 2 \hat{k} \times (2 \hat{j}) \]
This simplifies to:
\[ \mathbf{v} = 4 (\hat{k} \times \hat{i}) + 4 (\hat{k} \times \hat{j}) \]

### Step 5: Calculate the individual cross products
Using the right-hand rule for cross products:
- \(\hat{k} \times \hat{i} = \hat{j}\)
- \(\hat{k} \times \hat{j} = -\hat{i}\)

Substituting these results back into the equation:
\[ \mathbf{v} = 4 \hat{j} - 4 \hat{i} \]

### Step 6: Factor out the common term
We can factor out the common term:
\[ \mathbf{v} = 4 (\hat{j} - \hat{i}) \]

### Final Answer
The linear velocity perpendicular to the radius vector is:
\[ \mathbf{v} = 4 (\hat{j} - \hat{i}) \]