A particle of mass 0.2 kg moves along a path given by the relation : `vec(r)=2cos omegat hat(i)+3 sin omegat hat(j)`. Then the torque on the particle about the origin is :

Options

A. `sqrt(13) hat(k) Nm`
B. `sqrt((2)/(3)) hat (k) Nm`
C. `sqrt((3)/(2)) hat(k) Nm`
D. `0 hat(k)`

Explanation

To find the torque on the particle about the origin, we will follow these steps:

### Step 1: Identify the position vector
The position vector of the particle is given by:
\[ \vec{r} = 2 \cos(\omega t) \hat{i} + 3 \sin(\omega t) \hat{j} \]

### Step 2: Differentiate the position vector to find the velocity
To find the velocity, we differentiate the position vector with respect to time \( t \):
\[ \vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}(2 \cos(\omega t) \hat{i} + 3 \sin(\omega t) \hat{j}) \]
Using the chain rule:
\[ \vec{v} = -2\omega \sin(\omega t) \hat{i} + 3\omega \cos(\omega t) \hat{j} \]

### Step 3: Differentiate the velocity vector to find the acceleration
Next, we differentiate the velocity vector to find the acceleration:
\[ \vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}(-2\omega \sin(\omega t) \hat{i} + 3\omega \cos(\omega t) \hat{j}) \]
Using the chain rule again:
\[ \vec{a} = -2\omega^2 \cos(\omega t) \hat{i} - 3\omega^2 \sin(\omega t) \hat{j} \]

### Step 4: Calculate the force vector
The force on the particle can be calculated using Newton's second law \( \vec{F} = m \vec{a} \):
\[ \vec{F} = m \vec{a} = 0.2 \, \text{kg} \cdot (-2\omega^2 \cos(\omega t) \hat{i} - 3\omega^2 \sin(\omega t) \hat{j}) \]
\[ \vec{F} = -0.4\omega^2 \cos(\omega t) \hat{i} - 0.6\omega^2 \sin(\omega t) \hat{j} \]

### Step 5: Calculate the torque vector
The torque \( \vec{\tau} \) about the origin is given by the cross product of the position vector \( \vec{r} \) and the force vector \( \vec{F} \):
\[ \vec{\tau} = \vec{r} \times \vec{F} \]
Substituting the expressions we have:
\[ \vec{\tau} = (2 \cos(\omega t) \hat{i} + 3 \sin(\omega t) \hat{j}) \times (-0.4\omega^2 \cos(\omega t) \hat{i} - 0.6\omega^2 \sin(\omega t) \hat{j}) \]

### Step 6: Evaluate the cross product
Since the vectors \( \hat{i} \) and \( \hat{j} \) are perpendicular, the cross product of any vector with itself is zero:
\[ \vec{\tau} = \vec{r} \times \vec{F} = 0 \]

### Conclusion
The torque on the particle about the origin is:
\[ \vec{\tau} = 0 \]