A particle of mass 2 kg is moving such that at time t, its position, in meter, is given by `vecr(t)=5hati-2t^(2)hatj`. The angular momentum of the particle at t = 2s about the origin in kg `m^(-2)s^(-1)` is :

Options

A. `-80hatk`
B. `(10hati-16hatj)`
C. `-40hatk`
D. `40hatk`

Explanation

To solve the problem, we need to find the angular momentum of a particle with mass \( m = 2 \, \text{kg} \) at time \( t = 2 \, \text{s} \). The position vector of the particle is given by:

\[ \vec{r}(t) = 5 \hat{i} - 2t^2 \hat{j} \]

### Step 1: Find the velocity vector
The velocity vector \( \vec{v}(t) \) is the time derivative of the position vector \( \vec{r}(t) \).

\[ \vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}(5 \hat{i} - 2t^2 \hat{j}) = 0 \hat{i} - 4t \hat{j} = -4t \hat{j} \]

### Step 2: Calculate the velocity at \( t = 2 \, \text{s} \)
Substituting \( t = 2 \) into the velocity equation:

\[ \vec{v}(2) = -4(2) \hat{j} = -8 \hat{j} \, \text{m/s} \]

### Step 3: Find the position vector at \( t = 2 \, \text{s} \)
Now, substitute \( t = 2 \) into the position vector equation:

\[ \vec{r}(2) = 5 \hat{i} - 2(2^2) \hat{j} = 5 \hat{i} - 8 \hat{j} \]

### Step 4: Calculate the angular momentum
The angular momentum \( \vec{L} \) about the origin is given by:

\[ \vec{L} = m \vec{r} \times \vec{v} \]

Substituting the values we have:

\[ \vec{L} = 2 \, \text{kg} \cdot (5 \hat{i} - 8 \hat{j}) \times (-8 \hat{j}) \]

### Step 5: Compute the cross product
Using the properties of the cross product:

\[ \vec{L} = 2 \cdot (5 \hat{i} \times -8 \hat{j} + (-8 \hat{j}) \times -8 \hat{j}) \]

Since \( \hat{j} \times \hat{j} = 0 \):

\[ \vec{L} = 2 \cdot (5 \cdot -8)(\hat{i} \times \hat{j}) = 2 \cdot -40 \hat{k} = -80 \hat{k} \, \text{kg m}^2/\text{s} \]

### Final Answer
The angular momentum of the particle at \( t = 2 \, \text{s} \) about the origin is:

\[ \vec{L} = -80 \hat{k} \, \text{kg m}^2/\text{s} \]