A particle moves in space such that its position vector varies as `vec(r)=2thati+3t^(2)hatj`. If mass of particle is 2 kg then angular momentum of particle about origin at `t=2` sec is

Options

A. `12hatk`
B. `48hatk`
C. `36hatk`
D. `24hatk`

Answer

To solve the problem, we need to find the angular momentum of a particle about the origin at \( t = 2 \) seconds. The position vector of the particle is given as:

\[ \vec{r} = 2t \hat{i} + 3t^2 \hat{j} \]

The mass of the particle is \( m = 2 \) kg.

### Step 1: Find the position vector at \( t = 2 \) seconds

Substituting \( t = 2 \) into the position vector:

\[ \vec{r} = 2(2) \hat{i} + 3(2^2) \hat{j} = 4 \hat{i} + 12 \hat{j} \]

### Step 2: Find the velocity vector

The velocity vector \( \vec{v} \) is the time derivative of the position vector \( \vec{r} \):

\[ \vec{v} = \frac{d\vec{r}}{dt} \]

Differentiating \( \vec{r} \):

\[ \vec{v} = \frac{d}{dt}(2t \hat{i} + 3t^2 \hat{j}) = 2 \hat{i} + 6t \hat{j} \]

Now substituting \( t = 2 \):

\[ \vec{v} = 2 \hat{i} + 6(2) \hat{j} = 2 \hat{i} + 12 \hat{j} \]

### Step 3: Calculate the angular momentum

The angular momentum \( \vec{L} \) of the particle about the origin is given by the formula:

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

Substituting \( m = 2 \) kg, \( \vec{r} = 4 \hat{i} + 12 \hat{j} \), and \( \vec{v} = 2 \hat{i} + 12 \hat{j} \):

\[ \vec{L} = 2 \left( (4 \hat{i} + 12 \hat{j}) \times (2 \hat{i} + 12 \hat{j}) \right) \]

### Step 4: Calculate the cross product

Using the determinant method for the cross product:

\[ \vec{L} = 2 \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 12 & 0 \\ 2 & 12 & 0 \end{vmatrix} \]

Calculating the determinant:

\[ \vec{L} = 2 \left( \hat{k} \left( 4 \cdot 12 - 12 \cdot 2 \right) \right) \]

Calculating the values:

\[ = 2 \left( \hat{k} (48 - 24) \right) = 2 \left( 24 \hat{k} \right) = 48 \hat{k} \]

### Final Answer

Thus, the angular momentum of the particle about the origin at \( t = 2 \) seconds is:

\[ \vec{L} = 48 \hat{k} \, \text{kg m}^2/\text{s} \]

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