To solve the problem, we need to find the rate of change of angular momentum of a particle of mass 2 kg released from point A(2, 0, 0) m at time t = 2 seconds. The acceleration due to gravity is given as \( g = -10 \hat{j} \, \text{m/s}^2 \).
### Step-by-Step Solution:
1. **Identify the position vector at time t = 2 s:**
The particle is released from point A at \( (2, 0, 0) \) m. The only force acting on the particle is gravity, which acts downward. The position of the particle at time \( t = 2 \) seconds can be calculated using the equation of motion:
\[
\vec{r}(t) = \vec{r}_0 + \vec{v}_0 t + \frac{1}{2} \vec{a} t^2
\]
Here, \( \vec{r}_0 = (2, 0, 0) \), \( \vec{v}_0 = (0, 0, 0) \) (initially at rest), and \( \vec{a} = (0, -10, 0) \).
\[
\vec{r}(2) = (2, 0, 0) + (0, 0, 0) \cdot 2 + \frac{1}{2} (0, -10, 0) \cdot (2^2) = (2, 0, 0) + (0, -20, 0) = (2, -20, 0)
\]
2. **Calculate the velocity of the particle at t = 2 s:**
The velocity of the particle can be calculated using:
\[
\vec{v}(t) = \vec{v}_0 + \vec{a} t
\]
Substituting the values:
\[
\vec{v}(2) = (0, 0, 0) + (0, -10, 0) \cdot 2 = (0, -20, 0)
\]
3. **Calculate the angular momentum \( \vec{L} \):**
The angular momentum \( \vec{L} \) about the origin is given by:
\[
\vec{L} = \vec{r} \times m\vec{v}
\]
Here, \( m = 2 \, \text{kg} \) and \( \vec{r} = (2, -20, 0) \), \( \vec{v} = (0, -20, 0) \).
\[
\vec{L} = (2, -20, 0) \times (0, -20, 0)
\]
Using the determinant to compute the cross product:
\[
\vec{L} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
2 & -20 & 0 \\
0 & -20 & 0
\end{vmatrix} = \hat{i}(0) - \hat{j}(0) + \hat{k}(-40) = -40 \hat{k} \, \text{kg m}^2/\text{s}
\]
4. **Calculate the torque \( \tau \):**
The torque \( \tau \) about the origin due to the gravitational force \( \vec{F} = m\vec{g} = 2(-10 \hat{j}) = (0, -20, 0) \):
\[
\tau = \vec{r} \times \vec{F}
\]
\[
\tau = (2, -20, 0) \times (0, -20, 0)
\]
Using the determinant:
\[
\tau = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
2 & -20 & 0 \\
0 & -20 & 0
\end{vmatrix} = \hat{i}(0) - \hat{j}(0) + \hat{k}(-40) = -40 \hat{k} \, \text{N m}
\]
5. **Rate of change of angular momentum:**
The rate of change of angular momentum is equal to the torque:
\[
\frac{d\vec{L}}{dt} = \tau
\]
Thus, the rate of change of angular momentum about the origin at \( t = 2 \) seconds is:
\[
\frac{d\vec{L}}{dt} = -40 \hat{k} \, \text{kg m}^2/\text{s}^2
\]
### Final Answer:
The rate of change of angular momentum of the particle about the origin at time \( t = 2 \) s is \( -40 \hat{k} \, \text{kg m}^2/\text{s}^2 \).
