The position vector of a particle of mass m= 6kg is given as `vec(r)=[(3t^(2)-6t) hat(i)+(-4t^(3)) hat(j)] m`. Find:
(i) The force `(vec(F)=mvec(a))` acting on the particle.
(ii) The torque `(vec(tau)=vec(r)xxvec(F))` with respect to the origin, acting on the particle.
(iii) The momentum `(vec(p)=mvec(v))` of the particle.
(iv) The angular momentum `(vec(L)=vec(r)xxvec(p))` of the particle with respect to the origin.

Explanation

To solve the problem step by step, we will follow the instructions provided in the question:

### Given:
The position vector of the particle is given as:
\[ \vec{r} = (3t^2 - 6t) \hat{i} + (-4t^3) \hat{j} \text{ m} \]
The mass of the particle is:
\[ m = 6 \text{ kg} \]

### (i) Find the force \(\vec{F} = m\vec{a}\)

1. **Calculate the velocity \(\vec{v}\)**:
\[ \vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}[(3t^2 - 6t) \hat{i} + (-4t^3) \hat{j}] \]
Differentiate each component:
\[ \vec{v} = (6t - 6) \hat{i} + (-12t^2) \hat{j} \text{ m/s} \]

2. **Calculate the acceleration \(\vec{a}\)**:
\[ \vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}[(6t - 6) \hat{i} + (-12t^2) \hat{j}] \]
Differentiate each component:
\[ \vec{a} = 6 \hat{i} + (-24t) \hat{j} \text{ m/s}^2 \]

3. **Calculate the force \(\vec{F}\)**:
\[ \vec{F} = m\vec{a} = 6 \cdot (6 \hat{i} - 24t \hat{j}) = 36 \hat{i} - 144t \hat{j} \text{ N} \]

### (ii) Find the torque \(\vec{\tau} = \vec{r} \times \vec{F}\)

1. **Substitute \(\vec{r}\) and \(\vec{F}\)**:
\[ \vec{\tau} = \left[(3t^2 - 6t) \hat{i} + (-4t^3) \hat{j}\right] \times \left[36 \hat{i} - 144t \hat{j}\right] \]

2. **Calculate the cross product**:
Using the determinant method for cross products:
\[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3t^2 - 6t & -4t^3 & 0 \\ 36 & -144t & 0 \end{vmatrix} \]
This gives:
\[ \vec{\tau} = \hat{k} \left[(3t^2 - 6t)(-144t) - (-4t^3)(36)\right] \]
Simplifying:
\[ \vec{\tau} = \hat{k} \left[-432t^3 + 864t^2 + 144t^4\right] \]

### (iii) Find the momentum \(\vec{p} = m\vec{v}\)

1. **Calculate the momentum**:
\[ \vec{p} = m\vec{v} = 6 \cdot \left[(6t - 6) \hat{i} + (-12t^2) \hat{j}\right] \]
Simplifying:
\[ \vec{p} = (36t - 36) \hat{i} - 72t^2 \hat{j} \text{ kg m/s} \]

### (iv) Find the angular momentum \(\vec{L} = \vec{r} \times \vec{p}\)

1. **Substitute \(\vec{r}\) and \(\vec{p}\)**:
\[ \vec{L} = \left[(3t^2 - 6t) \hat{i} + (-4t^3) \hat{j}\right] \times \left[(36t - 36) \hat{i} - 72t^2 \hat{j}\right] \]

2. **Calculate the cross product**:
Using the determinant method for cross products:
\[ \vec{L} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3t^2 - 6t & -4t^3 & 0 \\ 36t - 36 & -72t^2 & 0 \end{vmatrix} \]
This gives:
\[ \vec{L} = \hat{k} \left[(3t^2 - 6t)(-72t^2) - (-4t^3)(36t - 36)\right] \]
Simplifying:
\[ \vec{L} = \hat{k} \left[-216t^4 + 432t^3 + 144t^4\right] \]
Final result:
\[ \vec{L} = \hat{k} \left[-72t^4 + 288t^3\right] \text{ kg m}^2/\text{s} \]

### Summary of Results:
1. Force: \(\vec{F} = 36 \hat{i} - 144t \hat{j} \text{ N}\)
2. Torque: \(\vec{\tau} = \hat{k} \left[-432t^3 + 864t^2 + 144t^4\right] \text{ N m}\)
3. Momentum: \(\vec{p} = (36t - 36) \hat{i} - 72t^2 \hat{j} \text{ kg m/s}\)
4. Angular Momentum: \(\vec{L} = \hat{k} \left[-72t^4 + 288t^3\right] \text{ kg m}^2/\text{s}\)