Step 1: Condition for angular momentum conservation
- Angular momentum \( \mathbf{L} \) is conserved if the net torque \( \mathbf{\tau} \) is zero.
- Torque is given by:
\[
\mathbf{\tau} = \mathbf{r} \times \mathbf{F}
\]
Step 2: Compute the cross-product
\[
\mathbf{\tau} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k}
2 & -6 & -12
\alpha & 3 & 6
\end{vmatrix}
\]
Expanding along the first row:
\[
\mathbf{\tau} = \hat{i} \left( (-6)(6) - (-12)(3) \right)
- \hat{j} \left( (2)(6) - (-12)(\alpha) \right)
+ \hat{k} \left( (2)(3) - (-6)(\alpha) \right)
\]
Step 3: Solve for \( \alpha \)
\[
\mathbf{\tau} = \hat{i} (-36 + 36) - \hat{j} (12 + 12\alpha) + \hat{k} (6 + 6\alpha)
\]
\[
\mathbf{\tau} = -\hat{j} (12 + 12\alpha) + \hat{k} (6 + 6\alpha)
\]
For \( \mathbf{\tau} = 0 \), the coefficients of \( \hat{j} \) and \( \hat{k} \) must be zero:
\[
12 + 12\alpha = 0 \quad \Rightarrow \quad \alpha = -1
\]