Question

A force $\mathbf{F} = \alpha \hat{i} + 3 \hat{j} + 6 \hat{k}$ is acting at a point $\mathbf{r} = 2 \hat{i} - 6 \hat{j} - 12 \hat{k}$. The value of $\alpha$ for which angular momentum about the origin is conserved is:

• A. $2$
• B. $0$
• C. $1$
• D. $-1$

Hint:

For angular momentum conservation:
- Torque must be zero: $\mathbf{\tau} = \mathbf{r} \times \mathbf{F} = 0$.
- If force acts along position vector, no torque is produced.

Answer 1

The Correct Option is D

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:
$$12 + 12\alpha = 0 \quad \Rightarrow \quad \alpha = -1$$