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:
\[ 12 + 12\alpha = 0 \quad \Rightarrow \quad \alpha = -1 \]