Question

The velocity of a particular mass \( m \) is \( \vec{v} = 5 \hat{i} + 4 \hat{j} + 6 \hat{k} \) when at \( \vec{r} = -2 \hat{i} + 4 \hat{j} + 6 \hat{k} \). The angular momentum of the particle about the origin is:

• A. \( m(42 \hat{i} - 28 \hat{k}) \)
• B. \( m(42 \hat{j} - 28 \hat{k}) \)
• C. \( m(42 \hat{i} + 28 \hat{j} + 28 \hat{k}) \)
• D. \( m(42 \hat{i} + 28 \hat{j} + 28 \hat{k}) \)

Hint:

The angular momentum is given by the cross product of the position vector and momentum vector. It represents the rotational effect of the object's motion.

Answer 1

The Correct Option is A

The angular momentum \( \vec{L} \) about the origin is given by: \[ \vec{L} = \vec{r} \times \vec{p} = \vec{r} \times m \vec{v} \] Substitute the given values: \[ \vec{r} = -2 \hat{i} + 4 \hat{j} + 6 \hat{k}, \quad \vec{v} = 5 \hat{i} + 4 \hat{j} + 6 \hat{k} \] The cross product \( \vec{r} \times \vec{v} \) is computed as: \[ \vec{L} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
-2 & 4 & 6
5 & 4 & 6 \end{vmatrix} \] Expanding the determinant: \[ \vec{L} = \hat{i} \begin{vmatrix} 4 & 6
4 & 6 \end{vmatrix} - \hat{j} \begin{vmatrix} -2 & 6
5 & 6 \end{vmatrix} + \hat{k} \begin{vmatrix} -2 & 4
5 & 4 \end{vmatrix} \] This simplifies to: \[ \vec{L} = \hat{i}(0) - \hat{j}(-12) + \hat{k}(42) = 12 \hat{j} + 42 \hat{k} \] Thus, the angular momentum is \( \vec{L} = m(42 \hat{i} - 28 \hat{k}) \).