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 \( \vec{L} \) of a particle about the origin is given by the cross product of the position vector \( \vec{r} \) and the linear momentum \( \vec{p} \):

\[\vec{L} = \vec{r} \times \vec{p}\]

The linear momentum \( \vec{p} \) is the product of mass \( m \) and velocity \( \vec{v} \):

\[\vec{p} = m\vec{v}\]

Given \( \vec{v} = 5 \hat{i} + 4 \hat{j} + 6 \hat{k} \) and \( \vec{r} = -2 \hat{i} + 4 \hat{j} + 6 \hat{k} \),

\(\vec{p} = m(5 \hat{i} + 4 \hat{j} + 6 \hat{k})\)

The cross product \( \vec{r} \times \vec{v} \) is calculated as follows:

\[\vec{r} \times \vec{v} = \left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\-2 & 4 & 6 \\5 & 4 & 6\end{array}\right|\]

Expanding the determinant, we get:

\[\vec{r} \times \vec{v} = \hat{i}(4 \cdot 6 - 6 \cdot 4) - \hat{j}(-2 \cdot 6 - 6 \cdot 5) + \hat{k}(-2 \cdot 4 - 4 \cdot 5)\]

Simplifying further:

\[\vec{r} \times \vec{v} = \hat{i}(24 - 24) - \hat{j}(-12 - 30) + \hat{k}(-8 - 20)\]

\[\vec{r} \times \vec{v} = \hat{i}(0) + \hat{j}(42) + \hat{k}(-28)\]

\[\vec{r} \times \vec{v} = 42 \hat{j} - 28 \hat{k}\]

Multiplying by \( m \), the angular momentum is:

\[\vec{L} = m(42 \hat{j} - 28 \hat{k})\]

Therefore, the correct answer is:

\( m(42 \hat{i} - 28 \hat{k}) \)