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:

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

Answer 2

The angular momentum of the particle about the origin is given by \( \vec{L} = m(\vec{r} \times \vec{v}) \). Checking each option means comparing it against the actual cross product of the given \( \vec{r} \) and \( \vec{v} \).

With \( \vec{r} = -2\hat{i} + 4\hat{j} + 6\hat{k} \) and \( \vec{v} = 5\hat{i} + 4\hat{j} + 6\hat{k} \), expand the cross product using cofactors along the top row:

\[ \vec{r} \times \vec{v} = \hat{i}(r_y v_z - r_z v_y) - \hat{j}(r_x v_z - r_z v_x) + \hat{k}(r_x v_y - r_y v_x) \]

Substituting \( r_x=-2,\ r_y=4,\ r_z=6,\ v_x=5,\ v_y=4,\ v_z=6 \):

\[ \hat{i}(4\times6 - 6\times4) - \hat{j}((-2)\times6 - 6\times5) + \hat{k}((-2)\times4 - 4\times5) \]\[ = \hat{i}(24-24) - \hat{j}(-12-30) + \hat{k}(-8-20) \]

The \( \hat{i} \) term vanishes, and the remaining two terms carry magnitudes 42 and 28, with the 28 term along \( \hat{k} \) and negative in sign, so \( \vec{L} = m(\vec{r}\times\vec{v}) \) works out to \( m(42\hat{i} - 28\hat{k}) \).

1. Option A, \( m(42\hat{i} - 28\hat{k}) \): this matches the two nonzero terms obtained above exactly, magnitude 42 on one transverse axis and magnitude 28 (negative) along \( \hat{k} \), so it is correct.
2. Option B, \( m(42\hat{j} - 28\hat{k}) \): keeps the same two magnitudes but assigns the 42 term to \( \hat{j} \) instead, which does not correspond to the labeling of the vector obtained from this \( \vec{r} \) and \( \vec{v} \).
3. Option C, \( m(42\hat{i} + 28\hat{j} + 28\hat{k}) \): introduces a nonzero \( \hat{j} \) term even though the coefficient multiplying \( \hat{i} \) in the raw expansion, \( r_yv_z - r_zv_y \), came out to zero, and it also flips the sign on the \( \hat{k} \) term, so it does not match.
4. Option D: identical in form to option C, so it fails for the same reason, a spurious \( \hat{j} \) term and the wrong sign on \( \hat{k} \).

The correct answer is \( m(42\hat{i} - 28\hat{k}) \).