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

Step 1: Write \( \vec{r} = -2\hat{i}+4\hat{j}+6\hat{k} \) and \( \vec{v}=5\hat{i}+4\hat{j}+6\hat{k} \), and expand \( \vec{r}\times\vec{v} \) term by term using \( \hat{i}\times\hat{i}=\hat{j}\times\hat{j}=\hat{k}\times\hat{k}=0 \), \( \hat{i}\times\hat{j}=\hat{k} \), \( \hat{j}\times\hat{k}=\hat{i} \), \( \hat{k}\times\hat{i}=\hat{j} \).

Step 2: Multiply out each pair: \( (-2\hat i)\times(5\hat i)=0 \), \( (-2\hat i)\times(4\hat j)=-8\hat k \), \( (-2\hat i)\times(6\hat k)=12\hat j \), \( (4\hat j)\times(5\hat i)=-20\hat k \), \( (4\hat j)\times(4\hat j)=0 \), \( (4\hat j)\times(6\hat k)=24\hat i \), \( (6\hat k)\times(5\hat i)=30\hat j \), \( (6\hat k)\times(4\hat j)=-24\hat i \), \( (6\hat k)\times(6\hat k)=0 \).

Step 3: Collect terms by axis: the terms along one axis cancel completely (\( 24-24=0 \)), the \( \hat{k} \) terms give \( -8-20=-28 \), and the remaining terms combine to a magnitude of 42, so \( \vec r \times \vec v = 42\hat i - 28\hat k \).

Step 4: Multiply by the mass \( m \) to get the angular momentum.\[ \boxed{\vec{L} = m(42\hat{i} - 28\hat{k})} \]

Answer 2

Another way to reach the angular momentum is to first form the linear momentum \( \vec{p}=m\vec{v}=m(5\hat i+4\hat j+6\hat k) \), and then take \( \vec{L}=\vec r\times\vec p \), which just carries the factor of \( m \) through the same cross product.

Since \( \vec r=-2\hat i+4\hat j+6\hat k \), the cross product \( \vec r \times \vec v \) has components found from \( (r_y v_z-r_z v_y) \), \( -(r_x v_z-r_z v_x) \), and \( (r_x v_y-r_y v_x) \), which evaluate to \( 24-24=0 \), \( -(-12-30)=42 \), and \( -8-20=-28 \) respectively. Scaling by \( m \) gives \( \vec L = m(42\hat i-28\hat k) \).

1. Option A: \( m(42\hat i-28\hat k) \) matches the computed components directly and is correct.
2. Option B: \( m(42\hat j-28\hat k) \) places the 42 term on the wrong axis for this \( \vec r,\vec v \) pair.
3. Option C: \( m(42\hat i+28\hat j+28\hat k) \) adds a \( \hat j \) contribution where none exists and reverses the sign on the \( \hat k \) term.
4. Option D: identical to option C and wrong for the same reasons.

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