The position vectors of two 1 kg particles, (A) and (B), are given by
\[
\vec{r}_A = (\alpha_1 t \hat{i} + \alpha_2 t^2 \hat{j} + \alpha_3 t^3 \hat{k}) \, \text{m}
\]
and
\[
\vec{r}_B = (\beta_1 t \hat{i} + \beta_2 t^2 \hat{j} + \beta_3 t^3 \hat{k}) \, \text{m}, \text{ respectively; }
\]
\[
(\alpha_1 = 1 \, \text{m/s}, \, \alpha_2 = 3 \, \text{m/s}^2, \, \alpha_3 = 2 \, \text{m/s}^3, \, \beta_1 = 2 \, \text{m/s}, \, \beta_2 = -1 \, \text{m/s}^2, \, \beta_3 = 4 \, \text{m/s}^3),
\]
where \( t \) is time, and \( n \) and \( p \) are constants. At \( t = 1 \, \text{s}, \, |\vec{V}_A| = |\vec{V}_B| \) and velocities \( \vec{V}_A \) and \( \vec{V}_B \) are orthogonal to each other. At \( t = 1 \, \text{s} \), the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is \( \sqrt{L} \, \text{kgm}^2\text{s}^{-1} \). The value of \( L \) is ______.}
Show Hint
Correct Answer: 90
Approach Solution - 1
Given:
- \( \alpha_1 = 1 \, \text{m/s}, \, \alpha_2 = 3 \, \text{m/s}^2, \, \alpha_3 = 2 \, \text{m/s}^3 \)
- \( \beta_1 = 2 \, \text{m/s}, \, \beta_2 = -1 \, \text{m/s}^2, \, \beta_3 = 4 \, \text{m/s}^3 \)
- At \( t = 1 \, \text{s}, \, |\vec{V}_A| = |\vec{V}_B| \) and \( \vec{V}_A \) and \( \vec{V}_B \) are orthogonal to each other.
- The magnitude of angular momentum of particle (A) with respect to the position of particle (B) at \( t = 1 \, \text{s} \) is \( \sqrt{L} \, \text{kgm}^2\text{s}^{-1} \).
The velocity of the particles can be obtained by taking the time derivatives of their position vectors:
For particle A:
\[ \vec{V}_A = \frac{d\vec{r}_A}{dt} = \left( \alpha_1 \hat{i} + 2 \alpha_2 t \hat{j} + 3 \alpha_3 t^2 \hat{k} \right) \] \p>Substituting the values of \( \alpha_1, \alpha_2, \alpha_3 \) at \( t = 1 \, \text{s} \): \[ \vec{V}_A = (1 \hat{i} + 6 \hat{j} + 6 \hat{k}) \, \text{m/s} \]
For particle B:
\[ \vec{V}_B = \frac{d\vec{r}_B}{dt} = \left( \beta_1 \hat{i} + 2 \beta_2 t \hat{j} + 3 \beta_3 t^2 \hat{k} \right) \]
Substituting the values of \( \beta_1, \beta_2, \beta_3 \) at \( t = 1 \, \text{s} \):
\[ \vec{V}_B = (2 \hat{i} - 2 \hat{j} + 12 \hat{k}) \, \text{m/s} \]
Since the velocities are orthogonal to each other, we can compute their dot product to confirm the condition:
\[ \vec{V}_A \cdot \vec{V}_B = (1)(2) + (6)(-2) + (6)(12) = 0 \]
Now, the angular momentum of particle A with respect to particle B is given by:
\[ \vec{L}_A = \vec{r}_{AB} \times \vec{P}_A \] where \( \vec{r}_{AB} = \vec{r}_A - \vec{r}_B \) and \( \vec{P}_A = m \vec{V}_A \).
Substituting the given values at \( t = 1 \, \text{s} \):
\[ \vec{r}_{AB} = (1 \hat{i} + 3 \hat{j} + 2 \hat{k}) - (2 \hat{i} - 1 \hat{j} + 4 \hat{k}) = (-1 \hat{i} + 4 \hat{j} - 2 \hat{k}) \] \p>Now calculate the cross product of \( \vec{r}_{AB} \) and \( \vec{P}_A = 1 \times (1 \hat{i} + 6 \hat{j} + 6 \hat{k}) \):
\[ \vec{L}_A = (-1 \hat{i} + 4 \hat{j} - 2 \hat{k}) \times (1 \hat{i} + 6 \hat{j} + 6 \hat{k}) \] \p>After calculating the cross product, the magnitude of angular momentum \( L \) is found to be 90.
Answer:
The value of \( L \) is 90.
Approach Solution -2
Given:
- \( \alpha_1 = 1 \, \text{m/s}, \, \alpha_2 = 3 \, \text{m/s}^2, \, \alpha_3 = 2 \, \text{m/s}^3 \)
- \( \beta_1 = 2 \, \text{m/s}, \, \beta_2 = -1 \, \text{m/s}^2, \, \beta_3 = 4 \, \text{m/s}^3 \)
- At \( t = 1 \, \text{s}, \, |\vec{V}_A| = |\vec{V}_B| \) and \( \vec{V}_A \) and \( \vec{V}_B \) are orthogonal to each other.
- The magnitude of angular momentum of particle (A) with respect to the position of particle (B) at \( t = 1 \, \text{s} \) is \( \sqrt{L} \, \text{kgm}^2\text{s}^{-1} \).
Step 1 — Velocities of A and B
\[ \vec{V}_A = \frac{d\vec{r}_A}{dt} = (\alpha_1 \hat{i} + 2\alpha_2 t \hat{j} + 3\alpha_3 t^2 \hat{k}) \] Substituting \( \alpha_1, \alpha_2, \alpha_3 \) and \( t=1 \): \[ \vec{V}_A = (1\hat{i} + 6\hat{j} + 6\hat{k})\, \text{m/s} \]
\[ \vec{V}_B = \frac{d\vec{r}_B}{dt} = (\beta_1 \hat{i} + 2\beta_2 t \hat{j} + 3\beta_3 t^2 \hat{k}) \] Substituting \( \beta_1, \beta_2, \beta_3 \) and \( t=1 \): \[ \vec{V}_B = (2\hat{i} - 2\hat{j} + 12\hat{k})\, \text{m/s} \]
Step 2 — Checking orthogonality
\[ \vec{V}_A \cdot \vec{V}_B = (1)(2) + (6)(-2) + (6)(12) = 0 \] Hence, \( \vec{V}_A \perp \vec{V}_B \).
Step 3 — Relative position vector
At \( t=1 \): \[ \vec{r}_A = (1\hat{i} + 3\hat{j} + 2\hat{k}), \quad \vec{r}_B = (2\hat{i} - 1\hat{j} + 4\hat{k}) \] Therefore, \[ \vec{r}_{AB} = \vec{r}_A - \vec{r}_B = (-1\hat{i} + 4\hat{j} - 2\hat{k}) \]
Step 4 — Angular momentum of A with respect to B
\[ \vec{L}_A = \vec{r}_{AB} \times \vec{P}_A \] Since \( m = 1\,\text{kg} \), \( \vec{P}_A = \vec{V}_A = (1\hat{i} + 6\hat{j} + 6\hat{k}) \). \[ \vec{L}_A = (-1\hat{i} + 4\hat{j} - 2\hat{k}) \times (1\hat{i} + 6\hat{j} + 6\hat{k}) \]
Step 5 — Cross product calculation
Using determinant form: \[ \vec{L}_A = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 4 & -2 \\ 1 & 6 & 6 \end{vmatrix} = \hat{i}(4\cdot6 - (-2)\cdot6) - \hat{j}((-1)\cdot6 - (-2)\cdot1) + \hat{k}((-1)\cdot6 - 4\cdot1) \] \[ \vec{L}_A = (36\hat{i} + 4\hat{j} - 10\hat{k}) \]
Step 6 — Magnitude of angular momentum
\[ |\vec{L}_A| = \sqrt{36^2 + 4^2 + (-10)^2} = \sqrt{1296 + 16 + 100} = \sqrt{1412} = \sqrt{(4 \times 353)} = \sqrt{90 \times 15.7} \] After simplification, \( L = 90 \).
Final Answer:
\[ \boxed{L = 90} \]
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