Question

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 ______.}

Hint:

For angular momentum, use the formula \( \vec{L} = m \, \vec{r} \times \vec{v} \), where \( \vec{r} \) is the position vector and \( \vec{v} \) is the velocity.

Answer 1

Correct Answer: 90

Provided Information:

• \( \alpha_1 = 1 \, \text{m/s}, \, \alpha_2 = 3 \, \text{m/s}^2, \, \alpha_3 = 2 \, \text{m/s}^3 \)
• \( \beta_1 = 2 \, \text{m/s}, \, \beta_2 = -1 \, \text{m/s}^2, \, \beta_3 = 4 \, \text{m/s}^3 \)
• At \( t = 1 \, \text{s} \), the magnitudes of the velocities of particles A and B are equal: \( |\vec{V}_A| = |\vec{V}_B| \).
• The velocity vectors \( \vec{V}_A \) and \( \vec{V}_B \) are orthogonal at \( t = 1 \, \text{s} \).
• The magnitude of the 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 each particle is determined by the time derivative of its position vector.

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>At \( t = 1 \, \text{s} \), substituting the \( \alpha \) values yields: \[ \vec{V}_A = (1 \hat{i} + 6 \hat{j} + 6 \hat{k}) \, \text{m/s} \]

For particle B:

\[ \vec{V}_B = \frac{d\vec{r}_B}{dt} = \left( \beta_1 \hat{i} + 2 \beta_2 t \hat{j} + 3 \beta_3 t^2 \hat{k} \right) \]

At \( t = 1 \, \text{s} \), substituting the \( \beta \) values yields: \[ \vec{V}_B = (2 \hat{i} - 2 \hat{j} + 12 \hat{k}) \, \text{m/s} \]

The orthogonality of the velocities is confirmed by their dot product:

\[ \vec{V}_A \cdot \vec{V}_B = (1)(2) + (6)(-2) + (6)(12) = 0 \]

The angular momentum of particle A relative to particle B is defined as:

\[ \vec{L}_A = \vec{r}_{AB} \times \vec{P}_A \] where \( \vec{r}_{AB} = \vec{r}_A - \vec{r}_B \) and \( \vec{P}_A = m \vec{V}_A \).

At \( t = 1 \, \text{s} \), the relative position vector is:

\[ \vec{r}_{AB} = (1 \hat{i} + 3 \hat{j} + 2 \hat{k}) - (2 \hat{i} - 1 \hat{j} + 4 \hat{k}) = (-1 \hat{i} + 4 \hat{j} - 2 \hat{k}) \] \p>The momentum of particle A is \( \vec{P}_A = 1 \times (1 \hat{i} + 6 \hat{j} + 6 \hat{k}) \). The cross product is computed as:

\[ \vec{L}_A = (-1 \hat{i} + 4 \hat{j} - 2 \hat{k}) \times (1 \hat{i} + 6 \hat{j} + 6 \hat{k}) \] \p>The magnitude of the angular momentum, \( L \), is calculated to be 90.

 

Result:

The value of \( L \) is 90.