Question

The position of a particle of mass 1 kg at time $t$ is given by $\mathbf{r} = t\hat{i} + \hat{j} + 2t^2\hat{k}$, where $t$ is in seconds and the coefficients have the proper units for $\mathbf{r}$ to be in metres. What is the component of the angular momentum (with respect to the origin) in kg m$^2$ s$^{-1}$ along the vector $(\hat{i} + \hat{j})$?

• A. $\frac{1}{\sqrt{2}}(4t - 2t^2)$
• B. $\frac{1}{\sqrt{2}}(4t + 6t^2)$
• C. $4t - 2t^2$
• D. $4t + 6t^2$

Hint:

To find the scalar component of any vector along a target vector, always normalize the target vector to a unit vector first, and then take the dot product.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Angular momentum \( \vec{L} \) is defined as \( \vec{r} \times \vec{p} \). To find a scalar component along a vector \( \vec{A} \), we take the dot product with the unit vector of \( \vec{A} \).

Step 2: Detailed Explanation:
1. Velocity Vector: \( \vec{v} = \frac{d\vec{r}}{dt} = \hat{i} + 0\hat{j} + 4t\hat{k} \).
2. Angular Momentum (\( m = 1 \text{ kg} \)):
\[ \vec{L} = \vec{r} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} t & 1 & 2t^{2} 1 & 0 & 4t \end{vmatrix} \]
\[ \vec{L} = \hat{i}(4t - 0) - \hat{j}(4t^{2} - 2t^{2}) + \hat{k}(0 - 1) = 4t\hat{i} - 2t^{2}\hat{j} - \hat{k} \]
3. Component along \( \vec{A} = \hat{i} + \hat{j} \):
The unit vector is \( \hat{u} = \frac{\hat{i} + \hat{j}}{\sqrt{2}} \).
\[ L_{\text{comp}} = \vec{L} \cdot \hat{u} = (4t\hat{i} - 2t^{2}\hat{j} - \hat{k}) \cdot \left( \frac{\hat{i} + \hat{j}}{\sqrt{2}} \right) \]
\[ L_{\text{comp}} = \frac{4t - 2t^{2}}{\sqrt{2}} \]

Step 3: Final Answer:
The component is \( \frac{1}{\sqrt{2}}(4t - 2t^{2}) \).