Question

The position vector of a particle of mass \(m\) moving with a constant velocity \(\mathbf{u}\) is given by \(\mathbf{r} = x(t)\hat{i} + b\hat{j}\), where \(b\) is a constant. At an instant, \(\mathbf{r}\) makes an angle \(\theta\) with the x-axis as shown in the figure. The variation of the angular momentum of the particle about the origin with \(\theta\) will be:

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

The angular momentum of a particle is given by the cross product of its position vector and momentum vector. The magnitude of angular momentum depends on the sine of the angle between the position and velocity vectors.

Answer 1

The Correct Option is A

Step 1: Angular momentum \( \mathbf{L} \) of a particle about the origin is defined by the cross product:

\[ \mathbf{L} = \mathbf{r} \times m \mathbf{u} \]

where \( \mathbf{r} \) is the position vector and \( \mathbf{u} \) is the velocity vector.

Step 2: The magnitude of the angular momentum is:

\[ L = |\mathbf{r}| \cdot m |\mathbf{u}| \cdot \sin \theta \]

where \( \theta \) is the angle between \( \mathbf{r} \) and \( \mathbf{u} \).

Step 3: Given that \( b \) is constant and the particle's path is a straight line, angular momentum varies with \( \theta \), so:

\[ L = |\mathbf{r}| \cdot |\mathbf{u}| \cdot \sin \theta. \]