A particle of mass m is moving on the XY-plane along a straight line y = d, with constant velocity v. Its angular momentum about the origin

Options

A. is zero
B. remains constan
C. increases with time
D. may increase or decrease depending on direction of motion of particle.

Explanation

To find the angular momentum of a particle of mass \( m \) moving along the line \( y = d \) with a constant velocity \( v \) in the XY-plane, we can follow these steps:

### Step 1: Understand the Motion of the Particle
The particle is moving along the line \( y = d \). This means that its y-coordinate is always \( d \), while its x-coordinate changes as it moves. The particle's position can be described as \( (x, d) \), where \( x \) is a function of time.

### Step 2: Define Linear Momentum
The linear momentum \( \vec{p} \) of the particle is given by:
\[ \vec{p} = m \vec{v} \]
where \( \vec{v} \) is the velocity vector of the particle. Since the particle is moving with a constant velocity \( v \), we can express the velocity vector as:
\[ \vec{v} = (v_x, v_y) = (v, 0) \]
because the particle is moving horizontally along the line \( y = d \).

### Step 3: Calculate Angular Momentum
The angular momentum \( \vec{L} \) of a particle about a point (in this case, the origin) is given by the cross product of the position vector \( \vec{r} \) and the linear momentum \( \vec{p} \):
\[ \vec{L} = \vec{r} \times \vec{p} \]
The position vector \( \vec{r} \) of the particle at any time \( t \) can be expressed as:
\[ \vec{r} = (x, d) \]
Thus, the linear momentum is:
\[ \vec{p} = m(v, 0) \]

### Step 4: Compute the Cross Product
The angular momentum can be calculated using the determinant of a matrix:
\[ \vec{L} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & d & 0 \\ mv & 0 & 0 \end{vmatrix} \]
Calculating this determinant, we find:
\[ \vec{L} = \hat{k}(x \cdot 0 - d \cdot mv) = -d mv \hat{k} \]
Thus, the magnitude of the angular momentum is:
\[ L = mvd \]
and it points in the negative z-direction (out of the XY-plane).

### Step 5: Conclusion
Since the particle is moving along the line \( y = d \) with constant velocity \( v \), both \( m \), \( v \), and \( d \) are constants. Therefore, the magnitude of the angular momentum \( L = mvd \) is also constant.

### Final Answer
The angular momentum of the particle about the origin is:
\[ \vec{L} = -mvd \hat{k} \]
and its magnitude is \( L = mvd \), which remains constant.

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