A particle having mass 2 kg is moving along straight line ` 3 x + 4 y = 5 ` with speed `8m//s` . Find angular momentum of the particle about origin, x and y are in meters.

Answer

To find the angular momentum of a particle about the origin, we follow these steps:

### Step 1: Determine the equation of the line and its parameters
The line given is \(3x + 4y = 5\). We can rearrange this into slope-intercept form to find the slope and y-intercept:
\[ y = -\frac{3}{4}x + \frac{5}{4} \]

### Step 2: Find the perpendicular distance from the origin to the line
The formula for the perpendicular distance \(d\) from a point \((x_0, y_0)\) to a line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our line, we can rewrite it as:
\[ 3x + 4y - 5 = 0 \quad \text{(where \(A = 3\), \(B = 4\), and \(C = -5\))} \]
Substituting the origin \((0, 0)\):
\[ d = \frac{|3(0) + 4(0) - 5|}{\sqrt{3^2 + 4^2}} = \frac{5}{\sqrt{9 + 16}} = \frac{5}{5} = 1 \text{ meter} \]

### Step 3: Calculate the linear momentum of the particle
The linear momentum \(p\) of a particle is given by:
\[ p = mv \]
Where \(m = 2 \, \text{kg}\) and \(v = 8 \, \text{m/s}\):
\[ p = 2 \, \text{kg} \times 8 \, \text{m/s} = 16 \, \text{kg m/s} \]

### Step 4: Calculate the angular momentum about the origin
The angular momentum \(L\) about a point is given by:
\[ L = r \times p \]
Where \(r\) is the perpendicular distance from the point to the line (which we found to be 1 meter) and \(p\) is the linear momentum:
\[ L = d \cdot p = 1 \, \text{m} \times 16 \, \text{kg m/s} = 16 \, \text{kg m}^2/\text{s} \]

### Final Answer
The angular momentum of the particle about the origin is:
\[ \boxed{16 \, \text{kg m}^2/\text{s}} \]

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