A body of mass \( 5 \, \text{kg} \) moving with a uniform speed \( 3\sqrt{2} \, \text{m/s}^{-1} \) in the \( X - Y \) plane along the line \( y = x + 4 \). The angular momentum of the particle about the origin will be ______ \( \text{kg m}^2 \text{s}^{-1} \).
Given: A particle of mass \(m=5\,\text{kg}\) moves along the line \(y=x+4\) with a speed of \(3\sqrt{2}\,\text{m s}^{-1}\).
1) Position and velocity. A general point on the line is represented as \((x,\,x+4)\), thus the position vector is \(\vec r=x\,\hat i+(x+4)\,\hat j.\) The direction of motion is along the line, with a slope \(dy/dx=1\). This implies a direction vector of \(\hat i+\hat j\). The unit direction vector is \(\dfrac{1}{\sqrt2}(\hat i+\hat j)\). Given the speed of \(3\sqrt2\), the velocity vector is \[ \vec v=3\sqrt2\cdot \frac{1}{\sqrt2}(\hat i+\hat j)=3\hat i+3\hat j. \]
2) Angular momentum. The angular momentum is calculated as \( \vec L=m(\vec r\times \vec v)\). The cross product \(\vec r\times \vec v\) is computed as:
\[ \vec r\times \vec v= \begin{vmatrix} \hat i & \hat j & \hat k\\ x & x+4 & 0\\ 3 & 3 & 0 \end{vmatrix} = \big(3x-3(x+4)\big)\hat k=-12\,\hat k. \]
Therefore, the angular momentum is \[ \vec L=5(-12)\hat k=-60\,\hat k\ \text{kg m}^2\text{s}^{-1}. \] The magnitude of the angular momentum is \( |\vec L|=60\ \text{kg m}^2\text{s}^{-1}\).
