Question

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 ______ \( \text{kg m}^2 \text{s}^{-1} \).

Answer 1

Correct Answer: 60

Given: Particle of mass \(m=5\,\text{kg}\) moves along the line \(y=x+4\) with speed \(3\sqrt{2}\,\text{m s}^{-1}\).

1) Position and velocity. 
Any point on the line: \((x,\,x+4)\Rightarrow \vec r=x\,\hat i+(x+4)\,\hat j.\)
Direction of motion is along the line; slope \(dy/dx=1\Rightarrow\) direction vector \(\hat i+\hat j\). Unit direction \(=\dfrac{1}{\sqrt2}(\hat i+\hat j)\). Given speed \(3\sqrt2\), the velocity is \[ \vec v=3\sqrt2\cdot \frac{1}{\sqrt2}(\hat i+\hat j)=3\hat i+3\hat j. \]

2) Angular momentum.
\[ \vec L=m(\vec r\times \vec v),\qquad \vec r\times \vec v= \begin{vmatrix} \hat i & \hat j & \hat k\\ x & x+4 & 0\\ 3 & 3 & 0 \end{vmatrix} = \big(3x-3(x+4)\big)\hat k=-12\,\hat k. \] Thus \[ \vec L=5(-12)\hat k=-60\,\hat k\ \text{kg m}^2\text{s}^{-1}. \] \[ |\vec L|=60\ \text{kg m}^2\text{s}^{-1}. \]