Question

The position vector of s particle of mass 2kg as a function of time is given by r= 6i+5tj, where r is in meters and t is in seconds. Determine the angular momentum of the particle about the origin as a function of time?

Answer 1

The speed is the first derivative of position, therefore, the velocity is v=5j.

The momentum is mass (2 kg) times velocity: p=10j

The angular momentum of a pint mass moving (rotating) distance r about the origin is

$$L=m\begin{vmatrix} i & j & k \\ 6 & 5t & 0 \\ 5 &10 & 0 \end{vmatrix}=\\ =m[(5t\cdot0-10\cdot0)i+(6\cdot0-0\cdot0)j+(6\cdot10-0\cdot5t)k]=\\ =60k\text{ J}\cdot\text{s}.$$