Question

A 1.55 kg particle moves in the xy plane with a velocity of v = ( 3.51 , -3.39 ) m/s. Determine the angular momentum of the particle about the origin when its position vector is r = ( 1.22 , 1.26 ) m. ?

Answer 1

Let,the velocity vector is `vec v=3.51 hat i - 3.39 hat j`

So,`m vec v=(5.43 hat i-5.24 hat j)`

And,position vector is `vec r=1.22 hat i +1.26 hat j`

So,angular momentum about origin is `vec r ×m vec v=(1.22hati +1.26hatj)×(5.43hati-5.24 hat j)=-6.4hatk-6.83hatk=-13.23hatk`

So,the magnitude is `13.23Kgm^2s^-1`