Question

A particle of mass m moves in the xy plane with a velocity of v = vxî + vyĵ. Determine the angular momentum of the particle about the origin when its position vector is r = xî + yĵ. (Use the following as necessary: x, y, vx, vy, and m.)

Answer 1

Answer:

$L=m(xv_y-yv_x)k$

Explanation:

It is given that,

Velocity of a particle, $v=v_xi+v_yj$

Position vector of a particle, $r=xi+yj$

We need to find the angular momentum of the particle. It is given by :

$L=r\times p$, p = linear momentum

$L=r\times (mv)$

$L=m(r\times v)$

$L=m((xi+yj)\times (v_xi+v_yj))$

$L=m(xv_y-yv_x)k$

So, the angular momentum of the particle is $(xv_y-yv_x)k$. Hence, this is the required solution.