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.)

Respuesta :

Answer:

[tex]L=m(xv_y-yv_x)k[/tex]

Explanation:

It is given that,

Velocity of a particle, [tex]v=v_xi+v_yj[/tex]

Position vector of a particle, [tex]r=xi+yj[/tex]

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

[tex]L=r\times p[/tex], p = linear momentum

[tex]L=r\times (mv)[/tex]

[tex]L=m(r\times v)[/tex]

[tex]L=m((xi+yj)\times (v_xi+v_yj))[/tex]

[tex]L=m(xv_y-yv_x)k[/tex]

So, the angular momentum of the particle is [tex](xv_y-yv_x)k[/tex]. Hence, this is the required solution.

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