Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components px, py and pz. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.
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Let a position vector r
P
r = xi + yj + zk
Momentum vector(P) = Pxi + Pyj + Pzk
Then,
Angular momentum(L) = r × P
= (xi + yj + zk)× ( Pxi +Pyj + Pzk)
= (yPz - zPy)i + (zPx-xPz)j + (xPy-yPx)k
Hence,
Lx = yPz - zPy
Ly = zPx - xPz
Lz = xPy - yPx
(B) particle moves only x-y plane
Means position vector = xi+yj
Hence, momentum also produced in x-y plane
And momentum = Pxi+Pyj
Angular momentum (L) =r × P
= (xi + yj + 0k)×(Pxi + Pyj)
= (xPy - yPx)k
Hence, angular momentum has only z- component.
P
r = xi + yj + zk
Momentum vector(P) = Pxi + Pyj + Pzk
Then,
Angular momentum(L) = r × P
= (xi + yj + zk)× ( Pxi +Pyj + Pzk)
= (yPz - zPy)i + (zPx-xPz)j + (xPy-yPx)k
Hence,
Lx = yPz - zPy
Ly = zPx - xPz
Lz = xPy - yPx
(B) particle moves only x-y plane
Means position vector = xi+yj
Hence, momentum also produced in x-y plane
And momentum = Pxi+Pyj
Angular momentum (L) =r × P
= (xi + yj + 0k)×(Pxi + Pyj)
= (xPy - yPx)k
Hence, angular momentum has only z- component.
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