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, pᵧ and pᵤ. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.
The angular momentum of a particle is defined as l = r × p. Using the cross product of position vector r = xî + yĵ + zk̂ and momentum p = pₓî + pᵧĵ + pᵤk̂, we get:
l = î(y pᵤ - z pᵧ) + ĵ(z pₓ - x pᵤ) + k̂(x pᵧ - y pₓ)
Thus, the components are: lₓ = y pᵤ - z pᵧ, lᵧ = z pₓ - x pᵤ, and lᵤ = x pᵧ - y pₓ.
When the particle moves only in the x-y plane, we have z = 0 and pᵤ = 0. Substituting these values:
- lₓ = y(0) - (0)pᵧ = 0
- lᵧ = (0)pₓ - x(0) = 0
- lᵤ = x pᵧ - y pₓ (non-zero in general)
Hence, for motion confined to the x-y plane, the angular momentum has only a z-component.
Explanation
The question tests understanding of vector cross product applied to angular momentum. From the textbook context, angular momentum is defined as l = r × p (Eq. 6.25a). Computing the cross product using determinant method gives the three components. For planar motion in x-y plane, both z-coordinate and z-component of momentum vanish, leaving only the z-component of angular momentum non-zero. This is consistent with the textbook's discussion of angular momentum direction being perpendicular to the plane of r and p.
Solution Steps
Step 1: Write angular momentum as l = r × p using cross product definition
Step 2: Express r and p in component form and compute cross product using determinant
Step 3: Identify the three components: lₓ = ypᵤ - zpᵧ, lᵧ = zpₓ - xpᵤ, lᵤ = xpᵧ - ypₓ
Step 4: For x-y plane motion, substitute z = 0 and pᵤ = 0
Step 5: Show that lₓ = 0, lᵧ = 0, while lᵤ remains non-zero