Question

Calculate the angular momentum L of a particle with radial vector r = (4, -5, 3) and linear momentum p = (1, 4, -2) and find the angle between the position and the momentum vectors.

Answer 1

The angular momentum L of the particle is (-2, 11, 21) and the angle between the position and momentum vectors is approximately 152.24 degrees.

Step-by-step explanation:

The angular momentum ℓ of a particle is calculated using the cross product ℓ = ℝ × p. Given the radial vector ℝ = (4, -5, 3) and the linear momentum vector p = (1, 4, -2), we calculate ℓ as follows:

• ℓ = (4, -5, 3) × (1, 4, -2)
• ℓ = ((-5)(-2) - (3)(4), (3)(1) - (4)(-2), (4)(4) - (-5)(1))
• ℓ = (10 - 12, 3 + 8, 16 + 5)
• ℓ = (-2, 11, 21)

To find the angle between the position and momentum vectors, we use the dot product and the magnitude of the vectors:

• ℝ · p = (4)(1) + (-5)(4) + (3)(-2) = 4 - 20 - 6 = -22
• |ℝ| = √(4² + (-5)² + 3²) = √(16 + 25 + 9) = √50
• |p| = √(1² + 4² + (-2)²) = √(1 + 16 + 4) = √21
• cos(θ) = (ℝ · p) / (|ℝ||p|)
• cos(θ) = -22 / (√50 × √21)
• θ = cos⁻¹(-22 / (√50 × √21)) ≈ 152.24°

The angular momentum vector is ℓ = (-2, 11, 21) and the angle θ between the position ℝ and momentum p vectors is approximately 152.24°.