Irodov Problem 1.149 – Angular Momentum of a Particle in Gravity

Problem Statement

Find the angular momentum of a particle of mass m thrown horizontally with velocity v₀ from height h above ground, at the moment it hits the ground.

Given Information

• Mass m, initial horizontal velocity v₀
• Initial height h

Physical Concepts & Formulas

$$L=m(\mathbf{r}\times\mathbf{v})$$

Step-by-Step Solution

Step 1: At ground: vₓ = v₀, vᵧ = −√(2gh).

Step 2: Position at impact: x = v₀√(2h/g), y = 0.

Step 3: L = m|r × v| = m·x·|vᵧ| = m·v₀√(2h/g)·√(2gh) = 2mv₀h.

Worked Calculation

L = 2mv₀h

Answer

$$\boxed{L=2mv_0h}$$

Physical Interpretation

Taking origin at launch point, angular momentum has magnitude mxvᵧ = 2mv₀h at impact. Gravity (passing through origin) exerts no torque, so L changes only due to the changing moment arm.