Rod and Clay Collision – Angular Momentum Conservation
Question:
A 75 g, 40-cm-long rod hangs vertically on a frictionless, horizontal axle passing through its center. A 12 g ball of clay traveling horizontally at 2.5 m/s hits and sticks to the very bottom tip of the rod.
Determine: The angular speed of the rod–clay system immediately after the collision.
Answer:
Step 1: Understand the System
- Rod mass: M = 75 g = 0.075 kg
- Rod length: L = 40 cm = 0.40 m
- Clay mass: m = 12 g = 0.012 kg
- Initial speed of clay: v = 2.5 m/s
Step 2: Moment of Inertia of the Rod
For a rod rotating about its center:
Irod = (1/12) × M × L²
Substituting values:
Irod = (1/12) × 0.075 × (0.40)² = 0.001 kg·m²
Step 3: Angular Momentum of the Clay Before Collision
The clay hits the bottom of the rod, i.e., r = L/2 = 0.20 m
Angular momentum:
Lclay = m × v × r = 0.012 × 2.5 × 0.20 = 0.006 kg·m²/s
Step 4: Moment of Inertia of the Clay (as a Point Mass)
Iclay = m × r² = 0.012 × (0.20)² = 0.00048 kg·m²
Step 5: Total Moment of Inertia After Collision
Itotal = Irod + Iclay = 0.001 + 0.00048 = 0.00148 kg·m²
Step 6: Apply Conservation of Angular Momentum
Angular momentum before = Angular momentum after
Linitial = Itotal × ω
Solving for ω:
ω = Linitial / Itotal = 0.006 / 0.00148 ≈ 4.05 rad/s