Final Angular Velocity and Energy Loss in a Rod-Particle Collision
📘 Question:
A 0.80 m long, 1.20 kg mass rod which can rotate about a pivot at one end hangs straight down. A 0.60 kg particle moving horizontally with a speed of 10 m/s collides with and sticks to the end of the rod. (a) What is the final angular velocity of the rod-particle system after the collision? (b) How much kinetic energy is lost during the collision?
🔷 Given Data:
- Rod mass, M = 1.20 kg
- Rod length, L = 0.80 m
- Particle mass, m = 0.60 kg
- Particle speed before collision, v = 10 m/s
- Type of collision: Completely inelastic (particle sticks to rod)
🔹 Step 1: Conservation of Angular Momentum
Initial angular momentum (before collision):
Linitial = m × v × L = 0.60 × 10 × 0.80 = 4.8 kg·m²/s
Moment of inertia after collision:
Irod = (1/3)ML² = (1/3)(1.20)(0.80)² = 0.256 kg·m²
Iparticle = m × L² = 0.60 × 0.80² = 0.384 kg·m²
Itotal = 0.256 + 0.384 = 0.64 kg·m²
Iparticle = m × L² = 0.60 × 0.80² = 0.384 kg·m²
Itotal = 0.256 + 0.384 = 0.64 kg·m²
Final angular velocity (ω):
ω = Linitial / Itotal = 4.8 / 0.64 = 7.5 rad/s
🔹 Step 2: Kinetic Energy Before and After Collision
Initial kinetic energy (only particle is moving):
KEinitial = (1/2)mv² = (1/2)(0.60)(10²) = 30 J
Final kinetic energy (rotational):
KEfinal = (1/2)Itotalω² = (1/2)(0.64)(7.5²) = 18 J
🔻 Step 3: Energy Lost During Collision
Energy lost = KEinitial − KEfinal = 30 − 18 = 12 J
✅ Final Answers:
- (a) Final angular velocity = 7.5 rad/s
- (b) Kinetic energy lost = 12 J
📊 Summary Table:
| Parameter | Value |
|---|---|
| Rod mass (M) | 1.20 kg |
| Rod length (L) | 0.80 m |
| Particle mass (m) | 0.60 kg |
| Initial particle speed | 10 m/s |
| Final angular velocity (ω) | 7.5 rad/s |
| Initial KE | 30 J |
| Final KE | 18 J |
| Energy lost | 12 J |