Hour 6 Conservation of Angular Momentum Physics 321 Hour 6 Conservation of Angular Momentum

Linear and Angular Relationships ℓ = 𝑟 × 𝑝 𝛤 = 𝑟 × 𝐹 𝑝=𝑚𝑣 𝐹=𝑚𝑎 𝐹= 𝑝 𝑣= 𝑥 𝑎= 𝑥 ℓ=𝐼𝜔 𝛤=𝐼𝛼 𝛤= ℓ 𝜔= 𝜃 𝛼= 𝜃 𝑥=𝑅𝜃 𝑣=𝑅𝜔 𝑎=𝑅𝛼

Angular Momentum in Linear Motion P We must calculate angular momentum about a point P. b 𝑟 𝑝 ℓ = 𝑟 × 𝑝 ℓ=𝑏𝑝

Central Force The torque is zero, so angular momentum is constant. 𝑟 𝐹

𝑑𝐴 𝑑𝑡 = 1 2 | 𝑟 × 𝑣 𝑑𝑡| 𝑑𝑡 = ℓ 2𝑚 Kepler’s 2nd Law The area of a trapezoid with sides 𝐴 and 𝐵 is Area = | 𝐴 × 𝐵 | 𝑣 𝑑𝑡 𝑟 𝑑𝐴 𝑑𝑡 = 1 2 | 𝑟 × 𝑣 𝑑𝑡| 𝑑𝑡 = ℓ 2𝑚

“Bomb” Problem An artillery shell explodes just as the shell reaches its maximum height. It breaks into three pieces. Ignore drag forces and the mass of the explosive itself. Describe what happens.

Center of Mass 𝐹 𝑒𝑥𝑡= 𝑖 𝑚 𝑖 𝑟 𝑖 ≡𝑀 𝑅

Center of Mass 𝑅 ≡ 1 𝑀 𝑟 𝜌( 𝑟 )𝑑 3 𝑟 𝑅 ≡ 1 𝑀 𝑖 𝑚 𝑖 𝑟 𝑖 𝑀≡ 𝜌( 𝑟 )𝑑 3 𝑟

Moment of Inertia 𝑑 𝑖 𝐼≡ 𝑖 𝑚 𝑖 𝑑𝑖 2 𝑟 𝑖

Moment of Inertia 𝐼≡ 𝑖 𝑚 𝑖 𝑑𝑖 2 𝐼≡ 𝑑2 𝜌( 𝑟 )𝑑 3 𝑟

The Rocket Problem Newton’s 2nd Law: 𝐹 = 𝑝 =𝑚 𝑣 + 𝑚 𝑣 𝐹 = 𝑝 =𝑚 𝑣 + 𝑚 𝑣 This is true – but momentum conservation is more transparent: Start with mass 𝑚 and 𝑣=0. Let 𝑣𝑒𝑥 be the exhaust velocity. 𝑚−|∆𝑚| ∆𝑣= ∆𝑚 𝑣𝑒𝑥−∆𝑣 𝑚∆𝑣= 𝑣 𝑒𝑥 |∆𝑚| 𝑚𝑎= 𝑣 𝑒𝑥 | 𝑚 | Adding any external forces: 𝐹=𝑚 𝑣 = 𝐹 𝑒𝑥𝑡 + 𝑣 𝑒𝑥 | 𝑚 | Thrust = 𝑣𝑒𝑥| 𝑚 |