Physics 321 Hour 6 Conservation of Angular Momentum
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Physics 321 Hour 6 Conservation of Angular Momentum Linear and Angular Relationships π = ππ£ πΉ = ππ πΉ=π π£=π₯ π=π₯ β=π×π π€ =π×πΉ π₯ = π π π£ = π π π = π πΌ β = πΌπ π€ = πΌπΌ π€=β π=π πΌ=π Angular Momentum in Linear Motion We must calculate angular momentum about a point P. P b π π β=π×π β = ππ Central Force The torque is zero, so angular momentum is constant. π πΉ Kepler’s 2nd Law The area of a trapezoid with sides π΄ and π΅ is Area = |π΄ × π΅| π£ππ‘ π ππ΄ 1 |π × π£ππ‘| β = = ππ‘ 2 ππ‘ 2π “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 πΉ ππ₯π‘ = ππ ππ ≡ ππ π Center of Mass 1 π ≡ π ππ ππ π 1 π ≡ π π≡ ππ(π)π3 π π(π)π3 π Moment of Inertia ππ ππ 2 πΌ≡ π ππ ππ Moment of Inertia πΌ≡ ππ ππ π 2 πΌ≡ π2π(π)π3 π The Rocket Problem Newton’s 2nd Law: πΉ = π = ππ£ + ππ£ This is true – but momentum conservation is more transparent: Start with mass π and π£ = 0. Let π£ππ₯ be the exhaust velocity. π − |βπ| βπ£ = βπ π£ππ₯ − βπ£ πβπ£ = π£ππ₯ |βπ| ππ = π£ππ₯ |π| Adding any external forces: πΉ = ππ£ = πΉππ₯π‘ + π£ππ₯ |π| Thrust = π£ππ₯|π|
