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Hour 6 Conservation of Angular Momentum
Physics 321 Hour 6 Conservation of Angular Momentum
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Linear and Angular Relationships
β = π Γ π π€ = π Γ πΉ π=ππ£ πΉ=ππ πΉ= π π£= π₯ π= π₯ β=πΌπ π€=πΌπΌ π€= β π= π πΌ= π π₯=π
π π£=π
π π=π
πΌ
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Angular Momentum in Linear Motion
P We must calculate angular momentum about a point P. b π π β = π Γ π β=ππ
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Central Force The torque is zero, so angular momentum is constant. π πΉ
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ππ΄ ππ‘ = 1 2 | π Γ π£ ππ‘| ππ‘ = β 2π Keplerβs 2nd Law
The area of a trapezoid with sides π΄ and π΅ is Area = | π΄ Γ π΅ | π£ ππ‘ π ππ΄ ππ‘ = 1 2 | π Γ π£ ππ‘| ππ‘ = β 2π
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βBombβ 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.
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Center of Mass πΉ ππ₯π‘= π π π π π β‘π π
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Center of Mass π
β‘ 1 π π π( π )π 3 π π
β‘ 1 π π π π π π πβ‘ π( π )π 3 π
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Moment of Inertia π π πΌβ‘ π π π ππ 2 π π
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Moment of Inertia πΌβ‘ π π π ππ 2 πΌβ‘ π2 π( π )π 3 π
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The Rocket Problem Newtonβs 2nd Law: πΉ = π =π π£ + π π£
πΉ = π =π π£ + π π£ This is true β but momentum conservation is more transparent: Start with mass π and π£=0. Let π£ππ₯ be the exhaust velocity. πβ|βπ| βπ£= βπ π£ππ₯ββπ£ πβπ£= π£ ππ₯ |βπ| ππ= π£ ππ₯ | π | Adding any external forces: πΉ=π π£ = πΉ ππ₯π‘ + π£ ππ₯ | π | Thrust = π£ππ₯| π |
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