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Hour 6 Conservation of Angular Momentum

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

2 Linear and Angular Relationships
β„“ = π‘Ÿ Γ— 𝑝 𝛀 = π‘Ÿ Γ— 𝐹 𝑝=π‘šπ‘£ 𝐹=π‘šπ‘Ž 𝐹= 𝑝 𝑣= π‘₯ π‘Ž= π‘₯ β„“=πΌπœ” 𝛀=𝐼𝛼 𝛀= β„“ πœ”= πœƒ 𝛼= πœƒ π‘₯=π‘…πœƒ 𝑣=π‘…πœ” π‘Ž=𝑅𝛼

3 Angular Momentum in Linear Motion
P We must calculate angular momentum about a point P. b π‘Ÿ 𝑝 β„“ = π‘Ÿ Γ— 𝑝 β„“=𝑏𝑝

4 Central Force The torque is zero, so angular momentum is constant. π‘Ÿ 𝐹

5 𝑑𝐴 𝑑𝑑 = 1 2 | π‘Ÿ Γ— 𝑣 𝑑𝑑| 𝑑𝑑 = β„“ 2π‘š Kepler’s 2nd Law
The area of a trapezoid with sides 𝐴 and 𝐡 is Area = | 𝐴 Γ— 𝐡 | 𝑣 𝑑𝑑 π‘Ÿ 𝑑𝐴 𝑑𝑑 = 1 2 | π‘Ÿ Γ— 𝑣 𝑑𝑑| 𝑑𝑑 = β„“ 2π‘š

6 β€œBomb” 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.

7 Center of Mass 𝐹 𝑒π‘₯𝑑= 𝑖 π‘š 𝑖 π‘Ÿ 𝑖 ≑𝑀 𝑅

8 Center of Mass 𝑅 ≑ 1 𝑀 π‘Ÿ 𝜌( π‘Ÿ )𝑑 3 π‘Ÿ 𝑅 ≑ 1 𝑀 𝑖 π‘š 𝑖 π‘Ÿ 𝑖 𝑀≑ 𝜌( π‘Ÿ )𝑑 3 π‘Ÿ

9 Moment of Inertia 𝑑 𝑖 𝐼≑ 𝑖 π‘š 𝑖 𝑑𝑖 2 π‘Ÿ 𝑖

10 Moment of Inertia 𝐼≑ 𝑖 π‘š 𝑖 𝑑𝑖 2 𝐼≑ 𝑑2 𝜌( π‘Ÿ )𝑑 3 π‘Ÿ

11 The Rocket Problem Newton’s 2nd Law: 𝐹 = 𝑝 =π‘š 𝑣 + π‘š 𝑣
𝐹 = 𝑝 =π‘š 𝑣 + π‘š 𝑣 This is true – but momentum conservation is more transparent: Start with mass π‘š and 𝑣=0. Let 𝑣𝑒π‘₯ be the exhaust velocity. π‘šβˆ’|βˆ†π‘š| βˆ†π‘£= βˆ†π‘š 𝑣𝑒π‘₯βˆ’βˆ†π‘£ π‘šβˆ†π‘£= 𝑣 𝑒π‘₯ |βˆ†π‘š| π‘šπ‘Ž= 𝑣 𝑒π‘₯ | π‘š | Adding any external forces: 𝐹=π‘š 𝑣 = 𝐹 𝑒π‘₯𝑑 + 𝑣 𝑒π‘₯ | π‘š | Thrust = 𝑣𝑒π‘₯| π‘š |


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