# Physics 321 Hour 6 Conservation of Angular Momentum

```Physics 321
Hour 6
Conservation of Angular Momentum
Linear and Angular Relationships
π = ππ£
πΉ = ππ
πΉ=π
π£=π₯
π=π₯
β=π&times;π
π€ =π&times;πΉ
π₯ = ππ
π£ = ππ
π = ππΌ
β = πΌπ
π€ = πΌπΌ
π€=β
π=π
πΌ=π
Angular Momentum in Linear Motion
We must
calculate angular
momentum
P
b
π
π
β=π&times;π
β = ππ
Central Force
The torque is zero, so
angular momentum is
constant.
π
πΉ
Kepler’s 2nd Law
The area of a trapezoid
with sides π΄ and π΅ is
Area = |π΄ &times; π΅|
π£ππ‘
π
ππ΄ 1 |π &times; π£ππ‘|
β
=
=
ππ‘ 2
ππ‘
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
πΉ ππ₯π‘ =
ππ ππ ≡ ππ
π
Center of Mass
1
π≡
π
ππ ππ
π
1
π≡
π
π≡
ππ(π)π3 π
π(π)π3 π
Moment of Inertia
ππ ππ 2
πΌ≡
π
ππ
ππ
Moment of Inertia
πΌ≡
ππ ππ
π
2
πΌ≡
π2π(π)π3 π
The Rocket Problem
Newton’s 2nd Law:
πΉ = π = ππ£ + ππ£
This is true – but momentum conservation is more
transparent:
Start with mass π and π£ = 0. Let π£ππ₯ be the exhaust
velocity.
π − |βπ| βπ£ = βπ π£ππ₯ − βπ£
πβπ£ = π£ππ₯ |βπ|
ππ = π£ππ₯ |π|
Adding any external forces: πΉ = ππ£ = πΉππ₯π‘ + π£ππ₯ |π|
Thrust = π£ππ₯|π|
```