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Angular Momentum. Vector Calculus Take any point as the origin of a coordinate system Vector from that point to a particle: r = (x, y) function of time.

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Presentation on theme: "Angular Momentum. Vector Calculus Take any point as the origin of a coordinate system Vector from that point to a particle: r = (x, y) function of time."— Presentation transcript:

1 Angular Momentum

2 Vector Calculus Take any point as the origin of a coordinate system Vector from that point to a particle: r = (x, y) function of time r(t) = (x(t), y(t) ) Time derivative dr/dt = v = (dx/dt, dy/dt)

3 Second Law and Momentum m a = F m dv/dt = F d (mv) /dt = F dp/dt = F change in momentum = Force

4 Vector Product (Review) Torque = F x d (Perpendicular Distance from Force) = d x F (Perpendicular Distance to Force) sin 90° = 1 Torque is out of the paper. sin 0° = 0 Right Hand Rule

5 Time Derivative of r x v d(r x v)/dt = dr/dt x v + r x dv/dt = v x v + r x a = r x a multiply by m: m d(r x v)/dt = m (r x a) d(r x mv)/dt = r x ma Define: angular momentum = L = r x mv dL/dt = r x ma = r x F = Torque compare dp/dt = F Change in Angular Momentum = Torque

6 Work When Turning Torque = τ = F x r W = F∙s For one revolution W = F∙2πr = 2π τ For less than a revolution W = F ∙ θr = θ τ

7 Comparison velocity: v = dx/dt = r x ω acceleration: a = dv/dt const acceleration: x = x 0 + v 0 t + ½at² momentum: p = mv Kinetic Energy: ½mv ² F = ma = dp/dt Work: F ⋅ s angular velocity: ω = dθ/dt angular acceleration: α = dω/dt θ = θ 0 + ω 0 t + ½ αt² angular momentum: L = r x mv ½mv ² = ½m(r x ω)² = ½mr²ω² = ½∙I∙ω² I = mr² T = dL/dt = d(r x m(r x ω) )/dt = mr² dω/dt = I α Work: T ∙ θ

8 Textbook Review Page 331


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