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Example 8-15 v1 = 2.4 m/s r1 = 0.8m r2 = 0.48 m v2 = ? I1ω1 = I2ω2

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Presentation on theme: "Example 8-15 v1 = 2.4 m/s r1 = 0.8m r2 = 0.48 m v2 = ? I1ω1 = I2ω2"— Presentation transcript:

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2 Example 8-15 v1 = 2.4 m/s r1 = 0.8m r2 = 0.48 m v2 = ? I1ω1 = I2ω2
 ω2 = ω1 (r2)2/(r1)2 , ω1 = (v1/r1) v2 = r2ω2 = r2(v1/r1)[(r2)2/(r1)2] = v1 (r1/r2) v2 = 4.0 m/s

3 Angular Quantities are Vectors!

4 Conservation of Vector Angular Momentum

5 Conceptual Example 8-17 Conservation of angular momentum!!
Linitial = Lfinal L = - L + Lperson  Lperson = 2 L Demonstration!

6 Translation-Rotation Analogues & Connections
Displacement x θ Velocity v ω Acceleration a α Force (Torque) F τ Mass (moment of inertia) m I Newton’s 2nd Law ∑F = ma ∑τ = Iα Kinetic Energy (KE) (½)mv (½)Iω2 Work (constant F,τ) Fd τθ Momentum mv Iω CONNECTIONS: v = rω, atan= rα aR = (v2/r) = ω2r , τ = rF , I = ∑(mr2)

7 Problem 58 Demonstration!!


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