OLDHW27 - oldhomewk 27 PAPAGEORGE MATT Due Apr 1 2008 4:00...

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oldhomewk 27 – PAPAGEORGE, MATT – Due: Apr 1 2008, 4:00 am 1 Question 1, chap 13, sect 4. part 1 of 3 10 points A 3 kg bicycle wheel rotating at a 2262 rev / min angular velocity has its shaft supported on one side, as shown in the Fg- ure. The wheel is a hoop of radius 0 . 4 m, and its shaft is horizontal. The distance from the center of the wheel to the pivot point is 0 . 6 m. When viewing from the left (from the posi- tive x -axes), one sees that the wheel is rotat- ing in a clockwise manner. Assume: All of the mass of the system is located at the rim of the bicycle wheel. The acceleration of gravity is 9 . 8 m / s 2 . y z x b mg ω R The magnitude of the angular momentum of the wheel is given by 1. b v L b = mR 2 ω 2 2. b v L b = 1 4 mR 2 ω 3. b v L b = mR 2 ω correct 4. b v L b = 1 2 mR 2 ω 5. b v L b = 1 4 mR 2 ω 2 6. b v L b = 1 2 mR 2 ω 2 Explanation: Basic Concepts: v τ = d v L dt Solution: The magnitude of the angular momentum I of the wheel is L = I ω = mR 2 ω , and is along the negative x -axis. Question 2, chap 13, sect 4. part 2 of 3 10 points Let : m = 3 kg , ω = 2262 rev / min b = 0 . 6 m , and R = 0 . 4 m . ±ind the change in the precession angle after a 1 . 7 s time interval. Correct answer: 15 . 1115 (tolerance ± 1 %). Explanation: Let : ω = 2262 rev / min = 2 π (2262 rev / min) (60 s / min) = 236 . 876 rad / s . Solution: The magnitude of the angular momentum I of the wheel is L = I ω = mR 2 ω , and is along the negative x -axis. ±rom the Fgure in Part 2, we get Δ φ = Δ L L . Using the relation, Δ L = τ Δ t, where τ is the torque, v τ = v b × mvg . The precession angle Δ
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OLDHW27 - oldhomewk 27 PAPAGEORGE MATT Due Apr 1 2008 4:00...

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