P112_Final_Exam_Solu - 2Mv r R ⇒ ω o = 2v r R OR Force torque acceleration& angular acceleration a = f s/M = μ k Mg/M = μ k g and α = τ/I

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P112 Final Exam Solutions — Spring 2004 1. D 2. (a) C (b) E 3. A [4 points for A, -1 for each additional incorrect answer, minimum = 0] 4. (a) A (b) B 5. [2 points for flat and 1 near θ max = 0 and then increasing, 2 points for going to as θ max 90°] 6. (a) 10 cm [-1 for no units] (b) 8 s [-1 for no units] (c) π s -1 = 3.1 s -1 [-1 for no units] (d) - π /2 [-1 for + π /2] (e) 4 π (f) A 7. (a) Σ F x = f s - F cos θ = 0 f s = F cos θ (b) Σ F y = N + F sin θ - w = 0 N = w - F sin θ (c) About pole's lower end: Σ τ = FD - w D 2 cos θ = 0 F = w 2 cos θ 8. (a) C (b) C (c) E (d) A (e) A θ max T T o 0 30° 60° 120° 90° 150° 180°
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9. (a) A (b) A (c) Cons. of Energy: using I = MR 2 and ω r = v r /R E = K + U g = 1 2 Mv r 2 + 1 2 I ω r 2 = 1 2 Mv r 2 + 1 2 Mv r 2 = 0 + MgH H = v r 2 g (d) Cons. of Angular Momentum about a point along hoop's path on surface: L = I ω o = Mv r R + I ω r MR 2 ω o = Mv r R + MR 2 v r R =
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Unformatted text preview: 2Mv r R ⇒ ω o = 2v r R OR Force, torque, acceleration & angular acceleration: a = f s /M = μ k Mg/M = μ k g and α = τ /I = f s R/MR 2 = μ k Mg/MR 2 = μ k g/R v(t) = at = μ k g t and ω (t) = ω o- α t = ω o- (μ k g/R) t Set v(t) = R ω (t) ⇒ μ k g t = R ω o- μ k g t ⇒ t = R ω o /2μ k g ⇒ v r = μ k g (R ω o /2μ k g) = R ω o /2 ⇒ ω o = 2v r R 10. (a) where F s = kx = kD sin θ (b) Σ τ = - 2F s D cos θ- mgD sin θ =- (2kD 2 cos θ- mgD) sin θ (c) Σ τ ≈- (2kD 2- mgD) θ = I α = MD 2 d 2 θ dt 2 ⇒ d 2 θ dt 2 ≈- 2k m + g D θ (d) ω 2 = 2k D + g D ⇒ f = ω 2 π = 1 2 π 2k m + g D || string ⊥ string T mg F s...
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This test prep was uploaded on 04/21/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Fall '07 term at Cornell University (Engineering School).

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P112_Final_Exam_Solu - 2Mv r R ⇒ ω o = 2v r R OR Force torque acceleration& angular acceleration a = f s/M = μ k Mg/M = μ k g and α = τ/I

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